PKDI/\\!Heat and Thermodynamics/ch1.ipynb{
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"cells": [
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"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 1 : Thermometry"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.11 Page No : 7"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"R0 = 5.5 # The resistance of a platinum wire at 0 degree centigrade in ohms\n",
"R100 = 7.5 # The resistance of a platinum wire at 100 degree centigrade in ohms\n",
"R444 = 14.5 # The resistance of a platinum wire at 444.5 degree centigrade in ohms\n",
"\n",
"# Calculations\n",
"# The value of beta in per degree centigrade square\n",
"b = ((900 - (2. * 444.6)) / (5.5 * 444.6 * 100 * 344.6))\n",
"# The value of alpha in per degree centigrade\n",
"a = (2 / (5.5 * 100)) - (100 * (b))\n",
"\n",
"# Output\n",
"print 'The values are b = %3.4g degree centigrade square \\n and a = %3g degree centigrade ' % (b, a)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The values are b = 1.282e-07 degree centigrade square \n",
" and a = 0.00362355 degree centigrade \n"
]
}
],
"prompt_number": 1
}
],
"metadata": {}
}
]
}PKDIܓw!Heat and Thermodynamics/ch2.ipynb{
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{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 2 : Expansion \n"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.1 Page No : 24"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"l = 1. # The thickness of the crystal in cm\n",
"w = 5890. - 8 # The wavelength of light used in cm\n",
"t2 = 50. # The final temperature of the crystal in degree centigrade\n",
"t1 = 20. # The initial temperature of the crystal in degree centigrade\n",
"p = 14. # The number of fringes that crossed the field of view\n",
"\n",
"# Calculations\n",
"t = t2 - t1 # The temperature difference in degree centigrade\n",
"# The coefficient of linear expansion of the crystal in per degree centigrade\n",
"a = (p * w) / (2 * l * t)\n",
"\n",
"# output\n",
"print 'The coefficient of linear expansion of the crystal is %3.4g degree centigrade' % (a)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The coefficient of linear expansion of the crystal is 1372 degree centigrade\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.2 Page No : 31"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"L = 500. # The length of a steel rod in cm\n",
"t = 40. # The increase in temperature in degree centigrade\n",
"y = 2. * 10**12 # The youngs modulus of elasticity of steel in dynes/cm**2\n",
"e = 12. * 10**-6 # The coefficient of linear expansion of steel in per degree centigrade\n",
"\n",
"# Calculations\n",
"S = y * e * t # The stress in the rod in dynes/cm**2\n",
"\n",
"# Output\n",
"print 'The stress in the rod is %3g dynes/cm^2' % (S)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The stress in the rod is 9.6e+08 dynes/cm^2\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.3 Page No : 36"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"L = 800. # The length of the wire in cm\n",
"r = 0.2 # The radius of the wire in cm\n",
"t = 10. # The temperature fall in degree centigrade\n",
"a = 12. * 10**-6 # The coefficient of linear expansion of steel wire in per degree centigrade\n",
"y = 2. * 10**12 # The youngs modulus of elasticity of steel in dynes/cm**2\n",
"pi = (22. / 7) # Mathematical constant pi\n",
"\n",
"# Calculations\n",
"I = y * a * t * pi * r**2 # The increase in tension in dynes\n",
"\n",
"# Output\n",
"print 'The increase in tension is %3g dynes' % (I)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The increase in tension is 3.01714e+07 dynes\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.4 Page No : 39"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"A = 2. * 10**-6 # The cross section area of a uniform rod in m**2\n",
"t = 20. # The change in temperature in degree centigrade\n",
"y = 10.**11 # The youngs modulus of the rod in newtons/m**2\n",
"a = 12. * 10**-6 # The coefficient of linear expansion of rod in per degree centigrade\n",
"\n",
"# Calculations\n",
"F = y * a * t * A # The force required to prevent it from expanding in newtons\n",
"E = (1. / 2) * y * a * t * a * t # The energy stored per unit volume in j/m**3\n",
"\n",
"# Output\n",
"print 'The force required to prevent the rod from expanding is %3.0f newtons \\\n",
"\\nThe Energy stored per unit volume is %3.0f j/m^3' % (F, E)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The force required to prevent the rod from expanding is 48 newtons \n",
"The Energy stored per unit volume is 2880 j/m^3\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.5 Page No : 41"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"d = 10.**-3 # The diameter of a steel wire in m\n",
"t = 20. # The difference in the temperature in degree centigrade\n",
"y = 2. * 10**11 # The youngs modulus of a steel wire in newtons/m**2\n",
"a = 12. * 10**-6 # The coefficient of linear expansion of steel wire in per degree centigrade\n",
"pi = (22. / 7) # Mathematical constant value\n",
"\n",
"# calculations\n",
"A = (pi * d**2) / 4 # The cross sectional area of the steel wire in m**2\n",
"# Force required to maintain the original length in kg wt\n",
"F = (y * a * t * A) / (9.8)\n",
"\n",
"# output\n",
"print 'Force required to maintain the original length is %3.3f kg wt ' % (F)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Force required to maintain the original length is 3.848 kg wt \n"
]
}
],
"prompt_number": 6
}
],
"metadata": {}
}
]
}PKDIjI;;!Heat and Thermodynamics/ch3.ipynb{
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{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 3 : Calorimetry\n"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.1 Page No : 49"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"T = 5. # Time taken for a liquid to cool from 80 to 50 degree centigrade in minutes\n",
"t11 = 80. # The initial temperature of the liquid in degree centigrade\n",
"t12 = 50. # The final temperature of the liquid in degree centigrade\n",
"t21 = 60. # If the initial temperature of the liquid in degree centigrade\n",
"t22 = 30. # If the final temperature of the liquid in degree centigrade\n",
"ts = 20. # The temperature of the surrounding in degree centigrade\n",
"\n",
"# Calculations\n",
"# The time taken for the liquid to cool from 60 to 30 degree centigrade in\n",
"# minutes\n",
"T1 = ((math.log((t22 - ts) / (t21 - ts))) /\n",
" (math.log((t12 - ts) / (t11 - ts)))) * T\n",
"\n",
"# Output\n",
"print 'The time taken for a liquid to cool from 60 to 30 degree centigrade is t = %3.0f minutes ' % (T1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The time taken for a liquid to cool from 60 to 30 degree centigrade is t = 10 minutes \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.2 Page No : 55"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"dw = 1. # The density of water in g/cm**3\n",
"da = 0.8 # The density of alcohol in g/cm**3\n",
"t1 = 100. # The time taken for the water to cool from 50 to 40 degree centigrade in seconds\n",
"t2 = 74. # The time taken for the alcohol to cool from 50 to 40 degree centigrade in seconds\n",
"V = 1. # Let the volume of either liquid be in cm**3\n",
"\n",
"# Calculations\n",
"m = V * dw # The mass of water in g\n",
"M = V * da # The mass of alcohol in g\n",
"w = V # Water equivalent of each calorimeter in cm**3\n",
"# The specific heat of alcohol in calorie/g-K\n",
"C = ((((m + w) * t2) / (M * t1)) - (w / M))\n",
"\n",
"# Output\n",
"print 'The specific heat of alcohol is C = %3.1f calorie/g-K' % (C)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The specific heat of alcohol is C = 0.6 calorie/g-K\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.3 Page No : 61"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"t = 5. # Time taken for a body to cool from 60 to 40 degree centigrade in minutes\n",
"t11 = 60. # The initial temperature of the body in degree centigrade\n",
"t12 = 40. # The final temperature of the body in degree centigrade\n",
"ts = 10. # The temperature of the surrounding in degree centigrade\n",
"\n",
"# Calculations\n",
"# The constant value for the first case at ts\n",
"K = math.log((t12 - ts) / (t11 - ts))\n",
"# The temperature after the next 5 minutes in degree centigrade\n",
"x = ((math.exp(K)) * (t12 - ts)) + ts\n",
"\n",
"# Output\n",
"print 'The temperature after the next 5 minutes is x = %3.0f degree centigrade ' % (x)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The temperature after the next 5 minutes is x = 28 degree centigrade \n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.4 Page No : 63"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"T = 4. # Time taken for a liquid to cool from 70 to 50 degree centigrade in minutes\n",
"t11 = 70. # The initial temperature of the liquid in degree centigrade\n",
"t12 = 50. # The final temperature of the liquid in degree centigrade\n",
"t21 = 50. # If the initial temperature of the liquid in degree centigrade\n",
"t22 = 40. # If the final temperature of the liquid in degree centigrade\n",
"ts = 25. # The temperature of the surrounding in degree centigrade\n",
"\n",
"# Calculations\n",
"# The time taken for the liquid to cool from 50 to 40 degree centigrade in\n",
"# minutes\n",
"T1 = ((math.log((t22 - ts) / (t21 - ts))) /\n",
" (math.log((t12 - ts) / (t11 - ts)))) * T\n",
"\n",
"# Output\n",
"print 'The time taken for a liquid to cool from 50 to 40 degree centigrade is t = %3.3f minutes ' % (T1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The time taken for a liquid to cool from 50 to 40 degree centigrade is t = 3.476 minutes \n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.5 Page No : 67"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"t = 6. # Time taken for a liquid to cool from 80 to 60 degree centigrade in minutes\n",
"T = 10. # To find the temperature after the time in minutes\n",
"t11 = 80. # The initial temperature of the liquid in degree centigrade\n",
"t12 = 60. # The final temperature of the liquid in degree centigrade\n",
"ts = 30. # The temperature of the surrounding in degree centigrade\n",
"\n",
"# Calculations\n",
"# The constant value for the first case at ts\n",
"K = (math.log((t12 - ts) / (t11 - ts))) / (-t)\n",
"# The temperature after the next 10 minutes in degree centigrade\n",
"x = ((math.exp(-T * K)) * (t12 - ts)) + ts\n",
"\n",
"# Output\n",
"print 'The temperature after the next 10 minutes is x = %3.2f degree centigrade ' % (x)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The temperature after the next 10 minutes is x = 42.80 degree centigrade \n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.6 Page No : 69"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"t = 5. # The time taken for a body to cool from 80 to 64 degree centigrade in minutes\n",
"t11 = 80. # The initial temperature of the body in degree centigrade\n",
"t12 = 64. # The final temperature of the body in degree centigrade\n",
"t21 = 52. # The temperature of the body after 10 minutes in degree centigrade\n",
"T = 10. # The time taken for a body to cool from 80 to 52 degree centigrade in minutes\n",
"T1 = 15. # To find the temperature after the time in minutes\n",
"\n",
"# Calculations\n",
"# The temperature of the surroundings in degree centigrade\n",
"ts = ((t21 * t11) - (t12**2)) / (t11 + t21 - (2 * t12))\n",
"# The constant value for the first case at ts\n",
"K = (math.log((t21 - ts) / (t12 - ts)))\n",
"# The temperature after the next 15 minutes in degree centigrade\n",
"x = ((math.exp(K)) * (t21 - ts)) + ts\n",
"\n",
"# Output\n",
"print '(1)The temperature of the surroundings is %3.0f degree centigrade \\n (2)The temperature after the 15 minutes is %3.0f degree centigrade ' % (ts, x)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(1)The temperature of the surroundings is 16 degree centigrade \n",
" (2)The temperature after the 15 minutes is 43 degree centigrade \n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.7 Page No : 76"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"t2 = 2. # The time taken for the liquid to cool from 50 to 40 degree centigrade in minutes\n",
"t11 = 50. # The initial temperature of the liquid in degree centigrade\n",
"t12 = 40. # The final temperature of the liquid in degree centigrade\n",
"t1 = 5. # The time taken for the water to cool from 50 to 40 degree centigrade in minutes\n",
"m = 100. # The mass of water in gms\n",
"M = 85. # The mass of liquid in gms\n",
"w = 10. # Water equivalent of the vessel in gms\n",
"\n",
"# Calculations\n",
"# The specific heat of a liquid in calories/g-K\n",
"C = (((m + w) * (t2 * 60)) / (M * (t1 * 60))) - (w / M)\n",
"\n",
"# Output\n",
"print 'The specific heat of a liquid is C = %3.1f calories/g-K' % (C)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The specific heat of a liquid is C = 0.4 calories/g-K\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.8 Page No : 79"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"V = 22400. # The volume of One gram molecule of a gas at N.T.P in cm**3\n",
"p = 76. # The pressure in cm of Hg\n",
"T = 273. # The temperature in K\n",
"\n",
"# Calculations\n",
"P = p * 13.6 * 981 # The pressure in dynes/cm**2\n",
"# The universal gas constant for one gram molecule of a gas in ergs/mole-K\n",
"R = (P * V) / T\n",
"\n",
"# Output\n",
"print 'The universal gas constant for one gram molecule of a gas is R = %3.4g ergs/mole-K' % (R)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The universal gas constant for one gram molecule of a gas is R = 8.32e+07 ergs/mole-K\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.9 Page No : 83"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"Cp = 0.23 # Specific heat of air at constant pressure\n",
"J = 4.2 * 10**7 # The amount of energy in ergs/cal\n",
"d = 1.293 # The density of air at N.T.P in g/litre\n",
"p = 76. # The pressure in cm of Hg\n",
"T = 273. # The temperature in K\n",
"\n",
"# Calculations\n",
"P = p * 13.6 * 980 # The pressure in dynes/cm**2\n",
"V = (1000 / d) # Volume of one gram of air at N.T.P in cm**3\n",
"r = (P * V) / T # The gas constant for one gram of a gas in ergs/g-K\n",
"Cv = Cp - (r / J) # Specific heat of air at constant volume\n",
"\n",
"# Output\n",
"print 'The specific heat of air at constant volume is Cv = %3.4f ' % (Cv)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The specific heat of air at constant volume is Cv = 0.1617 \n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.10 Page No : 88"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"w = 4. # The Molecular weight of helium\n",
"v = 22400. # The volume of one gram molecule of a gas at N.T.P in cm**3\n",
"p = 76. # The pressure in cm of Hg\n",
"T = 273. # The temperature in K\n",
"J = 4.2 * 10**7 # The amount of energy in ergs/cal\n",
"\n",
"# Calculations\n",
"V = (v / w) # The volume of one gram of helium at N.T.P in cm**3\n",
"P = p * 13.6 * 980 # The pressure in dynes/cm**2\n",
"r = (P * V) / T # The gas constant for one gram of a gas in ergs/g-K\n",
"C = r / J # The difference in the two specific heats of one gram of helium\n",
"\n",
"# Output\n",
"print 'The difference in the two specific heats of one gram of helium is Cp-Cv = %3.4f' % (C)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The difference in the two specific heats of one gram of helium is Cp-Cv = 0.4947\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.11 Page No : 94"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"V = 25. # Volume of gasoline consumed by an engine in litres/hour\n",
"cv = 6. * 10**6 # The calorific value of gasoline in calories/litre\n",
"P = 35. # The output of the engine in kilowatts\n",
"\n",
"# Calculations\n",
"h = V * cv # Total heat produced by gasoline in one hour in calories\n",
"H = h / 3600 # Heat produced per second in cal/s\n",
"I = H * 4.2 # Heat produced per second in joules/s or watts\n",
"E = ((P * 1000) / I) * 100 # The efficiency in percent\n",
"\n",
"# Output\n",
"print 'The efficiency of the engine is %3.0f percent ' % (E)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The efficiency of the engine is 20 percent \n"
]
}
],
"prompt_number": 11
}
],
"metadata": {}
}
]
}PKDIY522!Heat and Thermodynamics/ch4.ipynb{
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"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 4 : Change of State\n"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.1 Page No : 106"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"m = 1000. # Mass of Ice in gms\n",
"Sp = 0.5 # Specific heat of Ice in cal/g-K\n",
"t1 = -10 # Initial temperature of Ice in degree centigrade\n",
"t2 = 0. # The final temperature of Ice in degree centigrade\n",
"Li = 80. # Latent heat of fusion of ice in cals per gram\n",
"Ls = 540. # Latent heat of fusion of steam in cals per gram\n",
"\n",
"# Calculations\n",
"h1 = m * -t1 * Sp # Heat required to raise the temperature of Ice in cals\n",
"h2 = m * Li # Heat required to melt ice at 0 degree centigrade in cals\n",
"h3 = m * 100 # Heat required to raise the temperature of water from 0 to 100 degree centigrade in cals\n",
"h4 = m * Ls # Heat required to convert water into steam at 100 degree centigrade in cals\n",
"T = h1 + h2 + h3 + h4 # Total quantity of heat required in cals\n",
"\n",
"# Output\n",
"print 'Total quantity of heat required is %3.0f cals ' % (T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Total quantity of heat required is 725000 cals \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.2 Page No : 112"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"m = 1. # Mass of steam in gms\n",
"Ls = 537. # Latent heat of fusion of steam in cal per gram\n",
"mi = 100. # mass of ice in gms\n",
"Li = 80. # Latent heat of fusion of ice in cal per gram\n",
"\n",
"# Calculations\n",
"h1 = m * Ls # Heat given out by one gram of steam when converted from steam into water at 100 degree centigrade in cals\n",
"h2 = 1. * 100 # Heat given out by one gram of water when cooled from 100 to 0 degree centigrade in cals\n",
"h = h1 + h2 # Total quantity of heat given out by one gram of steam in cals\n",
"m = h / Li # The amount of Ice melted in gms\n",
"\n",
"# Output\n",
"print 'The amount of Ice melted is m = %3.2f gms ' % (m)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The amount of Ice melted is m = 7.96 gms \n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.3 Page No : 115"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"m = 100. # Mass of water in gms\n",
"tw = 40. # The temperature of water in degree centigrade\n",
"mi = 52. # Mass of Ice in gms\n",
"Lw = 100. # Latent heat of fusion of water in cals per gram\n",
"Li = 80. # Latent heat of fusion of Ice in cals per gram\n",
"\n",
"# Calculations\n",
"h = Lw * tw # Heat lost by water when its temperature falls from 40 to 0 degree centigrade in cals\n",
"hi = mi * Li # Heat gained by Ice in cals\n",
"hg = h # The amount of heat gained by Ice in cals\n",
"ml = (hg / Li) # The amount of Ice melted in gms\n",
"M = mi - ml # The amount of ice remaining in gms\n",
"W = m + (mi - M) # The amount of water in gms\n",
"\n",
"# Output\n",
"print 'The remaining Ice is %3.0f g \\n Hence the result will be %3.0f g of Ice and %3.0f g of water at 0 degree centigrade ' % (M, M, W)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The remaining Ice is 2 g \n",
" Hence the result will be 2 g of Ice and 150 g of water at 0 degree centigrade \n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.4 Page No : 121"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"m = 100. # Let the mass of water in gms\n",
"t = 15. # Time taken for an electric kettle to heat a certain quantity of water from 0 to 100 degree centigrade in minutes\n",
"T = 80. # Time taken to turn all the water at 100 degree centigrade into steam in minutes\n",
"Lw = 100. # Latent heat of fusion of water in cals per gram\n",
"\n",
"# Calculations\n",
"h1 = m * Lw # Heat required to raise its temperature from 0 to 100 degree centigrade in cals\n",
"h2 = h1 # Heat produced by electric kettle in 15 minutes in cals\n",
"h3 = h2 / 15 # Heat produced by electric kettle in 1 minute in cals\n",
"h4 = h3 * 80 # Heat produced by electric kettle in 80 minutes in cals\n",
"L = h4 / m # Latent heat of steam in cal/g\n",
"\n",
"# Output\n",
"print 'The latent heat of steam is L = %3.2f cal/g ' % (L)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The latent heat of steam is L = 533.33 cal/g \n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.5 Page No : 123"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"m = 50. # Mass of water in gms\n",
"t1 = 15. # Initial temperature in degree centigrade\n",
"t2 = -20 # Final temperature in degree centigrade\n",
"Sp = 0.5 # Specific heat of Ice in cal/g-K\n",
"Li = 80. # Latent heat of fusion of Ice in cals per gram\n",
"\n",
"# Calculations\n",
"h1 = m * 1 * t1 # Heat removed in cooling water from 15 to 0 degree centigrade in cal\n",
"h2 = m * Li # Heat removed in converting water into Ice at 0 degree centigrade in cal\n",
"h3 = m * Sp * -t2 # Heat removed in cooling ice from 0 to -20 degree centigrade in cal\n",
"H = h1 + h2 + h3 # Total heat removed in one hour in cal\n",
"H1 = H / 60 # Heat removed per minute in cal/minute\n",
"\n",
"# Output\n",
"print 'The Quantity of heat removed per minute is %3.1f cal/minute ' % (H1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Quantity of heat removed per minute is 87.5 cal/minute \n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.6 Page No : 128"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"M = 20. # Mass of the substance in g\n",
"t = 100. # The temperature of the substance in degree centigrade\n",
"a = 1. / 100 # Area of cross section in cm**2\n",
"l = 5. # The length of the coloumn through which liquid moves in cm\n",
"V1 = 1000. # The volume of water in cm**3\n",
"V2 = 1090. # The volume of Ice from the volume of water on freezing in cm**3\n",
"Li = 80. # Latent heat of Ice in cals per gram\n",
"\n",
"# Calculations\n",
"V = V2 - V1 # The decrease in volume of Ice in cm**3\n",
"Vi = V / 1000 # The decrease in volume when one gram of ice melts in cm**3\n",
"v = l * a # Decrease in volume in cm**3\n",
"# Specific heat of the substance incal/g degree centigrade\n",
"S = (Li * v) / (Vi * M * t)\n",
"\n",
"# Output\n",
"print 'The specific heat of the substance is %3.3f cal/g.degree centigrade ' % (S)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The specific heat of the substance is 0.022 cal/g.degree centigrade \n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.7 Page No : 132"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"M = 27. # The mass of the substance in g\n",
"t = 100. # The temperature of the substance in degree centigrade\n",
"a = 3. / 100 # Area of cross section in cm**2\n",
"l = 10. # The length of the coloumn through which liquid moves in cm\n",
"Li = 80. # Latent heat of Ice in cals per gram\n",
"V1 = 1000. # The volume of water in cm**3\n",
"V2 = 1090. # The volume of Ice from the volume of water on freezing in cm**3\n",
"\n",
"# Calculations\n",
"v = l * a # Decrease in volume in cm**3\n",
"V = V2 - V1 # The decrease in volume of Ice in cm**3\n",
"Vi = V / 1000 # The decrease in volume when one gram of ice melts in cm**3\n",
"# Specific heat of the substance incal/g degree centigrade\n",
"S = (Li * v) / (Vi * M * t)\n",
"\n",
"# Output\n",
"print 'The specific heat of the substance is %3.3f cal/g.degree centigrade ' % (S)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The specific heat of the substance is 0.099 cal/g.degree centigrade \n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.8 Page No : 134"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"t = 16.5 # The temperature of air in degree centigrade\n",
"d = 6.5 # The dew point in degree centigrade\n",
"s1 = 7.05 # S.V.P at 6 degree centigrade in mm\n",
"s2 = 7.51 # S.V.P at 7 degree centigrade in mm\n",
"s3 = 13.62 # S.V.P at 16 degree centigrade in mm\n",
"s4 = 14.42 # S.V.P at 17 degree centigrade in mm\n",
"\n",
"# Calculations\n",
"s5 = (s1 + s2) / 2 # S.V.P at 6.5 degree centigrade in mm\n",
"s6 = (s3 + s4) / 2 # S.V.P at 16.5 degree centigrade in mm\n",
"R = (s5 / s6) * 100 # Relative humidity of air in percent\n",
"\n",
"# Output\n",
"print 'The percentage relative humidity of air is R.H = %3.1f percent ' % (R)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The percentage relative humidity of air is R.H = 51.9 percent \n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.9 Page No : 140"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"R = 52. # The relative humidity of air in percent\n",
"t = 20. # The temperature of air in degree centigrade\n",
"s1 = 17.5 # S.V.P of water at 20 degree centigrade in mm\n",
"s2 = 9.2 # S.V.P of water at 10 degree centigrade in mm\n",
"s3 = 8.6 # S.V.P of water at 9 degree centigrade in mm\n",
"\n",
"# Calculations\n",
"s4 = (R / 100) * s1 # S.V.P at dew point in mm\n",
"s5 = s2 - s3 # S.V.P for 1 degree centigrade difference in mm\n",
"# The dew point temperature in degree centigrade\n",
"d = 9. + ((s4 - s3) / (s2 - s3))\n",
"\n",
"# Output\n",
"print 'The dew point temperature is %3.2f degree centigrade ' % (d)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The dew point temperature is 9.83 degree centigrade \n"
]
}
],
"prompt_number": 9
}
],
"metadata": {}
}
]
}PKDIO4tt!Heat and Thermodynamics/ch5.ipynb{
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 5 : Nature of Heat"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.1 Page No : 159"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"v = 480. # The velocity of a lead bullet in m/s\n",
"Sp = 0.03 # Specific heat of lead cal/g-K\n",
"\n",
"# Calculations\n",
"m = 10. # Let us assume the mass of bullet in gms\n",
"V = v * 100 # The velocity of the bullet in cm/s\n",
"W = (1. / 2) * m * (V**2) # The work done in ergs\n",
"J = 4.2 * 10**7 # The mechanical equivalent of heat in ergs/calorie\n",
"H = W / J # The amount of heat produced in cals\n",
"H1 = H / 2 # Half of the heat energy is used to raise the temperature of the bullet in cals\n",
"t = H1 / (m * Sp) # The rise in the temperature in degree centigrade\n",
"\n",
"# Output\n",
"print 'The rise in the temperature is t = %3.2f degree centigrade ' % (t)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The rise in the temperature is t = 457.14 degree centigrade \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.2 Page No : 162"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"t = 1. # The increase in the temperature of a piece of aluminium in degree centigrade\n",
"a = 6. * 10**23 # The number of atoms present in 27 g of aluminium in atoms\n",
"Sp = 0.22 # The specific heat of aluminium in cal/g-K\n",
"m = 27. # The amount of aluminium in g\n",
"J = 4.2 * 10**7 # The mechanical equivalent of heat in ergs/calorie\n",
"\n",
"# Calculations\n",
"H = m * Sp * t # Heat required to raise the temperature of 27 gms of aluminium by 1 degree centigrade in cals\n",
"E = m * Sp * J # Energy gained by atoms of aluminium in ergs\n",
"E1 = E / a # Increase in energy per atom of aluminium in ergs\n",
"\n",
"# Output\n",
"print 'The increase in energy per atom of aluminium is %3.4g ergs ' % (E1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The increase in energy per atom of aluminium is 4.158e-16 ergs \n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.3 Page No : 168"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"h = 50. # The height from which water falls in metres\n",
"m = 100. # Let us assume the mass of the water in gms\n",
"g = 980. # Gravitational constant in gms/s**2\n",
"J = 4.2 * 10**7 # The mechanical equivalent of heat in ergs/calorie\n",
"\n",
"# Calculations\n",
"h1 = h * 100 # The height from which water falls in cm\n",
"W = m * g * h1 # The work done in ergs\n",
"t = W / (J * m) # The rise in temperature of water in degree centigrade\n",
"\n",
"# Output\n",
"print 'The rise in temperature of water is t = %3.3f degree centigrade ' % (t)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The rise in temperature of water is t = 0.117 degree centigrade \n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.4 Page No : 174"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"v = 1. # The volume of oxygen at N.T.P in cm**3\n",
"d = 13.6 # The density of mercury in g/cm**3\n",
"r = 4.62 * 10**4 # The R.M.S velocity of oxygen molecules at 0 degree centigrade in cm/s\n",
"m = 52.8 * 10**-24 # Mass of one molecule of oxygen in g\n",
"g = 980. # Gravitational constant in gms/s**2\n",
"\n",
"# Calculations\n",
"P = 76. * g * d # The pressure in dynes/cm**2\n",
"n = ((3 * P) / (m * r**2)) # Number of molecules in 1 cc of oxygen at N.T.P\n",
"\n",
"# Output\n",
"print 'The number of molecules in 1 c.c of oxygen at N.T.P is n = %3.4g ' % (n)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The number of molecules in 1 c.c of oxygen at N.T.P is n = 2.696e+19 \n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.5 Page No : 177"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"t = -100 # The given temperature in degree centigrade\n",
"\n",
"# Calculations\n",
"T1 = t + 273 # The given temperature in K\n",
"m1 = 1. # number of hydrogen molecules\n",
"m2 = 16. # number of oxygen molecules\n",
"m = m2 / m1 # Number of oxygen molecules to the hydrogen molecules\n",
"T2 = (T1 * m) - 273 # The temperature in degree centigrade\n",
"\n",
"# Output\n",
"print 'The temperature at which the oxygen molecules have the same root mean square velocity as that of hydrogen molecules is T2 = %3.0f degree centigrade ' % (T2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The temperature at which the oxygen molecules have the same root mean square velocity as that of hydrogen molecules is T2 = 2495 degree centigrade \n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.6 Page No : 180"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"t = 27. # The given temperature in degree centigrade\n",
"d = 13.6 # The density of mercury in g/cm**3\n",
"g = 980. # Gravitational constant in gms/s**2\n",
"m1 = 16. # number of oxygen molecules\n",
"D = 0.000089 # The density of hydrogen at N.T.P in g/cc\n",
"T = 273. # The temperature at N.T.P in K\n",
"\n",
"# Calculations\n",
"P = 76. * g * d # The pressure in dynes/cm**2\n",
"p = m1 * D # The density of oxygen at N.T.P in g/cc\n",
"C = ((3 * P) / (p))**(1. / 2) # The RMS velocity of oxygen molecule in cm/s\n",
"T1 = t + T # The given temperature in K\n",
"# The RMS velocity of the molecules at 27 degree centigrade in cm/s\n",
"C1 = C * (T1 / T)**(1. / 2)\n",
"\n",
"# Output\n",
"print 'The RMS velocity of the oxygen molecules at 27 degree centigrade is C1 = %3.4g cm/s ' % (C1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The RMS velocity of the oxygen molecules at 27 degree centigrade is C1 = 4.843e+04 cm/s \n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.7 Page No : 186"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"d = 13.6 # The density of mercury in g/cm**3\n",
"g = 980. # Gravitational constant in gms/s**2\n",
"m = 3.2 # Mass of oxygen in gms\n",
"t = 27. # The given temperature in degree centigrade\n",
"p = 76. # The pressure in cm of Hg\n",
"R = 8.31 * 10**7 # The Universal gas constant in ergs/g mol-K\n",
"\n",
"# Calculations\n",
"P = p * g * d # The given pressure in dynes/cm**2\n",
"T = t + 273 # The given temperature in K\n",
"V = (T * R) / P # Volume per g mol of oxygen in cc per g mol\n",
"m1 = 32. # Molecular weight of Oxygen\n",
"V1 = V * (m / m1) # Volume of 3.2 g of oxygen in cc\n",
"\n",
"# Output\n",
"print 'The Volume occupied by 3.2 gms of Oxygen is V = %3.0f cc ' % (V1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Volume occupied by 3.2 gms of Oxygen is V = 2461 cc \n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.9 Page No : 193"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"v = 1. # The volume of an Ideal gas at N.T.P in m**3\n",
"d = 13.6 # The density of mercury in g/cm**3\n",
"g = 980. # Gravitational constant in gms/s**2\n",
"p = 76. # The pressure in cm of Hg\n",
"R = 8.31 * 10**7 # The Universal gas constant in ergs/g mol-K\n",
"N = 6.023 * 10**23 # The Avogadro number\n",
"T = 273. # The temperature at N.T.P in K\n",
"\n",
"# Calculations\n",
"P = p * g * d # The given pressure in dynes/cm**2\n",
"x = (P * N * 10**6) / (R * T) # Number of molecules in one cubic metre volume\n",
"\n",
"# Output\n",
"print 'The number of molecules in one cubic metre of an ideal gas at N.T.P is x = %3.4g ' % (x)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The number of molecules in one cubic metre of an ideal gas at N.T.P is x = 2.689e+25 \n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.10 Page No : 196"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"v = 1. # The volume of an ideal gas in litre\n",
"d = 13.6 # The density of mercury in g/cm**3\n",
"g = 980. # Gravitational constant in gms/s**2\n",
"p = 76. # The pressure in cm of Hg\n",
"R = 8.31 * 10**7 # The Universal gas constant in ergs/g mol-K\n",
"N = 6.023 * 10**23 # The Avogadro number\n",
"T = 273. # The temperature at N.T.P in K\n",
"t = 136.5 # The given temperature in degree centigrade\n",
"p1 = 3. # The given atmospheric pressure in atm pressure\n",
"\n",
"# Calculations\n",
"T1 = T + t # The given temperature in K\n",
"P = p * g * d # The given pressure in dynes/cm**2\n",
"x = (p1 * P * N * 10**3) / (R * T1) # Number of molecules in one litre volume\n",
"\n",
"# Output\n",
"print 'The number of molecules in one litre of an ideal gas volume is x = %3.4g ' % (x)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The number of molecules in one litre of an ideal gas volume is x = 5.378e+22 \n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.11 Page No : 203"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"v = 1. # The volume of a gas in cc\n",
"d = 13.6 # The density of mercury in g/cm**3\n",
"p2 = 10.**-7 # The pressure in cm of Hg\n",
"g = 980. # Gravitational constant in gms/s**2\n",
"p1 = 76. # The pressure in cm of Hg\n",
"R = 8.31 * 10**7 # The Universal gas constant in ergs/g mol-K\n",
"N = 6.023 * 10**23 # The Avogadro number\n",
"T = 273. # The temperature at N.T.P in K\n",
"n1 = 2.7 * 10**19 # The number of molecules per cc of gas at N.T.P\n",
"t2 = 0. # The given temperature in degree centigrade\n",
"t3 = 39. # The given temperature in degree centigrade\n",
"\n",
"# Calculations\n",
"P1 = p1 * g * d # The given pressure in dynes/cm**2\n",
"P2 = p2 * g * d # The given pressure in dynes/cm**2\n",
"# The number of molecules per cc of the gas at 0 degree centigrade\n",
"n2 = n1 * (P2 / P1)\n",
"T2 = t2 + 273 # The given temperature in K\n",
"T3 = t3 + 273 # The given temperature in K\n",
"# The number of molecules per cc of the gas at 398 degree centigrade\n",
"n3 = n2 * (T2 / T3)\n",
"\n",
"# Output\n",
"print 'The number of molecules per cc of the gas , \\n (1)at 0 degree centigrade and 10^-6 mm pressure of mercury is n2 = %3.4g \\n (2)at 39 degree centigrade and 10^-6 mm pressure of mercury is n3 = %3.4g' % (n2, n3)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The number of molecules per cc of the gas , \n",
" (1)at 0 degree centigrade and 10^-6 mm pressure of mercury is n2 = 3.553e+10 \n",
" (2)at 39 degree centigrade and 10^-6 mm pressure of mercury is n3 = 3.109e+10\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.12 Page No : 208"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"T = 300. # The given temperature in K\n",
"R = 8.3 * 10**7 # The Universal gas constant in ergs/g mol-K\n",
"\n",
"# Calculations\n",
"# The total random kinetic energy per gram -molecule of oxygen in joules\n",
"E = ((3. / 2) * (R * T)) / 10**7\n",
"\n",
"# Output\n",
"print 'The total random kinetic energy of one gm-molecule of oxygen at 300 K is K.E = %3.0f joules' % (E)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The total random kinetic energy of one gm-molecule of oxygen at 300 K is K.E = 3735 joules\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.13 Page No : 213"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"T = 300. # The given temperature in K\n",
"k = 1.38 * 10**-16 # Boltzmann constant in erg/molecule-deg\n",
"\n",
"# Calculations\n",
"E = (3. / 2) * k * T # The average Kinetic energy of a molecule in ergs\n",
"\n",
"# Output\n",
"print 'The Average Kinetic energy of a molecule of a gas at 300 K is K.E = %3.4g ergs ' % (E)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Average Kinetic energy of a molecule of a gas at 300 K is K.E = 6.21e-14 ergs \n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.14 Page No : 220"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"R = 8.32 # Universal gas constant in joules/mole-K\n",
"t = 727. # The given temperature in degree centigrade\n",
"N = 6.06 * 10**23 # The Avogadro number\n",
"\n",
"# Calculations\n",
"T = 273. + t # The given temperature in K\n",
"k = R / N # Boltzmann constant in joules/mol-K\n",
"E = (3. / 2) * k * T # Mean translational kinetic energy per molecule in joules\n",
"\n",
"# Output\n",
"print 'The mean translational kinetic energy per molecule is K.E = %3.4g joule ' % (E)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The mean translational kinetic energy per molecule is K.E = 2.059e-20 joule \n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.15 Page No : 224"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"T = 300. # The given temperature in K\n",
"M = 28. # Molecular weight of nitrogen in g\n",
"R = 8.3 * 10**7 # The Universal gas constant in ergs/g mol-K\n",
"\n",
"# Calculations\n",
"E = (3. / 2) * R * T # The total random kinetic energy of nitrogen in ergs\n",
"# The total random kinetic energy of one gram of nitrogen at 300 K in joule\n",
"E1 = E / (M * 10**7)\n",
"\n",
"# Output\n",
"print 'The total random kinetic energy of one gram of nitrogen at 300 K is K.E = %3.1f joule ' % (E1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The total random kinetic energy of one gram of nitrogen at 300 K is K.E = 133.4 joule \n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.16 Page No : 228"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"T = 200. # The given temperature in K\n",
"m = 2. # Given mass of Helium in g\n",
"M = 4. # Molecular weight of helium in g\n",
"R = 8.3 * 10**7 # The Universal gas constant in ergs/g mol-K\n",
"\n",
"# Calculations\n",
"# The energy for 2 g of helium in joules\n",
"E = (m * (3. / 2) * (R * T) / (M)) / 10**7\n",
"\n",
"# Output\n",
"print 'The total random kinetic energy of 2 g of helium at 200 K is K.E = %3.0f joules' % (E)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The total random kinetic energy of 2 g of helium at 200 K is K.E = 1245 joules\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.17 Page No : 233"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"T = 300. # The given temperature in K\n",
"R = 8.3 * 10**7 # The Universal gas constant in ergs/g mol-K\n",
"M = 221. # The molecular weight of mercury\n",
"\n",
"# Calculations\n",
"# The root mean square velocity of a molecule of mercury vapour at 300 K\n",
"# in cm/s\n",
"C = ((3 * R * T) / (M))**(1. / 2)\n",
"\n",
"# Output\n",
"print 'The root mean square velocity of a molecule of mercury vapour at 300 K is C = %3.4g cm/s ' % (C)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The root mean square velocity of a molecule of mercury vapour at 300 K is C = 1.839e+04 cm/s \n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.18 Page No : 239"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"T = 300. # The given temperature in K\n",
"M = 32. # Molecular weight of oxygen\n",
"R = 8.3 * 10**7 # The Universal gas constant in ergs/g mol-K\n",
"\n",
"# Calculations\n",
"# Total random kinetic energy of 1 g molecule of oxygen in ergs\n",
"E = (3. / 2) * R * T\n",
"# The required speed of one gram molecule of oxygen in cm/s\n",
"v = ((E) * (2 / M))**(1. / 2)\n",
"\n",
"# Output\n",
"print 'The required speed of one gram molecule of oxygen is v = %3.2g cm/s ' % (v)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The required speed of one gram molecule of oxygen is v = 4.8e+04 cm/s \n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.19 Page No : 242"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"v = 8. # The speed of the earths first satellite in km/s\n",
"R = 8.3 * 10**7 # The Universal gas constant in ergs/g mol-K\n",
"M = 2. # Molecular weight of hydrogen\n",
"\n",
"# Calculations\n",
"V = v * 10**5 # The speed of the earths first satellite in cm/s\n",
"T = (M * V**2) / (3 * R) # The temperature at which it becomes equal in K\n",
"\n",
"# Output\n",
"print 'The temperature at which the r.m.s velocity of a hydrogen molecule will be equal to the speed of earths first satellite is T = %3.4g K' % (T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The temperature at which the r.m.s velocity of a hydrogen molecule will be equal to the speed of earths first satellite is T = 5141 K\n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.20 Page No : 248"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"t1 = 0. # The given temperature in degree centigrade\n",
"\n",
"# Calculations\n",
"T1 = t1 + 273 # The given temperature in K\n",
"# The temperature at which the r.m.s velocity of a gas be half its value\n",
"# at 0 degree centigrade in K\n",
"T2 = (1. / 2)**2 * T1\n",
"T21 = T2 - 273 # The required temperature in degree centigrade\n",
"\n",
"# Output\n",
"print 'The required temperature is T2 = %3.2f K (or) %3.2f degree centigrade ' % (T2, T21)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The required temperature is T2 = 68.25 K (or) -204.75 degree centigrade \n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.21 Page No : 252"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"n = 1.66 * 10**-4 # The viscosity of the gas in dynes/cm**2\n",
"C = 4.5 * 10**4 # The R.M.S velocity of the molecules in cm/s\n",
"d = 1.25 * 10**-3 # The density of the gas in g/cc\n",
"N = 6.023 * 10**23 # The Avogadro number\n",
"V = 22400. # The volume of a gas at N.T.P in cc\n",
"pi = 3.142 # The mathematical constant of pi\n",
"\n",
"# Calculations\n",
"L = (3 * n) / (d * C) # The mean free path of the molecules of the gas in cm\n",
"F = (C / L) # The frequency collision in per sec\n",
"n = N / V # Number of molecules per cc\n",
"# Molecular diameter of the gas molecules in cm\n",
"D = 1. / ((1.414 * pi * n * L)**(1. / 2))\n",
"\n",
"# Output\n",
"print '(1)The mean free path of the molecules of the gas is %3.0g cm \\n (2)The frequency of collision is N = %3.0g /sec \\n (3)Molecular diameter of the gas molecules is d = %3.0g cm ' % (L, F, D)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(1)The mean free path of the molecules of the gas is 9e-06 cm \n",
" (2)The frequency of collision is N = 5e+09 /sec \n",
" (3)Molecular diameter of the gas molecules is d = 3e-08 cm \n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.22 Page No : 255"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"n = 2.25 * 10**-4 # The viscosity of the gas in dynes/cm**2\n",
"C = 4.5 * 10**4 # The RMS velocity of the molecules in cm/s\n",
"d = 10.**-3 # The density of the gas in g/cc\n",
"\n",
"# Calculations\n",
"L = (3 * n) / (d * C) # The mean free path of the molecules in cm\n",
"\n",
"# Output\n",
"print 'The mean free path of the molecules is %3g cm ' % (L)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The mean free path of the molecules is 1.5e-05 cm \n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.23 Page No : 261"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"d = 2. * 10**-8 # The molecular diameter in cm\n",
"n = 3. * 10**19 # The number of molecules per cc\n",
"pi = 3.14 # Mathematical constant of pi\n",
"\n",
"# Calculations\n",
"L = 1. / ((pi * (d)**2 * n)) # The mean free path of a gas molecule in cm\n",
"\n",
"# Output\n",
"print 'The mean free path of a gas molecule is %3.0g cm ' % (L)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The mean free path of a gas molecule is 3e-05 cm \n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.24 Page No : 265"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"p = 760. # The given pressure in mm of Hg\n",
"T = 273. # The temperature of the chamber in K\n",
"V = 22400. # The volume of the gas at N.T.P in cc\n",
"p1 = 10.**-6 # The pressure in the chamber in mm of mercury pressure\n",
"N = 6.023 * 10**23 # The Avogadro number\n",
"d = 2. * 10**-8 # Molecular diameter in cm\n",
"pi = 3.14 # Mathematical constant of pi\n",
"\n",
"# Calculations\n",
"# The number of molecules per cm**3 in the chamber in molecules/cm**3\n",
"n = (N * p1) / (V * p)\n",
"# The mean free path of the gas molecules in the chamber in cm\n",
"L = 1. / (pi * (d)**2 * n)\n",
"\n",
"# Output\n",
"print 'The mean free path of gas molecules in a chamber is %3.4g cm ' % (L)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The mean free path of gas molecules in a chamber is 2.25e+04 cm \n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.25 Page No : 270"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"Tc = 132. # The given temperature in K\n",
"Pc = 37.2 # The given pressure in atms\n",
"R = 82.07 # Universal gas constant in cm**3 atoms K**-1\n",
"\n",
"# Calculations\n",
"# Vander Waals constant in atoms cm**6\n",
"a = (27. / 64) * ((R)**2 * (Tc)**2) / Pc\n",
"b = ((R * Tc) / (8 * Pc)) # Vander Waals constant in cm**3\n",
"\n",
"# Output\n",
"print 'The Van der Waals constants are , \\n (1) a = %3.4g atoms cm^6 \\n (2) b = %3.2f cm^3 ' % (a, b)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Van der Waals constants are , \n",
" (1) a = 1.331e+06 atoms cm^6 \n",
" (2) b = 36.40 cm^3 \n"
]
}
],
"prompt_number": 24
}
],
"metadata": {}
}
]
}PKDI@~4!Heat and Thermodynamics/ch6.ipynb{
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 6 : Thermodynamics\n"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.1 Page No : 292"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"H = 80. # The Heat flows into the system in joules\n",
"W = 30. # The Work done by the system in joules\n",
"\n",
"# Calculations\n",
"U = H - W # The internal energy of the system in joules\n",
"W1 = 10. # The work done along the path ADB in joules\n",
"H1 = W1 + U # The heat flows into the system along the path ADB in joules\n",
"W2 = -20 # The work done on the system from B to A in joules\n",
"H2 = W2 - U # The heat liberated from B to A in joules\n",
"Ua = 0. # Internal energy at A in joules\n",
"Ud = 40. # Internal energy at D in joules\n",
"Wa = 10. # Work done from A to D in joules\n",
"Wd = 0. # Work done from D to B in joules\n",
"Uc = 50. # Internal energy at C in joules\n",
"Had = (Ud - Ua) + Wa # Heat absorbed in the process AD in joules\n",
"Hdb = Uc - Ud + Wd # Heat absorbed in the process DB in joules\n",
"\n",
"# Output\n",
"print '(a)Heat flows into the system along the path ADB is H = %3.0f joules \\n (b)The heat liberated by the system is H = %3.0f joules \\n (c)The heat absorbed in the process AD is H = %3.0f joules and \\n The heat absorbed in the process DB is H = %3.0f joules ' % (H1, H2, Had, Hdb)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)Heat flows into the system along the path ADB is H = 60 joules \n",
" (b)The heat liberated by the system is H = -70 joules \n",
" (c)The heat absorbed in the process AD is H = 50 joules and \n",
" The heat absorbed in the process DB is H = 10 joules \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.2 Page No : 296"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"p = 2. # Given Pressure of a motor car tyre in atms\n",
"t = 27. # The room temperature in degree centigrade\n",
"g = 1.4 # Adiabatic index\n",
"\n",
"# Calculations\n",
"P1 = p # The pressure of a motor car tyre in atms\n",
"T1 = t + 273 # The room temperature in K\n",
"P2 = 1. # The surrounding pressure in atms\n",
"T2 = ((P2 / P1)**((g - 1) / g)) * T1 # The resulting temperature in K\n",
"T21 = T2 - 273 # The resulting temperature in degree centigrade\n",
"\n",
"# Output\n",
"print 'The resulting temperature is T2 = %3.1f K (or) %3.1f degree centigrade ' % (T2, T21)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The resulting temperature is T2 = 246.1 K (or) -26.9 degree centigrade \n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.3 Page No : 302"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"t = 27. # The room temperature of air in degree centigrade\n",
"g = 1.4 # Adiabatic index\n",
"\n",
"# Calculations\n",
"V1 = 1. # Let the Original volume in cc\n",
"V2 = V1 / 2 # The final volume i.e half the original volume in cc\n",
"P1 = 1. # The atmospheric pressure in atms\n",
"P2 = P1 * (V1 / V2)**g # The final pressure in atms\n",
"T1 = t + 273 # The room temperature in K\n",
"T2 = T1 * (V1 / V2)**(g - 1) # The final temperature in K\n",
"T21 = T2 - 273 # The final temperature in degree centigrade\n",
"\n",
"# Output\n",
"print '(1)The final pressure is P2 = %3.3f atmospheres \\n (2)The final temperature is T2 = %3.1f K (or) %3.1f degree centigrade ' % (P2, T2, T21)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(1)The final pressure is P2 = 2.639 atmospheres \n",
" (2)The final temperature is T2 = 395.9 K (or) 122.9 degree centigrade \n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.4 Page No : 306"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"g = 1.4 # Adiabatic index\n",
"\n",
"# Calculations\n",
"V1 = 1. # Let the initial volume be in cc\n",
"V2 = V1 / 2 # The final volume is half the initial volume in cc\n",
"T1 = 1. # Let the initial temperature of air be in K\n",
"T2 = T1 * (V1 / V2)**(g - 1) # The final temperature of air in K\n",
"T = T2 - T1 # The change in temperature of air in K\n",
"\n",
"# Output\n",
"print 'The change in the temperature is %3.3fT1 K ' % (T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The change in the temperature is 0.320T1 K \n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.5 Page No : 308"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"g = (5. / 3) # Adiabatic index for monoatomic\n",
"t = 27. # The room temperature in degree centigrade\n",
"P1 = 1. # The initial pressure in atmosphere\n",
"P2 = 50. # The final pressure in atmosphere\n",
"\n",
"# Calculations\n",
"T1 = t + 273 # The room temperature in K\n",
"T2 = ((P2 / P1)**((g - 1) / g)) * T1 # The final temperature in K\n",
"T21 = T2 - 273 # The final temperature in degree centigrade\n",
"\n",
"# Output\n",
"print 'The Final temperature is T2 = %3.0f K (or) %3.0f degree centigrade ' % (T2, T21)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Final temperature is T2 = 1435 K (or) 1162 degree centigrade \n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6 Page No : 314"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"t = 27. # The temperature of dry air in degree centigrade\n",
"g = 1.4 # Adiabatic index\n",
"\n",
"# Calculations\n",
"V1 = 1. # Let us assume the initial volume in cc\n",
"V2 = V1 / 3 # Then the final volume is 1/3 of the initial volume in cc\n",
"T1 = t + 273 # The initial temperature of dry air in K\n",
"T2 = ((V1 / V2)**(g - 1)) * T1 # The final temperature of air in K\n",
"T21 = T2 - 273 # The final temperature of air in degree centigrade\n",
"T = T21 - t # The change in temperature in degree centigrade\n",
"\n",
"# Output\n",
"print '(1)When the process is slow the temperature of the system remains constant so, there is no change in the temperature \\n (2)When the compression is sudden then, \\n The temperature of the air increases by T = %3.1f degree centigrade (or) %3.1f K' % (T, T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(1)When the process is slow the temperature of the system remains constant so, there is no change in the temperature \n",
" (2)When the compression is sudden then, \n",
" The temperature of the air increases by T = 165.6 degree centigrade (or) 165.6 K\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.7 Page No : 316"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"g = 1.4 # Adiabatic index\n",
"\n",
"# Calculations\n",
"V1 = 1. # Let the initial volume of the gas in cc\n",
"V2 = 3. * V1 # Then the final volume of the gas is 3 times the initial volume of the gas in cc\n",
"T1 = 273. # Initial temperature of the gas at NTP in K\n",
"T2 = ((V1 / V2)**(g - 1)) * T1 # The resulting temperature in K\n",
"T21 = T2 - 273 # The resulting temperature in degree centigrade\n",
"P1 = 1. # The atmospheric pressure in atms\n",
"P2 = ((V1 / V2)**(g)) * P1 # The resulting atmospheric pressure in atmosphere\n",
"\n",
"# Output\n",
"print '(1)The resulting temperature is T2 = %3.0f K (or) %3.0f degree centigrade \\n (2)The resulting pressure is P2 = %3.4f atmosphere ' % (T2, T21, P2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(1)The resulting temperature is T2 = 176 K (or) -97 degree centigrade \n",
" (2)The resulting pressure is P2 = 0.2148 atmosphere \n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.8 Page No : 321"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"t1 = 100. # The temperature at steam point in degree centigrade\n",
"t2 = 0. # The temperature at ice point in degree centigrade\n",
"\n",
"# Calculations\n",
"T1 = t1 + 273 # The temperature at steam point in K\n",
"T2 = t2 + 273 # The temperature at ice point in K\n",
"n = (1 - (T2 / T1)) * 100 # The efficiency of the carnots engine in percent\n",
"\n",
"# Output\n",
"print 'The efficiency of the Carnot engine is %3.2f percent ' % (n)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The efficiency of the Carnot engine is 26.81 percent \n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.9 Page No : 325"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"t1 = 127. # The temperature at initial point in degree centigrade\n",
"t2 = 27. # The temperature at final point in degree centigrade\n",
"\n",
"# Calculations\n",
"T1 = t1 + 273 # The temperature at initial point in K\n",
"T2 = t2 + 273 # The temperature at final point in K\n",
"n = (1 - (T2 / T1)) * 100 # The efficiency of the carnots engine in percent\n",
"\n",
"# Output\n",
"print 'The efficiency of the Carnot engine is %3.0f percent ' % (n)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The efficiency of the Carnot engine is 25 percent \n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.10 Page No : 329"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"T1 = 400. # The temperature of the source in k\n",
"H1 = 200. # The amount of heat taken by the engine at T1 in calories\n",
"H2 = 150. # The amount of heat rejected by the engine to the sinkk in calories\n",
"\n",
"# Calculations\n",
"T2 = (H2 / H1) * T1 # The temperature of the sinkk in K\n",
"n = (1 - (T2 / T1)) * 100 # The efficiency of the engine in percent\n",
"\n",
"# output\n",
"print 'The temperature of the sinkk is T2 = %3.0f K The efficiency of the engine is %3.0f percent ' % (T2, n)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The temperature of the sinkk is T2 = 300 K The efficiency of the engine is 25 percent \n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.11 Page No : 333"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"T1 = 450. # The temperature of the source in k\n",
"H1 = 1000. # The amount of heat taken by the engine at T1 in calories\n",
"T2 = 350. # The temperature of the sinkk in K\n",
"\n",
"# Calculations\n",
"# The amount of heat rejected to the sinkk in each cycle in calories\n",
"H2 = (T2 / T1) * H1\n",
"n = (1 - (T2 / T1)) * 100 # The efficiency of the engine in percent\n",
"W = H1 - H2 # The work done by the engine in each cycle in calories\n",
"W1 = W * 4.2 # The work done by the engine in each cycle in joules\n",
"\n",
"# Output\n",
"print 'The amount of heat rejected to the sinkk in each cycle is H2 = %3.2f cals The efficiency of the engine is %3.2f percent The work done by the engine in each cycle is W = %3.2f joules' % (H2, n, W1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The amount of heat rejected to the sinkk in each cycle is H2 = 777.78 cals The efficiency of the engine is 22.22 percent The work done by the engine in each cycle is W = 933.33 joules\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.12 Page No : 337"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"T1 = 300. # The higher temperature of the reservoir in K\n",
"T2 = 260. # The lower temperature of the reservoir in K\n",
"H2 = 500. # The amount of heat from the reservoir at the lower temperature in calories\n",
"\n",
"# Calculations\n",
"# The amount of heat rejected to the reservoir at the higher temperature\n",
"# in calories\n",
"H1 = (T1 / T2) * H2\n",
"# The amount of work done in each cycle to operate the refrigerator in joules\n",
"W = (H1 - H2) * 4.2\n",
"\n",
"# Output\n",
"print 'The amount of heat rejected to the reservoir at the higher temperature is H1 = %3.2f cal The amount of work done in each cycle to operate the refrigerator is W = %3.2f joules ' % (H1, W)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The amount of heat rejected to the reservoir at the higher temperature is H1 = 576.92 cal The amount of work done in each cycle to operate the refrigerator is W = 323.08 joules \n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.13 Page No : 340"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"T2 = 273. # The lower temperature of the reservoir for a carnot refrigerator in K\n",
"T1 = 27. + 273 # The higher temperature of the reservoir for a carnot refrigerator in K\n",
"H2 = 1000. * 80 # The amount of heat from the reservoir to the lower temperature in cal\n",
"J = 4.2 # The one calorie in joules\n",
"\n",
"# Calculations\n",
"H1 = (T1 / T2) * H2 # The amount of heat discarded to the room in calories\n",
"W = J * (H1 - H2) # The work done by the refrigerator in joules\n",
"C = H2 / (H1 - H2) # The coefficient of performance\n",
"\n",
"# output\n",
"print 'The amount of heat discarded to the room is H1 = %3.0f cal The work done by the refrigerator is W = %3.4g joules The coefficient of performance of the machine is %3.2f ' % (H1, W, C)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The amount of heat discarded to the room is H1 = 87912 cal The work done by the refrigerator is W = 3.323e+04 joules The coefficient of performance of the machine is 10.11 \n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.14 Page No : 344"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"t2 = 7. # The lower temperature of the reservoir in degree centigrade\n",
"n = 50. # The efficiency of the carnot engine in percent\n",
"n1 = 70. # It is desired to increase the efficiency in percent\n",
"\n",
"# Calculations\n",
"T2 = t2 + 273 # The lower temperature of the reservoir in K\n",
"# The higher temperature of the reservoir for 50% efficiency of the engine in K\n",
"T1 = T2 / (1 - (n / 100))\n",
"# The higher temperature of the reservoir for 70% efficiency of the engine in K\n",
"T11 = T2 / (1 - (n1 / 100))\n",
"T = T11 - T1 # Increase in temperature for the change in efficiencies in K\n",
"\n",
"# Output\n",
"print 'The temperature of the high temperature reservoir should be increased by %3.0f K ' % (T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The temperature of the high temperature reservoir should be increased by 373 K \n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.15 Page No : 348"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"T1 = 600. # The higher temperature of the reservoir in K\n",
"T2 = 300. # The lower temperature of the reservoir in K\n",
"n1 = 52. # The efficiency claimed by the inventor in percent\n",
"\n",
"# Calculations\n",
"n = (1 - (T2 / T1)) * 100 # The efficiency of the carnot engine in percent\n",
"\n",
"# Output\n",
"print 'The efficiency of the carnot engine is %3.0f percent \\n The efficiency claimed is %3.0f percent \\n The efficiency of the engine is more than the efficiency of the carnot engine \\n .But no engine can have an efficiency more than a carnots engine, \\n so his claim is invalid' % (n, n1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The efficiency of the carnot engine is 50 percent \n",
" The efficiency claimed is 52 percent \n",
" The efficiency of the engine is more than the efficiency of the carnot engine \n",
" .But no engine can have an efficiency more than a carnots engine, \n",
" so his claim is invalid\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.16 Page No : 350"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"P = 10.**5 # The average pressure of the steam in a double acting steam engine in newtons/m**2\n",
"L = 1. # The length of the stroke in m\n",
"A = 0.15 # The area of the piston in m**2\n",
"N = 5. # Number of strokes in strokes per second\n",
"\n",
"# Calculations\n",
"P = (2 * P * L * A * N) / 1000 # The power of the engine in kilowatts\n",
"\n",
"# Output\n",
"print 'The power of the engine is %3.0f kilowatts ' % (P)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The power of the engine is 150 kilowatts \n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.17 Page No : 352"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"l = 80. # The latent heat of ice in calories per gram\n",
"V1 = 1.091 # The specific volume of 1 gram of ice at 0 degree centigrade in cm**3\n",
"V2 = 1.000 # The specific volume of 1 gram of water at 0 degree centigrade in cm**3\n",
"p = 1. # The pressure in atm\n",
"T = 273. # The temperature at 0 degree centigrade in K\n",
"\n",
"# Calculations\n",
"L = 80. * 4.2 * 10**7 # The latent heat of ice in ergs\n",
"P = 76. * 13.6 * 980 # The pressure in dynes/cm**2\n",
"# The depression in the melting point of ice produced by one atmosphere\n",
"# increase of pressure in K\n",
"T = (P * T * (V2 - V1)) / L\n",
"T1 = -T # The decrease in the melting point of ice with an increase in pressure of one atmosphere\n",
"\n",
"# Output\n",
"print 'The decrease in the melting point of ice with an increase, \\n in pressure of one atmosphere is %3.4f K (or) %3.4f degree centigrade ' % (T1, T1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The decrease in the melting point of ice with an increase, \n",
" in pressure of one atmosphere is 0.0075 K (or) 0.0075 degree centigrade \n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.18 Page No : 358"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# input data\n",
"p = 1. # The pressure in atm\n",
"V1 = 1.000 # The specific volume of one gram of water in cm**3\n",
"V2 = 1677. # The specific volume of one gram of steam in cm**3\n",
"l = 540. # Latent heat of vaporisation of steam in cal/gram\n",
"\n",
"# Calculations\n",
"P = 76. * 13.6 * 980 # The pressure in dynes/cm**2\n",
"T = 100. + 273 # The temperature at 100 degree centigrade in K\n",
"L = l * 4.2 * 10**7 # The latent heat of vapourisation in ergs\n",
"# The increase in the boiling point of water with an increase in pressure\n",
"# of one atmosphere in degree centigrade\n",
"T = (P * T * (V2 - V1)) / L\n",
"\n",
"# Output\n",
"print 'The increase in the boiling point of water with an increase , \\n in pressure of one atmosphere is %3.2f degree centigrade (or) %3.2f K ' % (T, T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The increase in the boiling point of water with an increase , \n",
" in pressure of one atmosphere is 27.92 degree centigrade (or) 27.92 K \n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.19 Page No : 361"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"l = 537. # Latent heat of steam in cal/g\n",
"V2 = 1674. # The specific volume of one gram of steam in cm**3\n",
"V1 = 1.000 # The specific volume of one gram of water in cm**3\n",
"p = 2.712 # The increase in the pressure in cm of Hg\n",
"t = 100. # The boiling point of water in degree centigrade\n",
"\n",
"# Calculations\n",
"T = t + 273 # The boiling point of water in K\n",
"P = p * 13.6 * 980 # The increase in the pressure in dynes/cm**2\n",
"L = l * 4.2 * 10**7 # Latent heat of steam in ergs\n",
"# The change in the temperature of the boiling water when the pressure is\n",
"# increased in K\n",
"T1 = (P * T * (V2 - V1)) / L\n",
"\n",
"# Output\n",
"print 'The change in temperature of boiling water is %3.0f K (or) %3.0f degree centigrade ' % (T1, T1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The change in temperature of boiling water is 1 K (or) 1 degree centigrade \n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.20 Page No : 368"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"l = 4563. # The latent heat of fusion of naphthalene in cal/mol\n",
"V = 18.7 # The increase in volume of fusion in cm**3/mol\n",
"p = 1. # The pressure in atm\n",
"t = 80. # The melting point of naphthalene in degree centigrade\n",
"\n",
"# Calculations\n",
"L = l * 4.2 * 10**7 # The latent heat of fusion of naphthalene in ergs/mol\n",
"T = t + 273 # The melting point of naphthalene in K\n",
"P = 76. * 13.6 * 980 # The pressure in dynes/cm**2\n",
"# The increase in the melting point of naphthalene with an increase in\n",
"# pressure of one atmosphere in K\n",
"T1 = (P * T * (V)) / L\n",
"\n",
"# Output\n",
"print 'The increase in the melting point of naphthalene with an increase,\\n in pressure of one atmosphere is %3.5f K (or) %3.5f degree centigrade ' % (T1, T1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The increase in the melting point of naphthalene with an increase,\n",
" in pressure of one atmosphere is 0.03489 K (or) 0.03489 degree centigrade \n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.21 Page No : 372"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"p1 = 80. # The under pressure of benzene in cm of Hg\n",
"t = 80. # The normal boiling point of benzene in degree centigrade\n",
"l = 380. # The latent heat of vapourisation in joules/g\n",
"d2 = 4. # Density of vapour at boiling point in g/litre\n",
"d1 = 0.9 # Density of liquid in g/cm**3\n",
"\n",
"# Calculations\n",
"p = p1 - 76 # The change in pressure in cm of Hg\n",
"P = p * 13.6 * 980 # The change in pressure in dynes/cm**2\n",
"T = t + 273 # The normal boiling point of benzene in K\n",
"L = l * 10**7 # Latent heat of vapourisation in ergs/g\n",
"V1 = 1. / d1 # The specific volume of liquid in cm**3\n",
"V2 = 1000. / d2 # The specific volume of vapour in cm**3\n",
"# The increase in the boiling point of benzene in K\n",
"T1 = (P * T * (V2 - V1)) / L\n",
"T2 = t + T1 # The boiling point of benzene at a pressure of 80 cm of Hg in degree centigrade\n",
"\n",
"# Output\n",
"print 'The boiling point of benzene at a pressure of 80 cm of Hg is %3.3f degree centigrade ' % (T2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The boiling point of benzene at a pressure of 80 cm of Hg is 81.233 degree centigrade \n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.22 Page No : 378"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"t = 100. # The boiling point of water in degree centigrade\n",
"p1 = 1. # Initial pressure in atm\n",
"p2 = 1.10 # Final pressure in atm\n",
"l = 537. # Latent heat of water at 100 degree centigrade in cal/g\n",
"V1 = 1. # The specific volume of one gram of water in cm**3\n",
"V2 = 1676. # The specific volume of one gram of steam in cm**3\n",
"\n",
"# Calculations\n",
"p = p2 - p1 # The change in pressure in atm\n",
"P = p * 76 * 13.6 * 980 # The change in pressure in dynes/cm**2\n",
"T = t + 273 # The boiling point of water in K\n",
"L = l * 4.2 * 10**7 # The latent heat of water at 100 degree centigrade in ergs/g\n",
"# The change in boiling point of water in K (or) degree centigrade\n",
"T1 = (P * T * (V2 - V1)) / L\n",
"\n",
"# Output\n",
"print 'The increase in the boiling point of water with an increase,\\n of 0.1 atmosphere pressure is %3.3f K (or) %3.3f degree centigrade ' % (T1, T1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The increase in the boiling point of water with an increase,\n",
" of 0.1 atmosphere pressure is 2.806 K (or) 2.806 degree centigrade \n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.23 Page No : 382"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"p1 = 1. # The atmospheric pressure in atm\n",
"p2 = 100. # The given pressure in atm\n",
"d1 = 0.917 # The density of ice in g/cm**3\n",
"l = 336. # The latent heat of ice in j/g\n",
"\n",
"# Calculations\n",
"p = p2 - p1 # The change in pressure in atms\n",
"P = p * 76 * 13.6 * 980 # The change in pressure in dynes/cm**2\n",
"L = l * 10**7 # The latent heat of ice in ergs/g\n",
"T = 273. # The temperature of melting point of ice in K\n",
"V2 = 1. # The specific volume of one gram of water in cm**3\n",
"V1 = 1. / d1 # The specific volume of ice in cm**3\n",
"T1 = (T * P * (V2 - V1)) / L # The change in the melting point of ice in K\n",
"# The decrease in the melting point of ice in K (or) degree centigrade\n",
"T2 = -T1\n",
"\n",
"# Output\n",
"print 'The decrease in the melting point of ice,\\n with a pressure of 100 atmospheres is %3.4f degree centigrade ' % (T2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The decrease in the melting point of ice,\n",
" with a pressure of 100 atmospheres is 0.7375 degree centigrade \n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.24 Page No : 385"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"l = 79.6 # latent heat of ice in cal/g\n",
"V2 = 1. # The specific volume of water at 0 degree centigrade in cm**3\n",
"V1 = 1.091 # The specific volume of ice at 0 degree centigrade in cm**3\n",
"p = 1.013 * 10**6 # One atmospheric pressure in dynes/cm**3\n",
"T = -1 # The change in temperature in K\n",
"T1 = 273. # The temperature of water at 0 degree centigrade in K\n",
"p1 = 1. # The atmospheric pressure in atm\n",
"\n",
"# Calculations\n",
"L = l * 4.18 * 10**7 # The latent heat of ice in ergs/g\n",
"P = ((L * T) / (T1 * (V2 - V1) * p)) # The change in pressure in atmospheres\n",
"P1 = P + p1 # The pressure required in atmospheres\n",
"\n",
"# Output\n",
"print 'The pressure required to lower melting point of ice,\\n by 1 degree centigrade is %3.1f atmospheres ' % (P1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The pressure required to lower melting point of ice,\n",
" by 1 degree centigrade is 133.2 atmospheres \n"
]
}
],
"prompt_number": 24
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.25 Page No : 392"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"t = 100. # The temperature at which water boils in degree centigrade\n",
"p2 = 787. # The pressure at which water boils in mm of Hg\n",
"J = 4.2 * 10**7 # Joule in ergs/cal\n",
"p1 = 760. # The atmospheric pressure in mm of Hg\n",
"V2 = 1601. # The specific volume of 1 g of water at 100 degree centigrade in cm**3\n",
"V1 = 1. # The specific volume of 1 g of water at 0 degree centigrade in cm**3\n",
"\n",
"# Calculations\n",
"T = t + 273 # The temperature at which water boils in K\n",
"T1 = 1. # The difference in the temperature in K\n",
"p = p2 - p1 # The difference in the pressure in mm of Hg\n",
"P = (p / 10) * 13.6 * 980 # The difference in the pressure in dynes/cm**2\n",
"L = (T * P * (V2 - V1)) / T1 # The latent heat of steam in ergs/g\n",
"L1 = L / J # The latent heat of steam in cal/g\n",
"\n",
"# Output\n",
"print 'The Latent heat of steam is L = %3.1f cal/g ' % (L1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Latent heat of steam is L = 511.3 cal/g \n"
]
}
],
"prompt_number": 25
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.26 Page No : 394"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"T = 600. # The melting point of lead in K\n",
"d1 = 11.01 # Initial density of the lead in g/cm**3\n",
"d2 = 10.65 # The final density of the lead in g/cm**3\n",
"l = 24.5 # The latent heat of fusion of lead in j/g\n",
"p1 = 1. # The atmospheric pressure in atmospheres\n",
"p2 = 100. # The given pressure in atmospheres\n",
"\n",
"# Calculations\n",
"p = p2 - p1 # The change in pressure in atmospheres\n",
"P = p * 76 * 13.6 * 980 # The change in pressure in dynes/cm**2\n",
"L = l * 10**7 # The latent heat of fusion of lead in ergs/g\n",
"V1 = 1. / d1 # The initial specific volume of the lead in cm**3\n",
"V2 = 1. / d2 # The final specific volume of the lead in cm**3\n",
"T1 = (T * P * (V2 - V1)) / L # The change in the temperature in K\n",
"T2 = T + T1 # Melting point of lead at 100 atmospheres pressure in K\n",
"\n",
"# Output\n",
"print 'The melting point of lead at a pressure of 100 atmospheres is %3.4f K ' % (T2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The melting point of lead at a pressure of 100 atmospheres is 600.7540 K \n"
]
}
],
"prompt_number": 26
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.28 Page No : 397"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"t2 = 120. # The given temperature for the water to boil in degree centigrade\n",
"t1 = 100. # The actual boiling point of water in degree centigrade\n",
"V = 1676. # The change in specific volume in cm**3\n",
"l = 540. # Latent heat of steam in cal/g\n",
"J = 4.2 * 10**7 # joule in ergs/cal\n",
"\n",
"# Calculations\n",
"T1 = t2 - t1 # The change in temperature in degree centigrade (or) K\n",
"T = t1 + 273 # The boiling point of water in K\n",
"L = l * J # The latent heat of steam in ergs/g\n",
"p = 1. # The atmospheric pressure in atmospheres\n",
"P = (L * T1) / (T * V) # The change in pressure in dynes/cm**2\n",
"P1 = P / 10**6 # The change in pressure in atmospheres\n",
"P2 = P1 + p # The required pressure in atmospheres\n",
"\n",
"# Output\n",
"print 'The required pressure is %3.4f atmospheres ' % (P2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The required pressure is 1.7256 atmospheres \n"
]
}
],
"prompt_number": 27
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.29 Page No : 400"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"l = 80. # Latent heat of ice in cal/g\n",
"m = 10. # Mass of ice in g\n",
"T = 273. # The temperature of ice in K\n",
"\n",
"# Calculations\n",
"H = m * l # Heat absorbed by 10 g of ice at 273 K when it is converted into water at 273 K in cal\n",
"S = H / T # The gain in entropy in cal/K\n",
"\n",
"# Output\n",
"print 'The gain in entropy is %3.2f cal/K' % (S)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The gain in entropy is 2.93 cal/K\n"
]
}
],
"prompt_number": 28
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.30 Page No : 404"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"m = 5. # Mass of water in kg\n",
"t = 100. # The temperature of water in degree centigrade\n",
"l = 540. # Latent heat of water at 100 degree centigrade in cal/g\n",
"\n",
"# Calculations\n",
"T = t + 273 # The temperature of water in K\n",
"M = m * 1000 # Mass of water in g\n",
"H = M * l # Heat absorbed by 5 kg of water at 100 degree centigrade when it is converted into steam at 100 degree centigrade in cal\n",
"S = H / T # The gain in entropy in cal/K\n",
"\n",
"# Output\n",
"print 'The gain in entropy is %3.0f cal/K ' % (S)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The gain in entropy is 7239 cal/K \n"
]
}
],
"prompt_number": 29
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.31 Page No : 411"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"m = 1. # mass of ice in g\n",
"t1 = -10 # The given temperature of ice in degree centigrade\n",
"t2 = 100. # The given temperature of steam in degree centigrade\n",
"S = 0.5 # Specific heat of ice\n",
"s = 1. # Specific heat of water\n",
"l1 = 80. # Latent heat of ice in cal/g\n",
"l2 = 540. # Latent heat of steam in cal/g\n",
"\n",
"# Calculations\n",
"T = 273. # The temperature of ice at 0 degree centigrade in K\n",
"T1 = t1 + 273 # The given temperature of ice in K\n",
"T2 = t2 + 273 # The given temperature of steam in K\n",
"# Increase in entropy when the temperature of 1 gram of ice increases from\n",
"# -10 to 0 degree centigrade in cal/K\n",
"S1 = m * S * 2.3026 * math.log10(T / T1)\n",
"S2 = l1 / T # Increase in entropy when 1 g of ice at 0 degree centigrade is converted into water at 0 degree centigrade in cal/K\n",
"# Increase in entropy when 1 g of water raised from 0 to 100 degree\n",
"# centigrade in cal/K\n",
"S3 = m * s * 2.3026 * math.log10(T2 / T)\n",
"S4 = l2 / T2 # Increase in entropy when 1g water at 100 degree centigrade is converted into steam at 100 degree centigrade in cal/K\n",
"S5 = S1 + S2 + S3 + S4 # Total increase in entropy in cal/K\n",
"\n",
"# Output\n",
"print 'The total increase in entropy is %3.5f cal/K' % (S5)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The total increase in entropy is 2.07153 cal/K\n"
]
}
],
"prompt_number": 30
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.32 Page No : 413"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"V1 = 1. # Let us assume the initial volume be one in cc\n",
"V2 = 4. * V1 # Then the final volume is four times the initial volume in cc\n",
"\n",
"# Calculations\n",
"# The gain in entropy in terms of the gas constant in cal/K\n",
"S = 2.3026 * (math.log10(V2 / V1))\n",
"\n",
"# Output\n",
"print 'The gain in entropy in terms of the gas constant is %3.3f (R/J)cal/K' % (S)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The gain in entropy in terms of the gas constant is 1.386 (R/J)cal/K\n"
]
}
],
"prompt_number": 31
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.33 Page No : 420"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"m1 = 50. # Mass of water at 0 degree centigrade in g\n",
"m2 = 50. # Mass of water at 83 degree centigrade in g\n",
"t1 = 0. # The temperature of water in degree centigrade\n",
"t2 = 83. # The temperature of water in degree centigrade\n",
"\n",
"# Calculations\n",
"T1 = t1 + 273 # Temperature of water in K\n",
"T2 = t2 + 273 # Tempearture of water in K\n",
"s = 1. # The specific heat of water\n",
"# The final temperature of the mixture in K\n",
"T = ((m2 * s * T2) + (m1 * s * T1)) / ((m1 + m2) * s)\n",
"# The change in entropy by 50 g of water when its temperature rises from\n",
"# 273 K to 313 K in cal/K\n",
"S1 = (m1 * s * math.log(T / T1))\n",
"# The change in entropy by 50 g of water when its temperature falls from\n",
"# 353 K to 313 K in cal/K\n",
"S2 = (m2 * s * math.log(T / T2))\n",
"S3 = S1 + S2 # The total gain in the entropy of the system in cal/K\n",
"\n",
"# Output\n",
"print 'The total gain in entropy of the system is %3.3f cal/K ' % (S3)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The total gain in entropy of the system is 0.878 cal/K \n"
]
}
],
"prompt_number": 32
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.34 Page No : 424"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"m1 = 50. # Mass of water at 15 degree centigrade in g\n",
"m2 = 80. # Mass of water at 40 degree centigrade in g\n",
"t1 = 15. # The temperature of water in degree centigrade\n",
"t2 = 40. # The temperature of water in degree centigrade\n",
"\n",
"# Calculations\n",
"T1 = t1 + 273 # Temperature of water in K\n",
"T2 = t2 + 273 # Tempearture of water in K\n",
"s = 1. # The specific heat of water\n",
"# The final temperature of the mixture in K\n",
"T = ((m2 * s * T2) + (m1 * s * T1)) / ((m1 + m2) * s)\n",
"# The change in entropy by 50 g of water when its temperature rises from\n",
"# 288 K to 303.4 K in cal/K\n",
"S1 = (m1 * s * math.log(T / T1))\n",
"# The change in entropy by 80 g of water when its temperature falls from\n",
"# 313 K to 303.4 K in cal/K\n",
"S2 = (m2 * s * math.log(T / T2))\n",
"S3 = S1 + S2 # The total gain in the entropy of the system in cal/K\n",
"\n",
"# Output\n",
"print 'The net increase in the entropy of the system is %3.3f cal/K ' % (S3)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The net increase in the entropy of the system is 0.106 cal/K \n"
]
}
],
"prompt_number": 33
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.35 Page No : 429"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"m1 = 10. # Mass of steam in g\n",
"t1 = 100. # The temperature of the steam in degree centigrade\n",
"m = 90. # mass of water in g\n",
"t2 = 0. # The temperature of water in degree centigrade\n",
"m2 = m + m1 # The total mass of water in g\n",
"l = 540. # The latent heat of steam in cal/g\n",
"\n",
"# Calculations\n",
"T1 = t1 + 273 # The temperature of the steam in K\n",
"T2 = t2 + 273 # The temperature of the water in K\n",
"T = ((m1 * l) + (m1 * T1) + (m2 * T2)) / \\\n",
" (m1 + m2) # The final temperature in K\n",
"# The change in entropy when the temperature of water and calorimeter\n",
"# rises from 273 K to 331.2 K in cal/K\n",
"S1 = m2 * math.log(T / T2)\n",
"# The change in entropy when 10 grams of steam at 373 K condenses to water\n",
"# at 373K in cal/K\n",
"S2 = -(m1 * l) / T1\n",
"# Change in entropy when 10 g of water at 373 K is cooled to water at\n",
"# 331.2 K in cal/K\n",
"S3 = m1 * math.log(T / T1)\n",
"S4 = S1 + S2 + S3 # Net change in entropy in cal/K\n",
"\n",
"# Output\n",
"print 'The net increase in the entropy of the system is %3.3f cal/K ' % (S4)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The net increase in the entropy of the system is 3.653 cal/K \n"
]
}
],
"prompt_number": 34
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.36 Page No : 431"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"m = 1. # Mass of water in g\n",
"t1 = 20. # The temperature of water in degree centigrade\n",
"t2 = -10 # The temperature of ice in degree centigrade\n",
"s1 = 4.2 # Heat capacity for one gram of water in J/g-K\n",
"s2 = 2.1 # Heat capacity for ice in J/g-K\n",
"li = 335. # Latent heat of fusion of ice at 0 degree centigrade in J/g\n",
"\n",
"# Calculations\n",
"T = 273. # The temperature of water at 0 degree centigrade in K\n",
"T1 = t1 + 273 # The temperature of water in K\n",
"T2 = t2 + 273 # The temperature of ice in K\n",
"# Change in entropy when the temperature of 1 g of water at 293 K falls to\n",
"# 273 K in J/K\n",
"S1 = m * s1 * math.log(T / T1)\n",
"# Change in entropy when 1 g of water at 273 K is converted into ice at\n",
"# 273 K in J/K\n",
"S2 = -(m * li) / T\n",
"# Change in entropy when the temperature of 1 g of ice at 273 K falls to\n",
"# 263 K in J/K\n",
"S3 = m * s2 * math.log(T2 / T)\n",
"S4 = S1 + S2 + S3 # The total change in entropy of the system in J/K\n",
"\n",
"# Output\n",
"print 'The total change in the entropy of the system is %3.5f J/K \\n (Negative sign indicates that there is decrease in the entropy of the system)' % (S4)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The total change in the entropy of the system is -1.60242 J/K \n",
" (Negative sign indicates that there is decrease in the entropy of the system)\n"
]
}
],
"prompt_number": 35
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.37 Page No : 435"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"M = 1. # Mass of water in kg\n",
"m = M * 1000 # Mass of water in g\n",
"T1 = 273. # The temperature of the water in K\n",
"T2 = 373. # The temperature of the heat reservoir in K\n",
"s = 1. # Specific heat of water\n",
"\n",
"# Calculations\n",
"# Increase in entropy when the temperature of 1000 g of water is raised\n",
"# from 273 K to 373 k in cal/K\n",
"S1 = m * s * math.log(T2 / T1)\n",
"S2 = -(m * s * (T2 - T1)) / T2 # Change in entropy of the reservoir in cal/K\n",
"S = S1 + S2 # Change in entropy of the universe in cal/K\n",
"\n",
"# Output\n",
"print '(1)The change in entropy of water when temperature reaches 373 K is %3.0f cal/K \\n (2) (i)The Change in entropy of the reservoir is %3.1f cal/K \\n (ii)The Change in entropy of the universe is %3.1f cal/K ' % (S1, S2, S)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(1)The change in entropy of water when temperature reaches 373 K is 312 cal/K \n",
" (2) (i)The Change in entropy of the reservoir is -268.1 cal/K \n",
" (ii)The Change in entropy of the universe is 44.0 cal/K \n"
]
}
],
"prompt_number": 36
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.42 Page No : 440"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"l = 540. # Latent heat of vapourisation of steam in cal/g\n",
"L = l * 4.2 * 10**7 # Latent heat of vapourisation of steam in ergs/g\n",
"V = 1676. # The change in specific volume when 1 g of water is converted into steam in cc\n",
"t1 = 100. # The actual boiling temperature of water in degree centigrade\n",
"t2 = 150. # The given temperature at which water must boil in degree centigrade\n",
"p = 1. # The atmospheric pressure in atmospheres\n",
"\n",
"# Calculations\n",
"T1 = t1 + 273 # The actual boiling temperature of water in K\n",
"T2 = t2 + 273 # The given temperature at which water must boil in K\n",
"T = T2 - T1 # The change in temperature in K\n",
"P = (L * T) / (T1 * V) # The pressure in dynes/cm**2\n",
"P1 = P / 10**6 # The pressure in atmospheres\n",
"P2 = P1 + p # The pressure at which water would boil at 150 degree centigrade in atmospheres\n",
"\n",
"# Output\n",
"print 'The pressure at which water would boil at 150 degree centigrade is %3.3f atmospheres ' % (P2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The pressure at which water would boil at 150 degree centigrade is 2.814 atmospheres \n"
]
}
],
"prompt_number": 37
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.43 Page No : 447"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"l = 80. # Latent heat of fusion of ice in cal/g\n",
"L = l * 4.2 * 10**7 # Latent heat of fusion in ergs/g\n",
"V = 0.091 # The change in specific volume when 1 g of water freezes into ice in cc\n",
"t1 = 0. # The actual freezing point of ice in degree centigrade\n",
"t2 = -1 # The given temperature at which ice must freeze in degree centigrade\n",
"p = 1. # The atmospheric pressure in atmospheres\n",
"\n",
"# Calculations\n",
"T1 = t1 + 273 # The actual freezing point of ice in K\n",
"T2 = t2 + 273 # The given temperature at which ice must freeze in K\n",
"T = T1 - T2 # The change in temperature in K\n",
"P = (L * T) / (V * T1) # The pressure in dynes/cm**2\n",
"P1 = P / 10**6 # The pressure in atmospheres\n",
"P2 = P1 + p # The pressure under which ice would freeze in atmospheres\n",
"\n",
"# Output\n",
"print 'The pressure under which ice would freeze at -1 degree centigrade is %3.1f atmospheres ' % (P2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The pressure under which ice would freeze at -1 degree centigrade is 136.2 atmospheres \n"
]
}
],
"prompt_number": 38
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.44 Page No : 451"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"t = 100. # The given temperature of water in degree centigrade\n",
"C1 = 1.01 # The specific heat of water at 100 degree centigrade in cal/g\n",
"L = -0.64 # The rate at which the latent heat of vapourisation decreases with rise in temperature in cal/K\n",
"l = 540. # The latent heat of vapourisation of steam in cal\n",
"\n",
"# Calculations\n",
"T = t + 273 # The given temperature of water in K\n",
"C2 = L - (l / T) + C1 # The specific heat of saturated steam in cal/g\n",
"\n",
"# Output\n",
"print 'The specific heat of satureted steam is %3.3f cal/g The specific heat of saturated steam is negative)' % (C2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The specific heat of satureted steam is -1.078 cal/g The specific heat of saturated steam is negative)\n"
]
}
],
"prompt_number": 39
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.45 Page No : 454"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"t = 100. # The temperature of saturated steam in degree centigrade\n",
"L1 = 545.25 # The latent heat of saturated steam at 90 degree centigrade in cal\n",
"L2 = 539.30 # The latent heat of saturated steam at 100 degree centigrade in cal\n",
"L3 = 533.17 # The latent heat of saturated steam at 110 degree centigrade in cal\n",
"C1 = 1.013 # The specific heat of water at 100 degree centigrade in cal/g\n",
"\n",
"# Calculations\n",
"T = t + 273 # The temperature of saturated steam in K\n",
"# The rate at which the latent heat of saturated steam decreases with rise\n",
"# in temperature in cal/K\n",
"L = (L3 - L1) / (110 - 90)\n",
"# The specific heat of saturated steam at 100 degree centigrade in cal/g\n",
"C2 = C1 + L - (L2 / T)\n",
"\n",
"# Output\n",
"print 'The specific heat of saturated steam at 100 degree centigrade is %3.3f cal/g' % (C2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The specific heat of saturated steam at 100 degree centigrade is -1.037 cal/g\n"
]
}
],
"prompt_number": 40
}
],
"metadata": {}
}
]
}PKDIQgR@R@!Heat and Thermodynamics/ch8.ipynb{
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 8 : Transmission of Heat"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.1 Page No : 462"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"l1 = 10. # Length of the copper rod in cm\n",
"l2 = 4. # Length of the iron rod in cm\n",
"K1 = 0.9 # The thermal conductivity of copper\n",
"\n",
"# Calculations\n",
"K2 = (l2**2 / l1**2) * K1 # The Thermal conductivity of iron\n",
"\n",
"# Output\n",
"print 'The thermal conductivity of iron is K2 = %3.3f ' % (K2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The thermal conductivity of iron is K2 = 0.144 \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.2 Page No : 469"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"K = 0.2 # The thermal conductivity of the plate\n",
"d = 0.2 # The thickness of the plate in cm\n",
"A = 20. # The area of the plate in cm**2\n",
"T = 100. # The temperature difference in degree centigrade\n",
"t = 60. # The given time in seconds\n",
"\n",
"# Calculations\n",
"# The quantity of heat that will flow through the plate in one minute in cal\n",
"Q = (K * A * T * t) / d\n",
"\n",
"# Output\n",
"print 'The quantity of heat that will flow through the plate in one minute is Q = %3.4g cal ' % (Q)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The quantity of heat that will flow through the plate in one minute is Q = 1.2e+05 cal \n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.3 Page No : 473"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"l = 30. # The length of the bar in cm\n",
"A = 5. # The uniform area of cross section of a bar in cm**2\n",
"ta = 200. # The temperature maintained at the end A in degree centigrade\n",
"tc = 0. # The temperature maintained at the end C in degree centigrade\n",
"Kc = 0.9 # The thermal conductivity of copper\n",
"Ki = 0.12 # The thermal conductivity of iron\n",
"\n",
"# Calculations\n",
"# The temperature after the steady state is reached in degree centigrade\n",
"T = ((Kc * A * ta) + (Ki * A * tc)) / ((Kc + Ki) * A)\n",
"# The rate of flow of heat along the bar when the steady state is reached\n",
"# in cal/sec\n",
"Q = (Kc * A * (ta - T)) / (l / 2)\n",
"\n",
"# Output\n",
"print 'The rate of flow of heat along the bar when the steady state is reached is Q = %3.2f cal/s ' % (Q)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The rate of flow of heat along the bar when the steady state is reached is Q = 7.06 cal/s \n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.4 Page No : 477"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"d1 = 1.75 # The thickness of the wood in cm\n",
"d2 = 3. # The thickness of the cork in cm\n",
"t2 = 0. # The temperature of the inner surface of the cork in degree centigrade\n",
"t1 = 12. # The temperature of the outer surface of the wood in degree centigrade\n",
"K1 = 0.0006 # The thermal conductivity of wood\n",
"K2 = 0.00012 # The thermal conductivity of cork\n",
"\n",
"# Calculations\n",
"# The temperature of the interface in degree centigrade\n",
"T = (((K1 * t1) / d1) + ((K2 * t2) / d2)) / ((K1 / d1) + (K2 / d2))\n",
"\n",
"# Output\n",
"print 'The temperature of the interface is T = %3.2f degree centigrade ' % (T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The temperature of the interface is T = 10.75 degree centigrade \n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.5 Page No : 483"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"x1 = 3. # The thickness of the ice layer on the surface of a pond in cm\n",
"x = 1. # The increase in the thickness of the ice when the temperature is maintained at -20 degree centigrade in mm\n",
"# The increased thickness of the ice layer on the surface of a pond in cm\n",
"x2 = x1 + (x / 10)\n",
"T = -20 # The temperature of the surrounding air in degree centigrade\n",
"d = 0.91 # The density of ice at 0 degree centigrade in g/cm**3\n",
"L = 80. # The latent heat of ice in cal/g\n",
"K = 0.005 # The thermal conductivity of ice\n",
"\n",
"# Calculations\n",
"# The time taken to increase its thickness by 1 mm in sec\n",
"t = ((d * L) / (2 * K * (-T))) * (x2**2 - x1**2)\n",
"t1 = t / 60 # The time taken to increase its thickness by 1 mm in min\n",
"\n",
"# Output\n",
"print 'The time taken to increase its thickness by 1 mm is t = %3.2f s' % (t)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The time taken to increase its thickness by 1 mm is t = 222.04 s\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.6 Page No : 485"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"x1 = 10. # The thickness of the ice layer on the surface of a pond in cm\n",
"x = 5. # The increase in the thickness of the ice when the temperature is maintained at -10 degree centigrade in cm\n",
"# The increased thickness of the ice layer on the surface of a pond in cm\n",
"x2 = x1 + (x)\n",
"T = -10 # The temperature of the surrounding air in degree centigrade\n",
"d = 0.90 # The density of ice at 0 degree centigrade in g/cm**3\n",
"L = 80. # The latent heat of ice in cal/g\n",
"K = 0.005 # The thermal conductivity of ice\n",
"\n",
"# Calculations\n",
"# The time taken to increase its thickness by 5 cm in sec\n",
"t = ((d * L) / (2 * K * (-T))) * (x2**2 - x1**2)\n",
"# The time taken to increase its thickness by 5 cm in hours\n",
"t1 = t / (60. * 60)\n",
"\n",
"# Output\n",
"print 'The time taken to increase its thickness by 5 cm is t = %3.0g s (or) %3.0f hours' % (t, t1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The time taken to increase its thickness by 5 cm is t = 9e+04 s (or) 25 hours\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.7 Page No : 490"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# input data\n",
"# The temperature maintained on one sphere (black body radiat(or) in K\n",
"T1 = 300.\n",
"# The temperature maintained on another sphere (black body radiat(or) in K\n",
"T2 = 200.\n",
"s = 5.672 * 10**-8 # Stefans constant in M.K.S units\n",
"\n",
"# Calculations\n",
"# The net rate of energy transfer between the two spheres in watts/m**2\n",
"R = s * (T1**4 - T2**4)\n",
"\n",
"# output\n",
"print 'The net rate of energy transfer between the two spheres is R = %3.2f watts/m^2' % (R)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The net rate of energy transfer between the two spheres is R = 368.68 watts/m^2\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.8 Page No : 495"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"T1 = 400. # The given temperature of a black body in K\n",
"T2 = 4000. # The given temperature of a black body in K\n",
"s = 5.672 * 10**-8 # Stefans constant in M.K.S units\n",
"\n",
"# Calculations\n",
"R1 = s * T1**4 # The radiant emittance of a black body at 400 k in watts/m**2\n",
"# The radiant emittance of a black body at 4000 k in kilo-watts/m**2\n",
"R2 = (s * T2**4) / 1000\n",
"\n",
"# Output\n",
"print 'The Radiant emittance of a black body at a temperature of ,\\n (i) 400 K is R = %3.0f watts/m^2 \\n (ii) 4000 K is R = %3.0f kilo-watts/m^2' % (R1, R2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Radiant emittance of a black body at a temperature of ,\n",
" (i) 400 K is R = 1452 watts/m^2 \n",
" (ii) 4000 K is R = 14520 kilo-watts/m^2\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.9 Page No : 500"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"e = 0.35 # The relative emittance of tungsten\n",
"A = 10.**-3 # The surface area of a tungsten sphere in m**2\n",
"T1 = 300. # The temperature of the walls in K\n",
"T2 = 3000. # The temperature to be maintained by the sphere in K\n",
"s = 5.672 * 10**-8 # Stefans constant in M.K.S units\n",
"\n",
"# Calculations\n",
"# The power input required to maintain the sphere at 3000 K in watts\n",
"R = s * A * e * (T2**4 - T1**4)\n",
"\n",
"# Output\n",
"print 'The power input required to maintain the sphere at 3000 K is R = %3.0f watts' % (R)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The power input required to maintain the sphere at 3000 K is R = 1608 watts\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.10 Page No : 507"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"e = 0.1 # The relative emittance of an aluminium foil\n",
"T1 = 300. # The temperature of one sphere in K\n",
"T2 = 200. # The temperature of another sphere in K\n",
"s = 5.672 * 10**-8 # Stefans constant in M.K.S units\n",
"\n",
"# Calculations\n",
"# The temperature of the foil after the steady state is reached in K\n",
"x = (((T1**4 + T2**4) / 2)**(1. / 4))\n",
"# The rate of energy transfer between one of the spheres and foil in watts/m**2\n",
"R = e * s * (T1**4 - x**4)\n",
"\n",
"# Output\n",
"print '1)The temperature of the foil after the steady state reached is x = %3.1f K \\\n",
"\\n2)The rate of energy transfer between the sphere and the foil is R = %3.1f watts/m^2' % (x, R)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"1)The temperature of the foil after the steady state reached is x = 263.9 K \n",
"2)The rate of energy transfer between the sphere and the foil is R = 18.4 watts/m^2\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.11 Page No : 513"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"A = 5. * 10**-5 # The surface area of the filament in m**2\n",
"e = 0.85 # The relative emittance of the filament\n",
"s = 5.672 * 10**-8 # Stefans constant in M.K.S units\n",
"t = 60. # The time in seconds\n",
"T = 2000. # The temperature of the filament of an incandescent lamp in K\n",
"\n",
"# Calculations\n",
"E = A * e * s * t * (T**4) # The energy radiated from the filament in joules\n",
"\n",
"# Output\n",
"print 'The energy radiated from the filament is E = %3.0f joules ' % (E)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The energy radiated from the filament is E = 2314 joules \n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.12 Page No : 520"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"E = 1.53 * 10**5 # The energy radiated from an iron furnace in calories per hour\n",
"A = 10.**-4 # The cross section area of an iron furnace in m**2\n",
"e = 0.8 # The relative emittance of the furnace\n",
"t = 3600. # The time in seconds\n",
"s = 1.36 * 10**-8 # Stefans constant in cal/m**2-s-K**4\n",
"\n",
"# Calculations\n",
"T = ((E) / (A * e * s * t))**(1. / 4) # The temperature of the furnace in K\n",
"\n",
"# Output\n",
"print 'The temperature of the furnace is T = %3.0f K ' % (T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The temperature of the furnace is T = 2500 K \n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.13 Page No : 524"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"S = 2.3 # Solar constant in cal/cm**2/minute\n",
"r = 7. * 10**10 # The radius of the sun in cm\n",
"R = 1.5 * 10**13 # The distance between the sun and the earth in cm\n",
"s = 1.37 * 10**-12 # Stefans constant in cal/cm**2/s\n",
"\n",
"# Calculations\n",
"E = (S / 60) * (R / r)**(2) # The energy radiated from the sun in cal/s\n",
"T = (E / s)**(1. / 4) # The black body temperature of the sun in K\n",
"\n",
"# Output\n",
"print 'The black body temperature of the sun is T = %3.0f K ' % (T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The black body temperature of the sun is T = 5987 K \n"
]
}
],
"prompt_number": 14
}
],
"metadata": {}
}
]
}PKDIJS S !Heat and Thermodynamics/ch9.ipynb{
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{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 9 : Statistical Thermodynamics\n"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.1 Page No : 542"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"N = 6000. # Number of particles in a system\n",
"e = 3. # The number of energy states with equal spacing\n",
"n1 = 3000. # Number of particles in the lower level\n",
"n2 = 2500. # Number of particles in the middle level\n",
"n3 = 500. # Number of particles in the upper level\n",
"n11 = 3001. # Number of particles in the lower level in the second case\n",
"n22 = 2498. # Number of particles in the middle level in the second case\n",
"n33 = 501. # Number of particles in the upper level in the second case\n",
"g = 1. # Let us assume the probability of locating a particle in a certain energy state is one\n",
"\n",
"# Calculations\n",
"P1 = 1. / (2500. * 2499) # The probability in the first case\n",
"P2 = 1. / (3001. * 501) # The probability in the second case\n",
"P = P2 / P1 # Comparing the relative probabilities\n",
"\n",
"# Output\n",
"print 'By comparing the relative probabilities P = %3.1f \\n (It means the transfer of one particle from the middle to the\\n upper and the lower state has changed the probability by a factor %3.1f \\n Hence both the distributions are not near the equilibrium state)' % (P, P)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"By comparing the relative probabilities P = 4.2 \n",
" (It means the transfer of one particle from the middle to the\n",
" upper and the lower state has changed the probability by a factor 4.2 \n",
" Hence both the distributions are not near the equilibrium state)\n"
]
}
],
"prompt_number": 1
}
],
"metadata": {}
}
]
}PKDI r55!Heat and Thermodynamics/chA.ipynb{
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{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Appendix Examples"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.1 Page No : 557"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"T = 300. # The given temperature in K\n",
"R = 8.31 # Universal gas constant in J/mole-K\n",
"\n",
"# Calculations\n",
"# The total random kinetic energy of one gram mole of oxygen in J\n",
"U = (3. / 2) * R * T\n",
"\n",
"# Output\n",
"print 'The total random kinetic energy of one gram mole of oxygen is U = %3.0f J ' % (U)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The total random kinetic energy of one gram mole of oxygen is U = 3740 J \n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.2 Page No : 560"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"a = 0.245 # Van der Waals constant in atoms-litre**2-mole**-2\n",
"b = 2.67 * 10**-2 # Van der Waals constant in litre-mole**-1\n",
"R = 8.314 * 10**7 # Universal gas constant in ergs/mole-K\n",
"\n",
"# Calculations\n",
"# Van der Waals constant in dynes-cm**4-mole**-2\n",
"a1 = a * 76 * 13.6 * 980 * 10**6\n",
"b1 = b * 10**3 # Van der Waals constant in cm**3mole**-1\n",
"Tc = (8. / 27) * (a1 / b1) * (1 / R) # The critical temperature in K\n",
"Tc1 = Tc - 273 # The critical temperature in degree centigrade\n",
"\n",
"# Output\n",
"print 'The critical temperature is Tc = %3.2f K (or) %3.2f degree centigrade ' % (Tc, Tc1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The critical temperature is Tc = 33.12 K (or) -239.88 degree centigrade \n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.3 Page No : 565"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"t = 0. # The given temperature in degree centigrade\n",
"E = 5.64 * 10**-21 # The mean kinetic energy of molecules of hydrogen in J\n",
"R = 8.32 # Universal gas constant in J/mole-K\n",
"\n",
"# Calculations\n",
"T = t + 273 # The given temperature in K\n",
"N = (3. / 2) * (R / E) * (T) # Avogadros number\n",
"\n",
"# Output\n",
"print 'The Avogadro number is N = %3.4g ' % (N)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Avogadro number is N = 6.041e+23 \n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.4 Page No : 570"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"d = 2. * 10**-8 # The diameter of the molecule of a gas in cm\n",
"k = 1.38 * 10**-23 # Boltzmanns constant in J/K\n",
"T = 273. # The temperature at NTP in K\n",
"pi = 3.14 # The mathematical constant of pi\n",
"\n",
"# Calculations\n",
"d1 = d / 100 # The diameter of the molecule of a gas in m\n",
"P = 0.76 * 13.6 * 9.8 * 1000 # The pressure at NTP\n",
"n = P / (k * T) # The number of molecules per cubic meter\n",
"l = 1. / (pi * d1**2 * n) # The mean free path in m\n",
"\n",
"# Output\n",
"print 'The mean free path at NTP is %3.4g m ' % (l)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The mean free path at NTP is 2.961e-07 m \n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.6 Page No : 577"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"n = 3. * 10**25 # The number of molecules per cubic metre\n",
"d = 3.6 * 10**-10 # The diameter of oxygen molecule in m\n",
"M = 32. # Molecular weight of oxygen\n",
"N = 6.023 * 10**26 # Avogadro number\n",
"k = 1.38 * 10**-23 # Boltzmans constant in J/K\n",
"T = 273. # The temperature at NTP in K\n",
"pi = 3.14 # The mathematical constant of pi\n",
"\n",
"# Calculations\n",
"m = M / N # The mass of oxygen atom in kg\n",
"# Average speed of oxygen molecule at 273K in m/s\n",
"V = ((8 * k * T) / (pi * m))**(1. / 2)\n",
"c = pi * d**2 * V * n # The collision frequency of the molecules\n",
"l = 1. / (pi * d**2 * n) # The mean free path in m\n",
"\n",
"# Output\n",
"print '(a)The collision frequency of the molecules is %3.2g collisions/second \\n (b)The mean free path is %3.4g m ' % (c, l)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)The collision frequency of the molecules is 5.2e+09 collisions/second \n",
" (b)The mean free path is 8.191e-08 m \n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.7 Page No : 580"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"d = 9000. # The density of copper in kg/m**3\n",
"w = 63.5 # The atomic weight of copper in kg\n",
"N = 6.023 * 10**26 # Avogadros number\n",
"pi = 3.14 # Mathematical constant of pi\n",
"h = 6.624 * 10**-34 # Planks constant in Js\n",
"\n",
"# Calculations\n",
"V = w / d # The volume of copper in m**3\n",
"Ef = ((h**2 / (8. * 9. * 10**-31)) * ((3 / pi) * (N / V)) **\n",
" (2. / 3)) / (1.6 * 10**-19) # The fermi energy in eV\n",
"# The pressure at absolute zero for copper in N/m**2\n",
"P = (2. / 3) * (N / V) * Ef\n",
"\n",
"# Output\n",
"print '(a)The Fermi energy is E = %3.3f eV \\n (b)The pressure at absolute zero for copper is P = %3.6g N/m^2 ' % (Ef, P)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)The Fermi energy is E = 7.163 eV \n",
" (b)The pressure at absolute zero for copper is P = 4.07656e+29 N/m^2 \n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.8 Page No : 586"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"p1 = 80. # The initial pressure of a gas in cm of Hg\n",
"p2 = 60. # The final pressure of a gas in cm of Hg\n",
"v2 = 1190. # The final volume occupied by a gas in cc\n",
"v1 = 1000. # The initial volume occupied by a gas in cc\n",
"\n",
"# Calculations\n",
"g = (math.log10(p1 / p2)) / (math.log10(v2 / v1)) # The adiabatic index\n",
"\n",
"# Output\n",
"print 'The adiabatic index is %3.3f ' % (g)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The adiabatic index is 1.654 \n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.9 Page No : 589"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"t = 27. # The given temperature in degree centigrade\n",
"R = 8.3 # Universal gas constant in J/deg mole\n",
"\n",
"# Calculations\n",
"T = t + 273 # The given temperature in K\n",
"v1 = 1. # Let the original volume be in cc\n",
"v2 = 2. * v1 # The final volume in cc\n",
"W = R * T * math.log(v2 / v1) # The work done in J\n",
"\n",
"# Output\n",
"print 'The work done is W = %3.1f J ' % (W)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The work done is W = 1725.9 J \n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.10 Page No : 594"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"t1 = 27. # The initial temperature of the gas in degree centigrade\n",
"T1 = t1 + 273 # The initial temperature of the gas in K\n",
"g = 1.5 # The adiabatic index\n",
"p = 8. # The ratio of final pressure to the initial pressure\n",
"\n",
"# Calculations\n",
"T2 = ((p)**((g - 1) / g)) * T1 # The final temperature of the gas in K\n",
"T21 = T2 - 273 # The final temperature of the gas in degree centigrade\n",
"\n",
"# Output\n",
"print 'The final temperature of the gas is T2 = %3.0f K (or) %3.0f degree centigrade ' % (T2, T21)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The final temperature of the gas is T2 = 600 K (or) 327 degree centigrade \n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.11 Page No : 600"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"n = 0.3 # The efficiency of a carnot engine\n",
"t = 27. # The temperature of the sinkk in degree centigrade\n",
"n1 = 0.5 # The increased efficiency of a carnot engine\n",
"\n",
"# Calculations\n",
"T2 = t + 273 # The temperature of the sinkk in K\n",
"T1 = T2 / (1 - n) # The temperature of the source for 0.3 efficiency in K\n",
"T11 = T2 / (1 - n1) # The temperature of the source for 0.5 efficiency in K\n",
"T = T11 - T1 # The increase in temperature in K\n",
"\n",
"# Output\n",
"print 'The increase in temperature is T = %3.2f K ' % (T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The increase in temperature is T = 171.43 K \n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.12 Page No : 602"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"T1 = 2100. # One of the operating temperature in K\n",
"T2 = 700. # One of the another operating temperature in K\n",
"n1 = 40. # The actual efficiency of the engine in percent\n",
"\n",
"# Calculations\n",
"n = (1 - (T2 / T1)) * 100 # The efficiency of the engine in percent\n",
"# The percentage of actual efficiency to the maximum possible efficiency\n",
"# in percent\n",
"n2 = (n1 / n) * 100\n",
"\n",
"# Output\n",
"print 'The percentage of actual efficiency to the maximum possible efficiency is %3.0f percent ' % (n2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The percentage of actual efficiency to the maximum possible efficiency is 60 percent \n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.13 Page No : 609"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"T1 = 600. # The working temperature of the engine in K\n",
"T2 = 300. # The another working temperature of the engine in K\n",
"n = 52. # Efficiency of the engine claimed by the inventor in percent\n",
"\n",
"# Calculations\n",
"n1 = (1 - (T2 / T1)) * 100 # The carnot efficiency of the engine in percent\n",
"\n",
"# Output\n",
"print 'The efficiency of the engine claimed by inventor is n = %3.0f percent\\nThe carnot efficiency of the engine is n = %3.0f percent \\n (The efficiency claimed is more than the carnots engine efficiency \\n No engine can have efficiency more than carnots efficiency \\n Hence the claim is invalid)' % (n, n1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The efficiency of the engine claimed by inventor is n = 52 percent\n",
"The carnot efficiency of the engine is n = 50 percent \n",
" (The efficiency claimed is more than the carnots engine efficiency \n",
" No engine can have efficiency more than carnots efficiency \n",
" Hence the claim is invalid)\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.14 Page No : 612"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"H1 = 10.**4 # The heat absorbed by a carnots engine in calories\n",
"t1 = 627. # The temperature from a reservoir in degree centigrade\n",
"t2 = 27. # The temperature of the sinkk in degree centigrade\n",
"\n",
"# Calculations\n",
"T1 = t1 + 273 # The temperature of the reservoir in K\n",
"T2 = t2 + 273 # The temperature of the sinkk in K\n",
"n = (1 - (T2 / T1)) * 100 # The efficiency of the engine in percent\n",
"H2 = H1 * (T2 / T1) # The heat rejected to the sinkk in calories\n",
"W = (H1 - H2) * 4.2 # The work done by the engine in J\n",
"\n",
"# Output\n",
"print 'The efficiency of the engine is n = %3.2f percent The work done by the engine is W = %3.2g J ' % (n, W)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The efficiency of the engine is n = 66.67 percent The work done by the engine is W = 2.8e+04 J \n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.15 Page No : 619"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"w = 100. # The given power of an engine in kW\n",
"t1 = 117. # The operating temperature of an engine in degree centigrade\n",
"t2 = 17. # The another operating temperature of an engine in degree centigrade\n",
"\n",
"# Calculations\n",
"T1 = t1 + 273 # The operating temperature of an engine in K\n",
"T2 = t2 + 273 # The another operating temperature of an engine in K\n",
"W = w * 1000 # The given power of an engine in J/s\n",
"n = (1 - (T2 / T1)) * 100 # The efficiency of an engine in percent\n",
"H = (T1 / T2) # The amount of heat absorbed to the amount of heat rejected\n",
"H2 = W / (H - 1) # The amount of heat rejected per second in J/s\n",
"H1 = H * H2 # The amount of heat absorbed per second in J/s\n",
"\n",
"# Output\n",
"print '(i)The amount of heat absorbed is %3.0g J/s \\n (ii)The amount of heat rejected is %3.0g J/s \\n (iii)The efficiency of the engine is %3.1f percent ' % (H1, H2, n)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(i)The amount of heat absorbed is 4e+05 J/s \n",
" (ii)The amount of heat rejected is 3e+05 J/s \n",
" (iii)The efficiency of the engine is 25.6 percent \n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.16 Page No : 626"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"m1 = 10. # The mass of water at 60 degree centigrade in g\n",
"m2 = 30. # The mass of water at 20 degree centigrade in g\n",
"t1 = 60. # The temperature of 10 g water in degree centigrade\n",
"t2 = 20. # The temperature of 30 g water in degree centigrade\n",
"\n",
"# Calculations\n",
"T1 = t1 + 273 # The temperature of 10g water in K\n",
"T2 = t2 + 273 # The temperature of 30g water in K\n",
"T = ((m1 * T1) + (m2 * T2)) / (m1 + m2) # The final temperature of water in K\n",
"# The change in entropy of 10g water from 333 to 303 K in cal/K\n",
"s1 = m1 * math.log(T / T1)\n",
"# The change in entropy of 30g water from 293 to 303 K in cal/K\n",
"s2 = m2 * math.log(T / T2)\n",
"s = s1 + s2 # The total gain in the entropy of the system in cal/K\n",
"\n",
"# Output\n",
"print 'The change in entropy is %3.4f cal/K ' % (s)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The change in entropy is 0.0627 cal/K \n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.17 Page No : 629"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# input data\n",
"m = 10. # The given amount of water in kg\n",
"t1 = 100. # The temperature of water in degree centigrade\n",
"L = 540. # The latent heat of vapourisation of steam in cal\n",
"\n",
"# Calculations\n",
"m1 = m * 1000 # The given amount of water in g\n",
"T1 = t1 + 273 # The temperature of water in K\n",
"S = (m1 * L) / T1 # The increase in entropy in cal/K\n",
"\n",
"# Output\n",
"print 'The increase in entropy is %3.0f cal/K ' % (S)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The increase in entropy is 14477 cal/K \n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.18 Page No : 634"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"m = 50. # The given amount of water in g\n",
"t1 = 10. # The initial temperature of water in degree centigrade\n",
"t2 = 90. # The final temperature of water in degree centigrade\n",
"\n",
"# Calculations\n",
"T1 = t1 + 273 # The initial temperature of water in K\n",
"T2 = t2 + 273 # The final temperature of water in K\n",
"S = m * math.log(T2 / T1) # The increase in entropy in cal/K\n",
"\n",
"# Output\n",
"print 'The increase in entropy is %3.3f cal/K ' % (S)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The increase in entropy is 12.448 cal/K \n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.19 Page No : 640"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"m = 10. # The given amount of ice in g\n",
"T1 = 273. # The initial temperature of ice in K\n",
"T2 = 373. # The final temperature of steam in K\n",
"L1 = 80. # The latent heat of ice in cal/g\n",
"L2 = 540. # The latent heat of vapourisation of steam in cal\n",
"\n",
"# Calculations\n",
"s1 = (m * L1) / T1 # Increase in entropy from ice at 273K to water at 273K in cal/K\n",
"# Increase in entropy from water at 273K to water at 373K in cal/K\n",
"s2 = (m) * math.log(T2 / T1)\n",
"# Increase in entropy from water at 373K to steam at 373K in cal/K\n",
"s3 = (m * L2) / T2\n",
"s = s1 + s2 + s3 # The total increase in entropy in cal/K\n",
"\n",
"# Output\n",
"print 'The total increase in entropy is %3.2f cal/K ' % (s)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The total increase in entropy is 20.53 cal/K \n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.20 Page No : 645"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"m = 1. # The given amount of nitrogen in g\n",
"t1 = 50. # The initial temperature of nitrogen in degree centigrade\n",
"t2 = 100. # The final temperature of nitrogen in degree centigrade\n",
"Cv = 0.18 # Molar specific heat of nitrogen\n",
"w = 28. # Molecular weight of nitrogen\n",
"\n",
"# Calculations\n",
"T1 = t1 + 273 # The initial temperature of nitrogen in K\n",
"T2 = t2 + 273 # The final temperature of nitrogen in K\n",
"S = (Cv / w) # The Specific heat of nitrogen\n",
"s = m * S * math.log(T2 / T1) # The change in entropy in cal/K\n",
"\n",
"# Output\n",
"print 'The change in entropy is %3.4g cal/K ' % (s)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The change in entropy is 0.0009252 cal/K \n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.21 Page No : 650"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"p = 135.2 # The given increase in the pressure in atmospheres\n",
"V = -0.091 # The given increase in the specific volume when 1g of water freezes into ice in cm**3\n",
"L = 80. # Latent heat of fusion of ice in cal/gram\n",
"T = 273. # The temperature of ice in K\n",
"\n",
"# Calculations\n",
"L1 = L * 4.18 * 10**7 # The latent heat of fusion of ice in ergs/g\n",
"P = p * 1.013 * 10**6 # The given increase in the pressure in dynes/cm**2\n",
"# The temperature at which ice will freeze in degree centigrade\n",
"t = (P * T * V) / L1\n",
"t1 = t + 273 # The temperature at which ice will freeze in K\n",
"\n",
"# Calculations\n",
"print 'The temperature at which ice will freeze is %3.0f degree centigrade (or) %3.0f K ' % (t, t1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The temperature at which ice will freeze is -1 degree centigrade (or) 272 K \n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.22 Page No : 655"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"m = 1. # The given amount of water in kg\n",
"s = 1000. # The specific heat of water in cal/kg-K\n",
"T1 = 273. # The initial temperature of water in K\n",
"T2 = 373. # The temperature of the heat reservoir in K\n",
"\n",
"# Calculations\n",
"S = m * s * math.log(T2 / T1) # The increase in entropy in cal/K\n",
"\n",
"# Output\n",
"print 'The increase in the entropy of water is %3.0f cal/K' % (S)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The increase in the entropy of water is 312 cal/K\n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.23 Page No : 661"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# input data\n",
"m = 0.0273 # The given amount of ice in kg\n",
"L = 80. # The latent heat of fusion of ice in cal/gram\n",
"T = 273. # The temperature of ice in K\n",
"\n",
"# Calculations\n",
"L1 = L * 1000 # The latent heat of fusion of ice in cal/kg\n",
"S = (m * L1 * 4.2) / T # The change in entropy in J/K\n",
"\n",
"# Output\n",
"print 'The change in entropy is %3.1f J/K' % (S)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The change in entropy is 33.6 J/K\n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.24 Page No : 667"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"t1 = 27. # The given initial temperature in degree centigrade\n",
"p = 50. # The reduce in the pressure in atmospheres\n",
"a = 13.2 * 10**-2 # Van der Waals constant in Nm**4mole**-2\n",
"b = 31.2 * 10**-6 # Van der Waals constant in mole**-1m**3\n",
"R = 8.3 # Universal gas constant in JK**-1(mole)**-1\n",
"Cp = 3.5 # The specific heat at constant pressure\n",
"M = 32. # Molecular weight of oxygen\n",
"\n",
"# Calculations\n",
"T = t1 + 273 # The given initial temperature in K\n",
"P = p * 0.76 * 13.6 * 1000 * 9.8 # The reduce in the pressure in N/m**2\n",
"# The drop in the temperature in K\n",
"T1 = ((P) / (4.2 * M * Cp * R)) * (((2 * a) / (R * T)) - b)\n",
"\n",
"# Output\n",
"print 'The drop in the temperature is %3.4f K ' % (T1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The drop in the temperature is 0.0971 K \n"
]
}
],
"prompt_number": 24
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.25 Page No : 674"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"T = 300. # The temperature of the metallic copper disc in K\n",
"Cp = 24.5 # The specific heat at constant pressure in J/mol K\n",
"a = 50.4 * 10**-6 # The coefficient of thermal expansion in K**-1\n",
"K = 7.78 * 10**-12 # Isothermal compressibility in N/m**2\n",
"V = 7.06 * 10**-6 # The specific volume in m**3/mol\n",
"\n",
"# Calculations\n",
"C = (T * V * a**2) / K # The change in specific heats in J/mol K\n",
"Cv = Cp - C # The specific heat at constant volume in J/mol K\n",
"\n",
"# Output\n",
"print 'The specific heat at constant volume is Cv = %3.4f J/mol-K ' % (Cv)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The specific heat at constant volume is Cv = 23.8085 J/mol-K \n"
]
}
],
"prompt_number": 25
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.26 Page No : 676"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"p = 50. # The reduced pressure in atmospheres\n",
"t = 27. # The initial temperature of the gas in degree centigrade\n",
"a = 1.32 * 10**12 # Van der Waal constant a in cm**4 dynes/mole**2\n",
"b = 31.2 # Van der Waal constant b in cm**3/mole\n",
"Cp = 7. # The specific heat at constant pressure in cal/mole-K\n",
"\n",
"# Calculations\n",
"P = p * 76 * 13.6 * 980 # The reduced pressure in dynes/cm**2\n",
"Cp1 = Cp * 4.2 * 10**7 # The specific heat at constant pressure in ergs/mole-K\n",
"T = t + 273 # The initial temperature of the gas in K\n",
"R = 8.31 * 10**7 # The real gas constant in ergs/mole-K\n",
"# The drop in temperature in K or degree centigrade\n",
"dT = (P / Cp1) * (((2 * a) / (R * T)) - b)\n",
"\n",
"# Output\n",
"print 'The drop in temperature produced by adiabatic throttling process is %3.3f K (or) %3.3f degree centigrade ' % (dT, dT)\n",
"\n",
"# Error . There is a change in the result compared to the textbook because\n",
"# the final calculations did in the textbook went wrong , so the final\n",
"# result varied from the textbook\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The drop in temperature produced by adiabatic throttling process is 12.868 K (or) 12.868 degree centigrade \n"
]
}
],
"prompt_number": 26
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.27 Page No : 678"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"t = 0. # The initial temperature of mercury in degree centigrade\n",
"p = 1. # The initial pressure of mercury in atmospheres\n",
"Cp = 28. # The specific heat at constant pressure in J/mol K\n",
"V = 1.47 * 10**-5 # The given specific volume in m**3/mol\n",
"b = 1.81 * 10**-6 # The given volume expansivity in K**-1\n",
"k = 3.89 * 10**-11 # The given compressibility in pa**-1\n",
"\n",
"# Calculations\n",
"T = t + 273 # The initial temperature of mercury in K\n",
"# The specific heat at constant volume in J/mol K\n",
"Cv = Cp - ((T * V * b**2) / k)\n",
"g = Cp / Cv # The adiabatic index\n",
"\n",
"# Output\n",
"print 'The adiabatic index is %3.0f ' % (g)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The adiabatic index is 1 \n"
]
}
],
"prompt_number": 27
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.28 Page No : 681"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"K = 24. * 10**-3 # The coefficient of thermal conductivity of an oxygen molecule in J/m.s.K\n",
"Cv = 20.9 * 10**3 # The specific heat at constant volume in J/kilo.mole.K\n",
"k = 1.38 * 10**-23 # The boltzmanns constant in J/K\n",
"m = 5.31 * 10**-26 # The mass of an oxygen molecule in kg\n",
"T = 273. # The temperature of the molecule in K\n",
"pi = 3.142 # Mathematical constant of pi\n",
"\n",
"# Calculations\n",
"C = ((3 * k * T) / m)**(1. / 2) # The velocity of the molecule in m\n",
"r = (((3 * k * T * m)**(1. / 2) * Cv) / (3. * 2.**(1. / 2) * pi * K)\n",
" )**(1. / 2) # The radius of an oxygen molecule in m\n",
"\n",
"# Output\n",
"print 'The radius of an oxygen molecule is %3.4g m ' % (r)\n",
"\n",
"# Error . There is a change in the result compared to the textbook because\n",
"# the final calculations did in the textbook went wrong , so the final\n",
"# result varied from the textbook\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The radius of an oxygen molecule is 1.265e-09 m \n"
]
}
],
"prompt_number": 28
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.29 Page No : 687"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"b = 0.3 # The given wiens constant in cm-K\n",
"l = 5500. # The given wavelength in A units\n",
"\n",
"# Calculations\n",
"L = l * 10**-8 # The given wavelength in cm\n",
"T = b / L # The temperature of the sun in K\n",
"\n",
"# Output\n",
"print 'The temperature of the sun is %3.0f K ' % (T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The temperature of the sun is 5455 K \n"
]
}
],
"prompt_number": 29
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.30 Page No : 692"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"R = 1. * 10**4 # The rate at which black body loses thermal energy in watts/m**2\n",
"s = 5.672 * 10**-8 # Stefans constant in SI units\n",
"\n",
"# Calculations\n",
"T = (R / s)**(1. / 4) # The temperature of the black body in K\n",
"\n",
"# Output\n",
"print 'The temperature of the black body is %3.0f K' % (T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The temperature of the black body is 648 K\n"
]
}
],
"prompt_number": 30
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.31 Page No : 697"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"T = 6174. # The temperature of the black body in K\n",
"l = 4700. # The wavelength of the black body emitting in amstrong units\n",
"l1 = 1.4 * 10**-5 # The wavelength to be emitted by the black body in m\n",
"\n",
"# Calculations\n",
"L = l * 10**-10 # The wavelength of the black body emitted at 6174 K in m\n",
"L1 = l1 # The wavelength to be emitted by the black body in m\n",
"T1 = (L * T) / L1 # The temperature to be maintained by the black body in K\n",
"\n",
"# Output\n",
"print 'The temperature to be maintained by the black body is %3.2f K ' % (T1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The temperature to be maintained by the black body is 207.27 K \n"
]
}
],
"prompt_number": 31
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.32 Page No : 701"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"T = 5800. # The temperature of the sun in K\n",
"\n",
"# Calculations\n",
"r = 7. * 10**8 # The radius of the sun in m\n",
"pi = 3.142 # The mathematical constant of pi\n",
"A = 4. * pi * r**2 # The surface area of the sun in m**2\n",
"s = 5.672 * 10**-8 # Stefans constant in SI units\n",
"U = A * s * T**4 # The total energy emitted by sun per second in J\n",
"r1 = 1.5 * 10**11 # The distance of the earths atmosphere from the sun in m\n",
"# Energy reaching the top of earths atmosphere in kW/m**2\n",
"R = (U / (4 * pi * r1**2)) / 1000\n",
"\n",
"# Output\n",
"print 'The total radiant energy emitted by sun per second is %3.4g J \\\n",
"\\nThe rate at which energy is reaching earths atmosphere is %3.1f kW/m^2 ' % (U, R)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The total radiant energy emitted by sun per second is 3.953e+26 J \n",
"The rate at which energy is reaching earths atmosphere is 1.4 kW/m^2 \n"
]
}
],
"prompt_number": 33
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.33 Page No : 706"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"n = 5. # The molecules of ozone in grams\n",
"t = 27. # The temperature of ozone in degree centigrade\n",
"R = 8.3 # The universal gas constant in J/g-mol/K\n",
"\n",
"# Calculations\n",
"T = t + 273 # The temperature of ozone in K\n",
"U = n * ((3. / 2) * R * T) # The energy of ozone in J\n",
"\n",
"# Output\n",
"print 'The energy of 5 gms molecules of ozone at 27 degree centigrade is %3.6g J ' % (U)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The energy of 5 gms molecules of ozone at 27 degree centigrade is 18675 J \n"
]
}
],
"prompt_number": 34
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.34 Page No : 708"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"t = -1 # The pressure required to lower the melting point of ice in K\n",
"l = 79.6 # The latent heat of ice in cal/g\n",
"V1 = 1. # The specific volumes of water at 0 degree centigrade in cm**2\n",
"V2 = 1.091 # The specific volumes of ice at 0 degree centigrade in cm**2\n",
"p = 1.013 * 10**6 # One atmospheric pressure in dyne/cm**2\n",
"\n",
"# Calculations\n",
"T = 273. # The temperature of water in K\n",
"L = l * 4.18 * 10**7 # The latent heat of ice in ergs/g\n",
"p1 = (L * t) / (T * (V1 - V2)) # The obtained pressure in dynes/cm**2\n",
"P = p1 / p # The obtained pressure in atmospheres\n",
"P1 = P + 1 # The required pressure in atmospheres\n",
"\n",
"# Output\n",
"print 'The pressure required is %3.2f atmospheres ' % (P1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The pressure required is 133.21 atmospheres \n"
]
}
],
"prompt_number": 35
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.35 Page No : 710"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"t1 = 127. # The temperature of the black body in degree centigrade\n",
"t2 = 27. # The temperature of the walls maintained in degree centigrade\n",
"s = 5.672 * 10**-8 # Stefans constant in SI units\n",
"\n",
"# Calculations\n",
"T1 = t1 + 273 # The temperature of the black body in K\n",
"T2 = t2 + 273 # The temperature of the walls maintained in K\n",
"R = s * (T1**4 - T2**4) # The net amount of energy lost by body in W/m**2\n",
"\n",
"# Output\n",
"print 'The net amount of energy lost by body per sec per unit area is %3.1f watts/m^2' % (R)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The net amount of energy lost by body per sec per unit area is 992.6 watts/m^2\n"
]
}
],
"prompt_number": 36
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.36 Page No : 713"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"t2 = 7. # The low temperature of reservoir in degree centigrade\n",
"n1 = 50. # The efficiency of the carnots engine in percentage\n",
"n2 = 70. # The increased efficiency of the carnots engine in percentage\n",
"\n",
"# Calculations\n",
"T2 = t2 + 273 # The low temperature of the reservoir in K\n",
"T1 = T2 / (1 - (n1 / 100)) # The temperature of the source reservoir in K\n",
"# The temperature to be maintained by the source reservoir in K\n",
"T11 = T2 / (1 - (n2 / 100))\n",
"T = T11 - T1 # The increase in temperature of the source in K or degree centigrade\n",
"\n",
"# Output\n",
"print 'The increase in temperature of the source is %3.1f K (or) %3.1f degree centigrade ' % (T, T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The increase in temperature of the source is 373.3 K (or) 373.3 degree centigrade \n"
]
}
],
"prompt_number": 37
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.37 Page No : 717"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"T1 = 6174. # The temperature of the black body in K\n",
"l1 = 4700. # The wavelength emitted by the black body in amstrong units\n",
"l2 = 1400. # The wavelength to be emitted by the black body in amstrong units\n",
"\n",
"# Calculations\n",
"T2 = (l1 * T1) / l2 # The temperature to be maintained by the black body in K\n",
"\n",
"# Output\n",
"print 'The temperature to be maintained by the black body is %3.0f K ' % (T2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The temperature to be maintained by the black body is 20727 K \n"
]
}
],
"prompt_number": 38
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.38 Page No : 723"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"e = 8.5 * 10**28 # The given energy density of electrons in copper in electrons/m**3\n",
"k = 1.38 * 10**-23 # The boltzmann constant in J/K\n",
"h = 6.62 * 10**-34 # Planks constant in J.s\n",
"m = 9.1 * 10**-31 # The given mass of electrons in kg\n",
"pi = 3.14 # The mathematical constant of pi\n",
"\n",
"# Calculations\n",
"E = (((3 * e) / pi)**(2. / 3)) * (h**2) * (1. / 8) * \\\n",
" (1 / m) # The fermi energy for copper in J\n",
"EF = E / (1.6 * 10**-19) # The fermi energy for copper in eV\n",
"\n",
"# Output\n",
"print 'The fermi energy for copper at absolute zero is %3.3f eV ' % (EF)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The fermi energy for copper at absolute zero is 7.056 eV \n"
]
}
],
"prompt_number": 39
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.39 Page No : 725"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"t1 = 100. # The temperature of the source in degree centigrade\n",
"t2 = 0. # The temperature of the sinkk in degree centigrade\n",
"P = 100. # The power of the engine in watts (or) J/s\n",
"l = 80. # The latent heat of ice in cal/g\n",
"\n",
"# Calculations\n",
"T1 = t1 + 273 # The temperature of the source in K\n",
"T2 = t2 + 273 # The temperature of the sinkk in K\n",
"L = l * 4.2 * 10**3 # The latent heat of ice in ergs/kg\n",
"W = P * 60 # The amount of work done in one minute in J\n",
"H2 = (W * T2) / (T1 - T2) # The amount of heat at the sinkk in J\n",
"m = (H2 / L) # The amount of ice melts in kg\n",
"\n",
"# Output\n",
"print 'The amount of ice that will melt in one minute is %3.5f kg ' % (m)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The amount of ice that will melt in one minute is 0.04875 kg \n"
]
}
],
"prompt_number": 40
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.40 Page No : 730"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"C1 = 1.84 # The RMS speed of molecules of hydrogen at NTP in km/s\n",
"p1 = 2. # The molecular weight of hydrogen\n",
"p2 = 32. # The molecular weight of oxygen\n",
"\n",
"# Calculations\n",
"C2 = C1 * (p1 / p2)**(1. / 2) # The RMS speed of oxygen at NTP in km/s\n",
"C21 = C2 * 1000 # The RMS speed of oxygen at NTP in m/s\n",
"\n",
"# Output\n",
"print 'The RMS speed of oxygen at NTP is %3.2f km/s (or) %3.0f m/s ' % (C2, C21)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The RMS speed of oxygen at NTP is 0.46 km/s (or) 460 m/s \n"
]
}
],
"prompt_number": 41
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.41 Page No : 737"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"t = 101. # The temperature at which water boils in degree centigrade\n",
"p = 787. # The pressure maintained at water boils in mm of Hg\n",
"t1 = 100. # Normal boiling point of water in degree centigrade\n",
"T = t1 + 273 # Normal boiling point of water in K\n",
"p1 = 760. # The normal maintained pressure in mm of Hg\n",
"V2 = 1601. # The specific volume of water evaporation in cm**3\n",
"V1 = 1. # The specific volume of water in cm**3\n",
"\n",
"# Calculations\n",
"V = V2 - V1 # The change in specific volume in cm**3\n",
"dT = t - t1 # The change in temperature in degree centigrade or K\n",
"dP = (p - p1) / 10 # The change in pressure in cm of Hg\n",
"L = (T * dP * 13.6 * 980 * V) / dT # Latent heat of steam in ergs/g\n",
"L1 = L / (4.2 * 10**7) # The latent heat of steam in cal/g\n",
"\n",
"# Output\n",
"print 'The latent heat of steam is %3.4g ergs/g (or) %3.2f cal/g ' % (L, L1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The latent heat of steam is 2.148e+10 ergs/g (or) 511.34 cal/g \n"
]
}
],
"prompt_number": 42
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.42 Page No : 742"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"d = 7.7 * 10**3 # The density of aluminium in kg/m**3\n",
"w = 27. # The atomic weight of Al in kg/k.mol\n",
"N = 6.023 * 10**26 # The number of free electrons in Al\n",
"k = 1.38 * 10**-23 # The boltzmann constant in J/K\n",
"h = 6.62 * 10**-34 # Planks constant in J.s\n",
"m = 9.1 * 10**-31 # The given mass of electrons in kg\n",
"pi = 3.14 # The testematical constant of pi\n",
"\n",
"# Calculations\n",
"V = w / d # The volume occupied by Al in m**3/k.mol\n",
"E = (((3 * (N / V)) / pi)**(2. / 3)) * (h**2) * (1. / 8) * \\\n",
" (1 / m) # The fermi energy for aluminium in J\n",
"EF = E / (1.6 * 10**-19) # The fermi energy for aluminium in eV\n",
"# The pressure of electrons in aluminium at absolute zero in N/m**2\n",
"p = (2. / 3) * (N / V) * (E)\n",
"\n",
"# Output\n",
"print 'The fermi energy for aluminium at absolute zero is %3.3f eV \\\n",
"\\nThe pressure of electrons in aluminium at absolute zero is %3.4g N/m^2' % (EF, p)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The fermi energy for aluminium at absolute zero is 11.278 eV \n",
"The pressure of electrons in aluminium at absolute zero is 2.066e+11 N/m^2\n"
]
}
],
"prompt_number": 44
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.43 Page No : 747"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"t2 = 20. # The temperature of room in degree centigrade\n",
"t1 = 37. # The skin temperature of the boy in degree centigrade\n",
"t = 10. # The given time in min\n",
"A = 3. # The surface area of the student in m**2\n",
"e = 0.9 # The emissivity of the student\n",
"\n",
"# Calculations\n",
"T2 = t2 + 273 # The temperature of the room in K\n",
"T1 = t1 + 273 # The skin temperature of the boy in K\n",
"t1 = t * 60 # The given time in sec\n",
"s = 5.67 * 10**-8 # Stefans constant in W/m**2-K**4\n",
"R = e * A * s * (T1**4 - T2**4) # Heat loss by the skin in one second in J/s\n",
"Q = R * t1 # Total heat loss by the skin in 10 minutes in J\n",
"\n",
"# Output\n",
"print 'The total heat loss by the skin in 10 minutes is %3.4g J ' % (Q)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The total heat loss by the skin in 10 minutes is 1.713e+05 J \n"
]
}
],
"prompt_number": 45
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.44 Page No : 751"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"t1 = 20. # The temperature of the air in the cylinder of a combustion engine in degree centigrade\n",
"p1 = 1. # The initial pressure of the air in atmospheres\n",
"V1 = 8. * 10**-4 # The initial volume of the air in m**3\n",
"V2 = 6. * 10**-5 # The final volume of the air in m**3\n",
"g = 1.4 # The adiabatic index\n",
"\n",
"# Calculations\n",
"T1 = t1 + 273 # The temperature of the air in K\n",
"p2 = p1 * (V1 / V2)**(g) # The final pressure of the gas in atmospheres\n",
"T2 = (p2 / p1) * (V2 / V1) * T1 # The final temperature of the gas in K\n",
"T21 = T2 - 273 # The final temperature of the gas in degree centigrade\n",
"\n",
"# Output\n",
"print 'The final pressure of the gas is %3.1f atmospheres \\\n",
"\\nThe final temperature of the gas is %3.1f K (or) %3.1f degree centigrade ' % (p2, T2, T21)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The final pressure of the gas is 37.6 atmospheres \n",
"The final temperature of the gas is 825.7 K (or) 552.7 degree centigrade \n"
]
}
],
"prompt_number": 46
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.47 Page No : 757"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"d = 2. * 10**-10 # The molecular diameter of an ideal gas in m\n",
"t = 20. # The temperature of the gas in degree centigrade\n",
"p = 1. # The pressure of the gas in atmosphere\n",
"pi = 3.142 # The mathematical constant of pi\n",
"\n",
"# Calculations\n",
"T = t + 273 # The temperature of the gas in K\n",
"P = 1.01 * 10**5 # The pressure of the gas in N/m**2\n",
"v = 511. # The velocity of the molecules at 20 degree centigrade in m/s\n",
"k = 1.38 * 10**-23 # Boltzman constant in J/K\n",
"n = P / (k * T) # The number of molecules per m**3\n",
"l = 1. / (1.414 * pi * d**2 * n) # The mean free path in m\n",
"f = v / l # The collision frequency in per second\n",
"\n",
"# Output\n",
"print '(a)The mean free path is %3.4g m \\n (b)The collision frequency is %3.4g per second ' % (l, f)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)The mean free path is 2.253e-07 m \n",
" (b)The collision frequency is 2.268e+09 per second \n"
]
}
],
"prompt_number": 47
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.48 Page No : 760"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"l = 1.876 * 10**-7 # The mean free path of the gas in m\n",
"v = 511. # The average speed of the molecule in m/s\n",
"\n",
"# Calculations\n",
"f = v / l # The collision frequency in per second\n",
"\n",
"# Output\n",
"print 'The collision frequency is %3.4g per second ' % (f)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The collision frequency is 2.724e+09 per second \n"
]
}
],
"prompt_number": 48
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.49 Page No : 764"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"s = 1. # The specific heat of water in k cal kg C\n",
"m = 1. # The mass of ice in kg\n",
"H = 80. # The latent heat of ice in kcal/kg\n",
"H1 = 540. # The latent heat of steam in kcal/kg\n",
"T = 273. # The temperature of the ice in K\n",
"T1 = 373. # The temperature of water at 100 degree centigrade in K\n",
"\n",
"# Calculations\n",
"S1 = H / T # The increase in entropy when 1 kg of ice at 273 K is converted into water at 273 K in kcal/K\n",
"# The increase in entropy when 1 kg of water at 273 K is converted into\n",
"# water at 373 K in kcal/K\n",
"S2 = m * s * math.log(T1 / T)\n",
"S3 = H1 / T1 # The increase in entropy when 1 kg of water at 373 K is converted into steam at 373 K in kcal/K\n",
"S = S1 + S2 + S3 # The total increase in entropy in kcal/K\n",
"\n",
"# Output\n",
"print 'The total increase in entropy is %3.3f kcal/K ' % (S)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The total increase in entropy is 2.053 kcal/K \n"
]
}
],
"prompt_number": 49
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.50 Page No : 767"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"t1 = 27. # The initial temperature of the gas in degree centigrade\n",
"g = 1.4 # The adiabatic index\n",
"p1 = 1. # Let the initial pressure in atmospheres\n",
"p2 = 2. * p1 # The final pressure in atmospheres\n",
"\n",
"# Calculations\n",
"T1 = t1 + 273 # The initial temperature of the gas in K\n",
"# The final temperature of the gas in K\n",
"T2 = (((p2 / p1)**(g - 1)) * (T1)**g)**(1 / g)\n",
"T = T2 - T1 # The rise in temperature of a gas in K or degree centigrade\n",
"\n",
"# Output\n",
"print 'The rise in temperature is %3.1f degree centigrade ' % (T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The rise in temperature is 65.7 degree centigrade \n"
]
}
],
"prompt_number": 50
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.51 Page No : 770"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"V1 = 10.**-3 # One litre of monoatomic perfect gas at NTP in m**3\n",
"V2 = (V1 / 2) # The final volume in m**3\n",
"g = 1.67 # The adiabatic index\n",
"\n",
"# Calculations\n",
"# The work done on the gas in J\n",
"W = (1 / (g - 1)) * ((1 / (V2)**(g - 1)) - (1 / (V1)**(g - 1)))\n",
"\n",
"# Output\n",
"print 'The work done on the gas is %3.1f J ' % (W)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The work done on the gas is 90.3 J \n"
]
}
],
"prompt_number": 51
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.52 Page No : 775"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"T1 = 1200. # The temperature at which first engine receives heat in K\n",
"T2 = 300. # The temperature at which second engine rejects to heat reservoir in K\n",
"\n",
"# Calculations\n",
"# The temperature when the work outputs of two engines are equal in K\n",
"Tw = (T1 + T2) / 2\n",
"# The temperature when the efficiency of two engines are equal in K\n",
"Te = (T1 * T2)**(1. / 2)\n",
"\n",
"# Output\n",
"print '(a)The temperature when the work outputs of two engines are equal is %3.0f K \\n (b)The temperature when the efficiency of two engines are equal is %3.0f K ' % (Tw, Te)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)The temperature when the work outputs of two engines are equal is 750 K \n",
" (b)The temperature when the efficiency of two engines are equal is 600 K \n"
]
}
],
"prompt_number": 57
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.53 Page No : 777"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"t1 = 27. # The temperature of the source in degree centigrade\n",
"t2 = -73 # The temperature of the sinkk in degree centigrade\n",
"H2 = 300. # The amount of heat released by the sinkk in cal\n",
"\n",
"# Calculations\n",
"T1 = t1 + 273 # The temperature of the source in K\n",
"T2 = t2 + 273 # The temperature of the sinkk in K\n",
"H1 = H2 * (T1 / T2) # The amount of heat released by the source in cal\n",
"W = H1 - H2 # The work performed per cycle in cal\n",
"W1 = W * 4.2 # The work performed per cycle in J\n",
"\n",
"# Output\n",
"print 'The work performed by the engine per cycle is %3.0f J ' % (W1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The work performed by the engine per cycle is 630 J \n"
]
}
],
"prompt_number": 56
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.54 Page No : 782"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Input data\n",
"m = 3. # The rate at which ice melts in kg/hour\n",
"t = 28. # The external temperature in degree centigrade\n",
"Li = 3.3 * 10**5 # Specific latent heat of ice fusion in Jkg**-1\n",
"s = 4.2 * 10**3 # The specific heat in Jkg**-1.C\n",
"\n",
"# Calculations\n",
"Q = (m * Li) + (m * s * t) # The heat taken by the ice to melt into water in J\n",
"P = Q / 3600 # To prevent melting of ice ,the refrigerator should have the power out in J/s\n",
"\n",
"# Output\n",
"print 'The minimum power output of the motor is %3.0f watts ' % (P)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The minimum power output of the motor is 373 watts \n"
]
}
],
"prompt_number": 55
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.55 Page No : 788"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"Li = 3.3 * 10**5 # Specific latent heat of ice fusion in Jkg**-1\n",
"V1 = 1.090 * 10**-3 # The specific volume of one kg of ice in m**3\n",
"V2 = 10.**-3 # The specific volume of one kg of water in m**3\n",
"T = 273. # The temperature maintained in K\n",
"dP = 1.01 * 10**5 # The increase in pressure in N/m**2\n",
"\n",
"# Calculations\n",
"# The depression in the melting point of ice in K (or) degree centigrade\n",
"dT = -(dP * T * (V2 - V1)) / Li\n",
"\n",
"# Output\n",
"print 'The depression of melting point of ice is %3.2g K (or) %3.2g degree centigrade ' % (dT, dT)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The depression of melting point of ice is 0.0075 K (or) 0.0075 degree centigrade \n"
]
}
],
"prompt_number": 58
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.56 Page No : 791"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"dp = 100. # The change in mercury pressure in cm of Hg\n",
"v2 = 1601. # Specific volume of steam in cm**3/gram\n",
"v1 = 1. # Specific volume of water in cm**3/gram\n",
"l = 536. # Latent heat in cal/gram\n",
"t = 100. # The temperature of the steam in degree centigrade\n",
"\n",
"# calculations\n",
"dP = 1. * 13.6 * 10**3 * 9.8 # The change in mercury pressure in N/m**2\n",
"V2 = v2 * 10**-3 # Specific volume of steam in m**3/kg\n",
"V1 = v1 * 10**-3 # Specific volume of water in m**3/kg\n",
"L = l * 4.2 * 10**3 # Latent heat in J/kg\n",
"T = t + 273 # The temperature of the steam in K\n",
"# The increase in boiling point of water in K or degree centigrade\n",
"dT = (dP * T * (V2 - V1)) / L\n",
"\n",
"# Output\n",
"print 'The increase in boiling point of water is %3.2f K (or) %3.2f degree centigrade ' % (dT, dT)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The increase in boiling point of water is 35.33 K (or) 35.33 degree centigrade \n"
]
}
],
"prompt_number": 59
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.57 Page No : 796"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"L = 80. # The latent heat of fusion of ice in cal/gm\n",
"Li = 3.3 * 10**5 # Specific latent heat of ice fusion in Jkg**-1\n",
"dp = 1. # The increase in pressure in atmospheres\n",
"t = 0. # The given temperature in degree centigrade\n",
"v = -0.1 # The change in specific volume in cm**3/gm\n",
"\n",
"# Calculations\n",
"dP = 0.76 * 13.6 * 10**3 * 9.8 # The increase in pressure in N/m**2\n",
"V = v * 10**-3 # The change in specific volume in m**3/kg\n",
"T = t + 273 # The given temperature in K\n",
"# The decrease in the melting point of ice with increase in the pressure\n",
"# of one atmosphere in K\n",
"dT = -(dP * T * (V)) / Li\n",
"\n",
"# Output\n",
"print 'The decrease in melting point of ice is %3.4f K (or) %3.4f degree centigrade ' % (dT, dT)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The decrease in melting point of ice is 0.0084 K (or) 0.0084 degree centigrade \n"
]
}
],
"prompt_number": 60
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.58 Page No : 803"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"R = 8.4 # The universal gas constant in J.mol**-1.K**-1\n",
"Cv = 21. # The spacific heat at constant volume in J.mol**-1.K**-1\n",
"P1 = 2. * 10**5 # The initial pressure of gas in N/m**2\n",
"V1 = 20. # The initial volume of the gas occupied in litres\n",
"P2 = 5. * 10**5 # The final pressure of the gas in N/m**2\n",
"V2 = 50. # The final volume of the gas occupied in litres\n",
"\n",
"# Calculations\n",
"# The ratio of final temperature to the initial temperature for perfect gas\n",
"T = (P2 * V2) / (P1 * V1)\n",
"V = V2 / V1 # The ratio of final volume to the initial volume for perfect gas\n",
"S = (Cv * math.log(T)) + (R * math.log(V)) # The change of entropy in J/K\n",
"\n",
"# Output\n",
"print 'The increase in entropy is %3.2f J/K ' % (S)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The increase in entropy is 46.18 J/K \n"
]
}
],
"prompt_number": 61
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.59 Page No : 807"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"s = 4.2 * 10**3 # The specific heat of water is J/kg.C\n",
"m1 = 0.1 # The mass of water at 15 degree centigrade in kg\n",
"m2 = 0.16 # The mass of water at 40 degree centigrade in kg\n",
"t1 = 15. # The temperature of the first water in degree centigrade\n",
"t2 = 40. # The temperature of the second water in degree centigrade\n",
"\n",
"# Calculations\n",
"T1 = t1 + 273 # The temperature of the first water in K\n",
"T2 = t2 + 273 # The temperature of the second water in K\n",
"T = ((m1 * T1) + (m2 * T2)) / (m1 + m2) # The final mixed temperature in K\n",
"# The change in entropy for 0.1 kg of water in J/K\n",
"s1 = m1 * s * 2.3026 * math.log10(T / T1)\n",
"# The change in entropy for 0.16 kg of water in J/K\n",
"s2 = m2 * s * 2.3026 * math.log10(T / T2)\n",
"S = s1 + s2 # The net change in the entropy of the system in J/K\n",
"\n",
"# Output\n",
"print 'The net increase in entropy is %3.2f J/K ' % (S)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The net increase in entropy is 0.89 J/K \n"
]
}
],
"prompt_number": 62
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.60 Page No : 811"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"m = 12.5 * 10**-3 # The amount of ice in kg\n",
"li = 80. # Latent heat of ice in cal/gram\n",
"l = 536. # Latent heat of steam in cal/gram\n",
"si = 0.5 # Specific heat of ice in cal/gram-K\n",
"sw = 1. # Specific heat of water in cal/gram-K\n",
"T1 = -24 + 273 # The initial temperature of ice in K\n",
"T2 = 0. + 273 # The final temperature of ice in K\n",
"T3 = 100. + 273 # The final temperature of water in K\n",
"\n",
"# Calculations\n",
"Li = li * 10**3 * 4.2 # The latent heat of ice in J/kg\n",
"Ls = l * 10**3 * 4.2 # The latent heat of water in J/kg\n",
"Si = si * 10**3 * 4.2 # The specific heat of ice in J/kg-K\n",
"Sw = sw * 10**3 * 4.2 # The specific heat of water in J/kg-K\n",
"# The increase in entropy of ice from 249 K to 273 K in J/K\n",
"s1 = m * Si * math.log(T2 / T1)\n",
"s2 = (m * Li) / T2 # The increase in entropy from 273 K ice to 273 K water in J/K\n",
"# The increase in entropy of water from 273 K to 373 K in J/K\n",
"s3 = m * Sw * math.log(T3 / T2)\n",
"# The increase in entropy from water at 373 K to steam at 373 K in J/K\n",
"s4 = (m * Ls) / T3\n",
"S = s1 + s2 + s3 + s4 # The total increase in entropy in J/K\n",
"\n",
"# Output\n",
"print 'The total increase in entropy is %3.2f J/K ' % (S)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The total increase in entropy is 109.63 J/K \n"
]
}
],
"prompt_number": 63
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.61 Page No : 813"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"x1 = 20. # The initial thickness of the layer in cm\n",
"x2 = 30. # The final thickness of the layer in cm\n",
"t1 = -15 # The temperature of the surroundings in degree centigrade\n",
"L = 80. # The latent heat of ice in cal/gram\n",
"d = 0.9 # The given density of ice in g/cm**3\n",
"K = 0.005 # The coefficient of thermal conductivity in C.G.S units\n",
"\n",
"# Calculations\n",
"t = ((d * L) / (2 * K * t1)) * (x1**2 - x2**2) # The time taken in sec\n",
"\n",
"# Output\n",
"print 'The time taken for a layer of ice to increase the thickness is %3.2g sec ' % (t)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The time taken for a layer of ice to increase the thickness is 2.4e+05 sec \n"
]
}
],
"prompt_number": 64
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.62 Page No : 815"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"t1 = 121. # The temperature of solid copper sphere in degree centigrade\n",
"dt1 = 2.6 # The rate of cooling of copper sphere in degree centigrade per minute\n",
"t2 = 195. # The temperature of another solid sphere in degree centigrade\n",
"t = 30. # The surrounding temperature in degree centigrade\n",
"\n",
"# Calculations\n",
"T1 = t1 + 273 # The temperature of solid copper sphere in K\n",
"T2 = t2 + 273 # The temperature of another solid copper sphere in K\n",
"T0 = t + 273 # The surrounding temperature in K\n",
"R1 = 1. # Let the radius of the first sphere in m\n",
"R2 = 2. * R1 # The radius of the second sphere in m\n",
"# The rate at which solid copper sphere cools in degree centigrade per minute\n",
"dt2 = (dt1) * (R1 / R2) * ((T2**4 - T0**4) / (T1**4 - T0**4))\n",
"\n",
"# Output\n",
"print 'The rate at which solid copper sphere cools is %3.3f degree centigrade per minute ' % (dt2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The rate at which solid copper sphere cools is 3.281 degree centigrade per minute \n"
]
}
],
"prompt_number": 65
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.63 Page No : 820"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Input data\n",
"dt = 250. # The temperature gradient of an insulated copper rod in degree centigrade per metre\n",
"x = 0.05 # The distance between the two points in m\n",
"K = 384. # The thermal conductivity of copper in W.m**-1.K**-1\n",
"A = 1. # The surface area of the copper rod in m**2\n",
"t = 1. # The given time in seconds\n",
"\n",
"# Calculations\n",
"T = dt * x # The temperature difference in degree centigrade\n",
"# The amount of heat crossed per unit area per sec in J/s\n",
"Q = K * A * (dt) * t\n",
"\n",
"# Output\n",
"print '(1)The difference in temperature between two points seperated by 0.05m is %3.1f degree centigrade \\n (2)The amount of heat crossing per second per unit area normal to the rod is %3.2g J/s ' % (T, Q)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(1)The difference in temperature between two points seperated by 0.05m is 12.5 degree centigrade \n",
" (2)The amount of heat crossing per second per unit area normal to the rod is 9.6e+04 J/s \n"
]
}
],
"prompt_number": 66
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.64 Page No : 826"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"T1 = 200. # The first temperature of the black body in K\n",
"T2 = 2000. # The second temperature of the black body in K\n",
"s = 5.672 * 10**-8 # Stefans constant in M.K.S units\n",
"\n",
"# Calculations\n",
"# The comparision of radiant emittance of a black body for given\n",
"# temperatures\n",
"R = (s * T1**4) / (s * T2**4)\n",
"\n",
"# Output\n",
"print 'The comparision of radiant emittance of a black body at 200 K and 2000 K is %3.0g ' % (R)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The comparision of radiant emittance of a black body at 200 K and 2000 K is 0.0001 \n"
]
}
],
"prompt_number": 67
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example A.65 Page No : 828"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Input data\n",
"d = 0.08 # The diameter of the black sphere in m\n",
"T = 500. # The temperature of the black sphere in K\n",
"T0 = 300. # The temperature of the surroundings in K\n",
"s = 6. * 10**-8 # The stefans constant in W m**-2 K**-4\n",
"pi = 3.14 # The mathematical constant of pi\n",
"\n",
"# Calculations\n",
"A = pi * d**2 # The area of the black sphere in m**2\n",
"e = 1. # The emittance of the black body\n",
"# The rate at which energy is radiated in J/s or watts\n",
"R = s * A * e * (T**4 - T0**4)\n",
"\n",
"# Output\n",
"print 'The rate at which energy is radiated R = %3.2f J/s (or) %3.2f watts' % (R, R)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The rate at which energy is radiated R = 65.59 J/s (or) 65.59 watts\n"
]
}
],
"prompt_number": 68
}
],
"metadata": {}
}
]
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