{
"metadata": {
"name": "",
"signature": "sha256:981d9d414fb3ccab1f09efa00cc14135798c564e57ce294278994d4eafe77df4"
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"worksheets": [
{
"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": {}
}
]
}