# Chapter 8 : Transmission of Heat¶

## Example 8.1 Page No : 462¶

In :
# 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)

The thermal conductivity of iron is K2 = 0.144


## Example 8.2 Page No : 469¶

In :
# 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)

The quantity of heat that will flow through the plate in one minute is Q = 1.2e+05 cal


## Example 8.3 Page No : 473¶

In :
# 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)

The rate of flow of heat along the bar when the steady state is reached is Q = 7.06 cal/s


## Example 8.4 Page No : 477¶

In :
# 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)

The temperature of the interface is T = 10.75 degree centigrade


## Example 8.5 Page No : 483¶

In :
# 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)

The time taken to increase its thickness by 1 mm is t = 222.04 s


## Example 8.6 Page No : 485¶

In :
# 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)

The time taken to increase its thickness by 5 cm is t = 9e+04 s (or)  25 hours


## Example 8.7 Page No : 490¶

In :
# 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)

The net rate of energy transfer between the two spheres is R = 368.68 watts/m^2


## Example 8.8 Page No : 495¶

In :
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)

The Radiant emittance of a black body at a temperature of ,
(i) 400 K  is  R = 1452 watts/m^2
(ii) 4000 K  is  R = 14520 kilo-watts/m^2


## Example 8.9 Page No : 500¶

In :
# 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)

The power input required to maintain the sphere at 3000 K is R = 1608 watts


## Example 8.10 Page No : 507¶

In :
# 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)

1)The temperature of the foil after the steady state reached is x = 263.9 K
2)The rate of energy transfer between the sphere and the foil is R = 18.4 watts/m^2


## Example 8.11 Page No : 513¶

In :
# 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)

The energy radiated from the filament is E = 2314 joules


## Example 8.12 Page No : 520¶

In :
# 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)

The temperature of the furnace is T = 2500 K


## Example 8.13 Page No : 524¶

In :
# 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)

The black body temperature of the sun is T = 5987 K