Problem 1: A 10 cm thick block of ice with a temperature of 0 °C lies on
the upper surface of 2400 cm2 slab of stone. The slab is steam-exposed on
the lower surface at a temperature of 100 °C. Find the heat conductivity of
stone if 4000 g of ice is melted in one hour given that the latent heat of
fusion of ice is 80 cal ⁄ gm.
Solution:
Area of slab, A = 2400 cm2
Thickness of ice, d = 10 cm
Temperature difference, Th – Tc = 100 °C – 0 °C = 100 °C
Time of heat transfer, t = 1 hr = 3600 s
Amount of heat transfer, Q = m L = 4000 × 80 = 320000 cal
Heat transfer rate, q = Q ⁄ t = 320000 cal ⁄ 3600 s = 89 cal ⁄ s
The formula for heat transfer rate is given as:
q = K A (Th – Tc) ⁄ d
Rearrange the above formula in terms of K.
K = q d ⁄ A (Th – Tc)
= (89 × 10) ⁄ (2400 × 100) cal ⁄ cm s °C
= 3.7 × 10-3 cal ⁄ cm s °C
Hence, the thermal conductivity of stone is 3.7 × 10-3 cal ⁄ cm s °C.
Problem 2: A metal rod 0.4 m long & 0.04 m in diameter has one end at 373
K & another end at 273 K. Calculate the total amount of heat conducted in
1 minute. (Given K = 385 J ⁄ m s °C)
Solution:
Given:
Thermal conductivity, K = 385 J ⁄ m s °C
Length of rod, d = 0.4 m
Diameter of rod, D = 0.04 m
Area of slab, A = π D2 ⁄ 4 = 0.001256 m2
Temperature difference, Th – Tc = 373 K – 273 K = 100 K
Time of heat transfer, t = 1 min = 60 s
The formula for heat transfer rate is given as:
Q ⁄ t = K A (Th – Tc) ⁄ d
Q = K A t (Th – Tc) ⁄ d
= (385 × 0.001256 × 60 × 100) ⁄ 0.4 J
= 7.25 × 103 J
Hence, the total amount of heat transfer is 7.25 × 103 J.
Problem 3: An aluminium rod and a copper rod of equal length 2.0 m and
cross-sectional area 2 cm2 are welded together in parallel. One end is kept
at a temperature of 10 °C and the other at 30 °C. Calculate the amount of
heat taken out per second from the hot end. (Thermal conductivity of
aluminium is 200 W ⁄ m °C and of copper is 390 W ⁄ m °C).
Solution:
Given:
Thermal conductivity of aluminium, KAl = 200 W ⁄ m °C
Thermal conductivity of aluminium, KCu = 390 W ⁄ m °C
Combined thermal conductivity for parallel combination, K = 200 W ⁄ m °C + 390
W ⁄ m °C = 590 W ⁄ m °C
Length of rod, d = 2 m
Area of rod, A = 2 cm2 = 2 × 10-4 m2
Temperature difference, Th – Tc = 30 °C – 10 °C = 20 °C
The formula for heat transfer rate is given as:
q = K A (Th – Tc) ⁄ d
= (590 × 2 × 10-4 × 20) ⁄ 2 W
= 1.18 W
Hence, the total amount of heat transfer is 1.18 W.
Problem 4: The average rate at which energy is conducted outward
through the ground surface at a place is 50.0 mW ⁄ m2, and the average
thermal conductivity of the near-surface rocks is 2.00 W ⁄ m K. Assuming
surface temperature of 20.0 °C, find the temperature at a depth of 25.0 km.
Solution:
Given:
Average thermal conductivity, K = 2.00 W ⁄ m K
Depth, d = 25.0 km = 2.50 × 104 m
Surface temperature, Tc = 20.0 °C = (20 + 273) K = 293 K
Heat transfer rate per unit area, q ⁄ A = 50.0 mW ⁄ m2 = 50.0 × 10-3 W ⁄ m2
The formula for heat transfer rate is given as:
q = K A (Th – Tc) ⁄ d
Rearrange the above formula in terms of Th.
Th = q d ⁄ KA + Tc
= ((50.0 × 10-3 × 2.00 × 104) ⁄ 2.00) + 293
= (500 + 293) K
= 893 – 273 K
= 520 °C
Hence, the temperature at depth of 25.0 km is 520 °C.
Problem 5: The energy lost from a 10 cm thick slab of steel is 50 W.
Assuming the temperature difference of 10.0 K, find the area of the slab.
(Thermal conductivity of steel = 45 W ⁄ m K).
Solution:
Given:
Thermal conductivity, K = 45 W ⁄ m K
Thickness of slab, d = 10 cm = 0.1 m
Temperature difference, Th – Tc = 10.0 K
Energy lost per sec, q = 50 W
The formula for heat transfer rate is given as:
q = K A (Th – Tc) ⁄ d
Rearrange the above formula in terms of A.
A = q d ⁄ K (Th – Tc)
= (50 × 0.1) ⁄ (45 × 10.0) m2
= 0.011 m2
Hence, the area of the slab is 0.011 m2.
Problem 6: One face of an aluminium cube of edge 5 meters is maintained
at 60 ºC and the other end is maintained at 0 ºC. All other surfaces are
covered by adiabatic walls. Find the amount of heat flowing through the
cube in 2 seconds. (Thermal conductivity of aluminium is 209 W ⁄ m ºC).
Solution:
Given:
Edge length of cube, d = 5 m
Surface area of cube, A = d2 = (5 m)2 = 25 m2
Temperature difference, Th – Tc = 60 ºC – 0 ºC = 60 ºC
Thermal conductivity, K = 209 W ⁄ m ºC
Heat transfer time, t = 2 sec
The formula for heat transfer rate is given as:
q = K A (Th – Tc) ⁄ d
= (209 × 25 × 60) ⁄ 5 J
= 62700 J
= 62.7 KJ
Hence, the amount of heat that flows through the cube is 62.7 KJ.
Problem 7: An aluminium rod and a copper rod of equal length 2.0 m and
cross-sectional area 2 cm2 are welded together in series. One end is kept
at a temperature of 10 °C and the other at 30 °C. Calculate the amount of
heat taken out per second from the hot end. (Thermal conductivity of
aluminium is 200 W ⁄ m °C and of copper is 390 W ⁄ m °C).
Solution:
Given:
Thermal conductivity of aluminium, KAl = 200 W ⁄ m °C
Thermal conductivity of aluminium, KCu = 390 W ⁄ m °C
Combined thermal conductivity for parallel combination, 1 ⁄ K = 1 ⁄ 200 W ⁄ m °C
+ 1 ⁄ 390 W ⁄ m °C
K = (200 × 390) ⁄ (200 + 390) W ⁄ m °C
= 132.2 W ⁄ m °C
Length of rod, d = 2 m
Area of rod, A = 2 cm2 = 2 × 10-4 m2
Temperature difference, Th – Tc = 30 °C – 10 °C = 20 °C
The formula for heat transfer rate is given as:
q = K A (Th – Tc) ⁄ d
= (132.2 × 2 × 10-4 × 20) ⁄ 2 W
= 0.2644 W
Hence, the total amount of heat transfer is 0.2644 W.
