In [1]:

```
# Variables
m = 20; #calorimeter of water equivalent in gm
n = 1030; #weight of water in gm
p = 2; #no.of paddles
a = 10; #weight of each paddle in kg
s = 80; #dismath.tance between paddles in m
g = 980; #accelaration due to gravity in cm/sec**2
# Calculations
E = (p*a*1000*g*s*100); #potential energy in dyne cm
T = (E)/(1050*4.18*10**7); #rise in temperature in deg.C
# Result
print 'the rise in temperature of water is %3.2f deg.C'%(T)
```

In [2]:

```
# Variables
cp = 0.1; #specific heat of copper in kj/kg-K
w = 120; #weight of copper calorimeter in gm
a = 1400; #weight of paraffin oil in gm
cp1 = 0.6; #specific of parafin oil in kj/kg-K
b = 10**8; #force to rotate the paddle in dynes
T = 16; #rise in temperature in deg.C
n = 900; #no.of revolutions stirred
pi = 3.14; #value of pi
# Calculations
c = 2*pi*b; #work done by a rotating paddle per rotation in dyne cm per rotation
d = c*n; #total work done in dyne cm
hc = w*cp*16; #heat gained by calorimeter in calories
hp = a*cp1*16; #heat gaained by paraffin oil in calories
J = d/(hc+hp); #mecanical equivalent of heat in erg/cal
# Result
print 'mecanical equivalent of heat is %.2e erg/cal'%(J)
```

In [3]:

```
# Variables
cp = 0.12; #specific heat of iron in kj/kg-K
m = 25; #mass of iron in lb
h = 0.4; #horse power developed in 3 min
t = 3; #time taken to develop the horse power in min
T = 17; #raise in temp in deg.C
# Calculations
w = h*33000*t; #total work done in ft-lb
H = m*cp*T; #aount of heat developed in B.Th.U
J = (w)/H; #the value of mechanical equivalent of heat
# Result
print 'the mechanical equivalent of water is %3.1f ft-lb/B.Th.U'%(J)
```

In [4]:

```
# Variables
n = 2.; #no.of lead blocks
m = 210.; #mass of each lead block in gm
v = 20000.; #velocity of block relative to earth in cm/sec
J = 4.2*10**7; #mechanical equivalent of heat in ergs/calorie
cp = 0.03; #specific heat of lead in kj/kg-K
# Calculations
E = (m*v**2)/2; #kinetic energy of each block in ergs
E2 = n*E; #total kinetic energy in ergs
T = E2/(J*m*n*cp); #mean rise in temperature in T
# Result
print 'the mean rise in temperature is %3.1f deg.C'%(T)
```

In [5]:

```
# Variables
h = 150; #height froom which ball fallen in ft
cp = 0.03; #specific heat of lead in kj/kg-K
J = 778; #mechanical equivalent of heat in ft lb/B.Th.U
# Calculations
#work done in falling is equal to heat absorbed by the ball
T = 160./(J*cp)*(5./9); #the raise in temperature in T
# Result
print 'the raise in temperature is %3.1f deg.C'%(T)
```

In [7]:

```
import math
# Variables
w = 26.6; #work done one horse in to raise the temperature in lb
T1 = 32.; #temperature at initial in deg.F
T2 = 212.; #temperature at final in deg.F
t = 2.5; #time to raise the tmperature in hrs
p = 25.; #percentage of heat lossed
# Calculations
#only 75% of heat is utillised
x = w*180.*100.*778./((100-p)*150); #the rate at which horse worked
# Result
print 'the rate at which horse worked is %3.0f ft-lb wt/min'%(x)
print "Note : Answer in book is rounded off, Please calculate manually. This answer is accurate."
```

In [8]:

```
# Variables
l = 100.; #length of glass tube in cm
m = 500.; #mass of mercury in glass tube in gm
n = 20.; #number of times inverted i succession
cp = 0.03; #specific heat of mercury in cal/gm/deg.C
J = 4.2; #joule's equivalent in j/cal
g = 981.; #accelaration due to gravity in cm/s**2
# Calculations
PE = m*g*l; #potential energy for each time in ergs
TE = PE*n; #total loss in ergs
T = TE/(m*cp*J*10**7); #rise in temperature in deg.C
#if T is the rise in temperature,then heat devoloped is m*cp*T
# Result
print 'the rise in temperature is %3.2f deg.C'%(T)
```

In [11]:

```
# Variables
d = 0.02; #diameter of the copper wire in cm
i = 1; #current in amp
T = 100; #maximum steady temperature in deg.C
r = 2.1; #resistance of the wire in ohm cm
J = 4.2; #mechanical equivalent of heat in j/cal
a = 3.14*d**2/4; #area of the copper wire in sq.cm
a2 = 1; #area of the copper surface in sq.cm
# Calculations
l = 1/(2*3.14*d/2); #length corresponding to the area in cm
R = r*l/a; #resistance of the copper wire in ohm
w = R*a2**2; #work done in joule
h = w/J; #heat devoleped in cal
# Result
print 'the heat developed is %.f calories'%(round(h,-1))
```

In [15]:

```
import math
# Variables
h = 10000; #vertical height of water fall in cm
v = 5; #volume disharged per sec in litres
J = 4.18; #joule's constant in j/cal
g = 981; #accelaration due to gravity in cm/sec**2
# Calculations
m = v*1000; #mass of water disharged per sec in gm
w = m*h*g; #work done in falling through 100m in erg
H = (v*10**7 *g)/(J*10**7); #quantity of heat produced in cal
T = H/m; #rise in temperature in deg.C
# Result
print 'the quantity of heat produced is %3f cal \
\nthe rise in temperature is %3.2f deg.C'%(H,T)
print "Note : Answer for part A in book is wrong. Please calculate manually."
```

In [16]:

```
# Variables
cp = 0.03; #specific heat of lead in kj/kg.k
v = 10000; #initial velocity of bullet in cm/sec
J = 4.2*10**7; #joules constant in ergs/cal
# Calculations
ke = (v**2)/2; #kinetic energy of the bullet per unit mass in (cm/sec)**2
T = ke*95/(cp*J*100); #rise in temperature in deg.C
# Result
print 'the rise in temperature is %3.1f deg.C'%(T)
```

In [17]:

```
# Variables
h = 5000.; #height of the niagara falls in cm
J = 4.2*10**7; #joules constant in ergs per cal
g = 981; #accelaration due to gravity in cm/sec**2
#CALCULATIONS
w = h*g; #work done per unit mass in ergs/gn
T = w/J; #rise in temperature in deg.C
# Result
print 'the rise in temperature is %3.2f deg.C'%(T)
```

In [18]:

```
import math
# Variables
E1 = 3.75; #potential difference in v
E2 = 3.; #potential differnce in v
i1 = 2.5; #current in amp
i2 = 2; #current in amp
T = 2.7; #the rise in temperature of the water in deg.C
m1 = 48.; #water flow rate at 3 volts in gm/min
m2 = 30.; #water flow rate at 3.75volts in gm/min
s = 1; #specific heat of the water kj/kg-K
# Calculations
J = (E1*i1-E2*i2)/(s*T*(m1-m2)/60); #the mechanical equivalent in j/cal
# Result
print 'the mechanical equivalent is %3.3f j/cal'%(J)
```

In [6]:

```
# Variables
R = 64*10**7; #mean radius of the earth in cm
cp = 0.15; #specific heat of earth in kj/kg-K
J = 4.2*10**7; #joules consmath.tant in erg/cal
# Calculations
i = 2./5*R**2; #moment of inertia of the earth per unit mass in joules
w = (2*3.14)/(24*60*60); #angular velocity of the earth in rad/sec
T = (i*w**2)/(2*J*cp); #rise in temperature in deg.C
# Result
print 'the rise in the temperature is %.1f deg C'%(T)
```

In [20]:

```
# Variables
cp = 1.25; #specific heat of helium inkj/kg-K
v = 1000; #volume of the gas in ml
w = 0.1785; #mass of the gas at N.T.P in gm
p = 76*13.6*981; #pressure of the gas at N.T.P in dynes
T = 273; #temperature at N.T.P in K
# Calculations
V = 1000/w; #volume occupied by the 1gm of helium gas in cc
cv = cp/1.66; #specific heat at constant volume it is monatomuc gas kj/kg-K
r = p*V/T; #gas constant in cm**3.atm./K.mol
J = r/(cp-cv); #mechanical equivalent of heat in erg/cal
# Result
print 'the mechanical equivalent of heat is %.2e ergs/calories'%(J)
print "Note: answer slightly different because of rounding error."
```

In [22]:

```
# Variables
n = 1./273; #coefficent of expaaansion of air
a = 0.001293; #density of air in gm/cc
cp = 0.2389; #specific heat at consmath.tant pressure in kj/kg.K
p = 76*13.6*981; #pressure at 0 deg.C in dynes
# Calculations
J = (p*n)/(a*(cp-(cp/1.405))); #mechanical equivalent of heat
# Result
print 'mechanical equivalent of heat is %.2e ergs/cal'%(J)
print "Note: answer slightly different because of rounding error."
```

In [24]:

```
import math
# Variables
r = 120./60; #rate of flow of water in gm/sec
T1 = 27.30; #temperature at initial in deg.C
T2 = 33.75; #temperature at final in deg.C
v = 12.64; #potential drop in volts
s = 1.; #specific heat of water in kj/kg-K
i = 4.35; #current through the heating element in amp
# Calculations
J = (v*i)/(r*s*(T2-T1)); #the mechanical equivalent of heat in joule/calorie
# Result
print 'the mechanical equivalent of heat is %3.2f j/cal'%(J)
print "Note: answer slightly different because of rounding error."
```

In [26]:

```
# Variables
cp = 6.865; #molar specific heat of hydrogen at consmath.tant pressure in kj/kg-K
cv = 4.880; #molar specific heat of hydrogen at consmath.tant volume in kj/kg-K
p = 1.013*10**6; #atmospheric pressure in dynes/cm**2
v = 22.4*10**3; #gram molar volume in ml
T = 273; #temperature at N.T.P in kelvins
# Calculations
J = (p*v)/(T*(cp-cv)); #mechanical equivalent of heat
# Result
print 'the mechanical equivalent of heat is %.2e ergs/cal'%(J)
```

In [1]:

```
import math
# Variables
v = 1000.; #volume of hydrogen in ml
t = 273.; #tempature of hydrogen in kelvin
p = 76.; #pressure of hydrogen in mm of hg
w = 0.0896; #weigh of hydrogen in gm
cp = 3.409; #specific heat of hydogen in kj/kg-K
cv = 2.411; #specific heat of hydrogen in kj/kg-K
g = 981.; #accelaration due to gravity in cm/sec**2
a = 13.6; #density of mercury in gm/cm**2
# Calculations
J = (p*v*g*a)/(w*t*(cp-cv)); #mechanical equivalent of heat in ergs/cals
# Result
print 'mechanical equivalent of heat is %.2e ergs/calorie'%(J)
```

In [29]:

```
# Variables
cp = 0.23; #specific heat at constant pressure in kj/kg-K
a = 1.18; #density of air in gm/lit
J = 4.2*10**7; #mechanical equivalent of heat in ergs/cal
t = 300; #temperature of air in kelvin
p = 73*13.6*981; #pressure of air in dynes
#cp-cv = (r/J) = pv/(tj)
#CALCULATON
cv = cp-(p*1000/(a*t*J)); #specific heat at constant volume in calories
# Result
print 'the specific heat at constant volume is %.4f calories'%(cv)
print "Note: answer slightly different because of rounding error."
```

In [30]:

```
# Variables
t1 = 0; #temperature of water in deg.C
t2 = 0; #temperature of ice in deg.C
J = 4.18*10**7; #the joules thomson coefficent in erg/cal
l = 80; #latent heat og fusion kj/kg
g = 981; #accelaration due to gravity in cm/sec**2
# Calculations
h = l*J/(15*g); #height from which ice has fallen
# Result
print 'the height from which ice has fallen is %.2e cm'%(h)
```

In [31]:

```
# Variables
T = 80; #temperature of bullet in deg.C
cp = 0.03; #specific heat of lead in kj/kg-K
J = 4.2; #mechanical equivalent of heat in j/cal
# Calculations
h = T*cp; #heat developed per unit mass in calorie
v = (J*10**7*h*2/0.9)**0.5; #velocity of bullet in cm/sec
# Result
print 'the velocity of bullet is %.1e cm/sec'%(v)
```

In [32]:

```
# Variables
w = 5.0; #weight of lead ball in lb
cp = 0.032; #specific heat of lead in Btu/lbdeg.F
h = 50; #height at which ball thrown in feets
v = 20; #vertical speed in ft/sec
g = 32; #accelararion due to gravity in ft/sec**2
# Calculations
u = (v**2)+2*g*h
ke = (w/2*(u)); #kinetic energy of the ball at ground
T = ke/(2*32*778*w*cp); #rise of temperature in deg.F
# Result
print 'the rise in temperature is %.1f deg.F'%(T)
```