In [1]:

```
# Input data
l1 = 10. # Length of the copper rod in cm
l2 = 4. # Length of the iron rod in cm
K1 = 0.9 # The thermal conductivity of copper
# Calculations
K2 = (l2**2 / l1**2) * K1 # The Thermal conductivity of iron
# Output
print 'The thermal conductivity of iron is K2 = %3.3f ' % (K2)
```

In [2]:

```
# Input data
K = 0.2 # The thermal conductivity of the plate
d = 0.2 # The thickness of the plate in cm
A = 20. # The area of the plate in cm**2
T = 100. # The temperature difference in degree centigrade
t = 60. # The given time in seconds
# Calculations
# The quantity of heat that will flow through the plate in one minute in cal
Q = (K * A * T * t) / d
# Output
print 'The quantity of heat that will flow through the plate in one minute is Q = %3.4g cal ' % (Q)
```

In [3]:

```
# Input data
l = 30. # The length of the bar in cm
A = 5. # The uniform area of cross section of a bar in cm**2
ta = 200. # The temperature maintained at the end A in degree centigrade
tc = 0. # The temperature maintained at the end C in degree centigrade
Kc = 0.9 # The thermal conductivity of copper
Ki = 0.12 # The thermal conductivity of iron
# Calculations
# The temperature after the steady state is reached in degree centigrade
T = ((Kc * A * ta) + (Ki * A * tc)) / ((Kc + Ki) * A)
# The rate of flow of heat along the bar when the steady state is reached
# in cal/sec
Q = (Kc * A * (ta - T)) / (l / 2)
# Output
print 'The rate of flow of heat along the bar when the steady state is reached is Q = %3.2f cal/s ' % (Q)
```

In [4]:

```
# Input data
d1 = 1.75 # The thickness of the wood in cm
d2 = 3. # The thickness of the cork in cm
t2 = 0. # The temperature of the inner surface of the cork in degree centigrade
t1 = 12. # The temperature of the outer surface of the wood in degree centigrade
K1 = 0.0006 # The thermal conductivity of wood
K2 = 0.00012 # The thermal conductivity of cork
# Calculations
# The temperature of the interface in degree centigrade
T = (((K1 * t1) / d1) + ((K2 * t2) / d2)) / ((K1 / d1) + (K2 / d2))
# Output
print 'The temperature of the interface is T = %3.2f degree centigrade ' % (T)
```

In [5]:

```
# Input data
x1 = 3. # The thickness of the ice layer on the surface of a pond in cm
x = 1. # The increase in the thickness of the ice when the temperature is maintained at -20 degree centigrade in mm
# The increased thickness of the ice layer on the surface of a pond in cm
x2 = x1 + (x / 10)
T = -20 # The temperature of the surrounding air in degree centigrade
d = 0.91 # The density of ice at 0 degree centigrade in g/cm**3
L = 80. # The latent heat of ice in cal/g
K = 0.005 # The thermal conductivity of ice
# Calculations
# The time taken to increase its thickness by 1 mm in sec
t = ((d * L) / (2 * K * (-T))) * (x2**2 - x1**2)
t1 = t / 60 # The time taken to increase its thickness by 1 mm in min
# Output
print 'The time taken to increase its thickness by 1 mm is t = %3.2f s' % (t)
```

In [6]:

```
# Input data
x1 = 10. # The thickness of the ice layer on the surface of a pond in cm
x = 5. # The increase in the thickness of the ice when the temperature is maintained at -10 degree centigrade in cm
# The increased thickness of the ice layer on the surface of a pond in cm
x2 = x1 + (x)
T = -10 # The temperature of the surrounding air in degree centigrade
d = 0.90 # The density of ice at 0 degree centigrade in g/cm**3
L = 80. # The latent heat of ice in cal/g
K = 0.005 # The thermal conductivity of ice
# Calculations
# The time taken to increase its thickness by 5 cm in sec
t = ((d * L) / (2 * K * (-T))) * (x2**2 - x1**2)
# The time taken to increase its thickness by 5 cm in hours
t1 = t / (60. * 60)
# Output
print 'The time taken to increase its thickness by 5 cm is t = %3.0g s (or) %3.0f hours' % (t, t1)
```

In [7]:

```
# input data
# The temperature maintained on one sphere (black body radiat(or) in K
T1 = 300.
# The temperature maintained on another sphere (black body radiat(or) in K
T2 = 200.
s = 5.672 * 10**-8 # Stefans constant in M.K.S units
# Calculations
# The net rate of energy transfer between the two spheres in watts/m**2
R = s * (T1**4 - T2**4)
# output
print 'The net rate of energy transfer between the two spheres is R = %3.2f watts/m^2' % (R)
```

In [8]:

```
import math
# Input data
T1 = 400. # The given temperature of a black body in K
T2 = 4000. # The given temperature of a black body in K
s = 5.672 * 10**-8 # Stefans constant in M.K.S units
# Calculations
R1 = s * T1**4 # The radiant emittance of a black body at 400 k in watts/m**2
# The radiant emittance of a black body at 4000 k in kilo-watts/m**2
R2 = (s * T2**4) / 1000
# Output
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)
```

In [9]:

```
# Input data
e = 0.35 # The relative emittance of tungsten
A = 10.**-3 # The surface area of a tungsten sphere in m**2
T1 = 300. # The temperature of the walls in K
T2 = 3000. # The temperature to be maintained by the sphere in K
s = 5.672 * 10**-8 # Stefans constant in M.K.S units
# Calculations
# The power input required to maintain the sphere at 3000 K in watts
R = s * A * e * (T2**4 - T1**4)
# Output
print 'The power input required to maintain the sphere at 3000 K is R = %3.0f watts' % (R)
```

In [11]:

```
# Input data
e = 0.1 # The relative emittance of an aluminium foil
T1 = 300. # The temperature of one sphere in K
T2 = 200. # The temperature of another sphere in K
s = 5.672 * 10**-8 # Stefans constant in M.K.S units
# Calculations
# The temperature of the foil after the steady state is reached in K
x = (((T1**4 + T2**4) / 2)**(1. / 4))
# The rate of energy transfer between one of the spheres and foil in watts/m**2
R = e * s * (T1**4 - x**4)
# Output
print '1)The temperature of the foil after the steady state reached is x = %3.1f K \
\n2)The rate of energy transfer between the sphere and the foil is R = %3.1f watts/m^2' % (x, R)
```

In [12]:

```
# Input data
A = 5. * 10**-5 # The surface area of the filament in m**2
e = 0.85 # The relative emittance of the filament
s = 5.672 * 10**-8 # Stefans constant in M.K.S units
t = 60. # The time in seconds
T = 2000. # The temperature of the filament of an incandescent lamp in K
# Calculations
E = A * e * s * t * (T**4) # The energy radiated from the filament in joules
# Output
print 'The energy radiated from the filament is E = %3.0f joules ' % (E)
```

In [13]:

```
# Input data
E = 1.53 * 10**5 # The energy radiated from an iron furnace in calories per hour
A = 10.**-4 # The cross section area of an iron furnace in m**2
e = 0.8 # The relative emittance of the furnace
t = 3600. # The time in seconds
s = 1.36 * 10**-8 # Stefans constant in cal/m**2-s-K**4
# Calculations
T = ((E) / (A * e * s * t))**(1. / 4) # The temperature of the furnace in K
# Output
print 'The temperature of the furnace is T = %3.0f K ' % (T)
```

In [14]:

```
# Input data
S = 2.3 # Solar constant in cal/cm**2/minute
r = 7. * 10**10 # The radius of the sun in cm
R = 1.5 * 10**13 # The distance between the sun and the earth in cm
s = 1.37 * 10**-12 # Stefans constant in cal/cm**2/s
# Calculations
E = (S / 60) * (R / r)**(2) # The energy radiated from the sun in cal/s
T = (E / s)**(1. / 4) # The black body temperature of the sun in K
# Output
print 'The black body temperature of the sun is T = %3.0f K ' % (T)
```