# Sample Problems on Heat Conduction

Heat may be transferred via any substance made up of atoms and molecules. At any one time, the atoms are in many states of motion. Heat or thermal energy is produced by the motion of molecules and atoms, and it is present in all matter.

The more molecules move, the more heat energy is released. When it comes to heat transfer, however, it simply refers to the act of transferring heat from a high-temperature body to a low-temperature body. Heat may move from one location to another in a variety of ways. Meanwhile, if the two systems have a temperature difference, heat will find a method to flow from the upper to the lower system.

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The following are the heat transmission modes: Conduction, Convection and Radiation.

### What is Conduction?

The process of heat transfer from things with higher temperatures to items with lower temperatures is known as conduction.

Thermal energy is transferred from a higher kinetic energy area to a lower kinetic energy area. When high-speed particles collide with slow-moving particles, the kinetic energy of the slow-moving particles increases. Conduction can occur in solids, liquids and gases.

When heat is transmitted from one molecule to another by conduction, the heat energy is generally transported from one molecule to another because they are in direct touch. The location of the molecules, however, remains unchanged. They just resonate with one another.

### Conduction Equation

When it comes to conduction, the coefficient of thermal conductivity reveals that a metal body transmits heat better.

The following equation can be used to calculate the rate of conduction:

q = Q ⁄ t = K A (T_{h}– T_{c}) ⁄ dwhere K is the thermal conductivity, q is heat transfer rate, t is the time of transfer, Q is the amount of heat transfer,A is the area of surfaced is the thickness of the body, T

_{h}is the temperature of the hot region and T_{c}is the temperature of the cold region.

### Sample Problems

**Problem 1: A 10 cm thick block of ice with a temperature of 0 °C lies on the upper surface of 2400 cm ^{2} 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:**

Given:

Area of slab, A = 2400 cm

^{2}Thickness of ice, d = 10 cm

Temperature difference, T

_{h}– T_{c}= 100 °C – 0 °C = 100 °CTime 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 (T

_{h}– T_{c}) ⁄ dRearrange the above formula in terms of K.

K = q d ⁄ A (T

_{h}– T_{c})= (89 × 10) ⁄ (2400 × 100) cal ⁄ cm s °C

= 3.7 × 10

^{-3}cal ⁄ cm s °CHence, 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 D

^{2}⁄ 4 = 0.001256 m^{2}Temperature difference, T

_{h}– T_{c}= 373 K – 273 K = 100 KTime of heat transfer, t = 1 min = 60 s

The formula for heat transfer rate is given as:

Q ⁄ t = K A (T

_{h}– T_{c}) ⁄ dQ = K A t (T

_{h}– T_{c}) ⁄ d= (385 × 0.001256 × 60 × 100) ⁄ 0.4 J

= 7.25 × 10

^{3}JHence, the total amount of heat transfer is

7.25 × 10.^{3}J

**Problem 3: An aluminium rod and a copper rod of equal length 2.0 m and cross-sectional area 2 cm ^{2} 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, K

_{Al}= 200 W ⁄ m °CThermal conductivity of aluminium, K

_{Cu}= 390 W ⁄ m °CCombined 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 cm

^{2}= 2 × 10^{-4}m^{2}Temperature difference, T

_{h}– T_{c}= 30 °C – 10 °C = 20 °CThe formula for heat transfer rate is given as:

q = K A (T

_{h}– T_{c}) ⁄ 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 ⁄ m ^{2}, 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 × 10

^{4}mSurface temperature, T

_{c}= 20.0 °C = (20 + 273) K = 293 KHeat transfer rate per unit area, q ⁄ A = 50.0 mW ⁄ m

^{2}= 50.0 × 10^{-3}W ⁄ m^{2}The formula for heat transfer rate is given as:

q = K A (T

_{h}– T_{c}) ⁄ dRearrange the above formula in terms of T

_{h}.T

_{h}= q d ⁄ KA + T_{c}= ((50.0 × 10

^{-3}× 2.00 × 10^{4}) ⁄ 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, T

_{h}– T_{c}= 10.0 KEnergy lost per sec, q = 50 W

The formula for heat transfer rate is given as:

q = K A (T

_{h}– T_{c}) ⁄ dRearrange the above formula in terms of A.

A = q d ⁄ K (T

_{h}– T_{c})= (50 × 0.1) ⁄ (45 × 10.0) m

^{2}= 0.011 m

^{2}Hence, the area of the slab is

0.011 m.^{2}

**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 = d

^{2}= (5 m)^{2}= 25 m^{2}Temperature difference, T

_{h}– T_{c}= 60 ºC – 0 ºC = 60 ºCThermal conductivity, K = 209 W ⁄ m ºC

Heat transfer time, t = 2 sec

The formula for heat transfer rate is given as:

q = K A (T

_{h}– T_{c}) ⁄ 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 cm ^{2} 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, K

_{Al}= 200 W ⁄ m °CThermal conductivity of aluminium, K

_{Cu}= 390 W ⁄ m °CCombined 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 cm

^{2}= 2 × 10^{-4}m^{2}Temperature difference, T

_{h}– T_{c}= 30 °C – 10 °C = 20 °CThe formula for heat transfer rate is given as:

q = K A (T

_{h}– T_{c}) ⁄ d= (132.2 × 2 × 10

^{-4}× 20) ⁄ 2 W= 0.2644 W

Hence, the total amount of heat transfer is

0.2644 W.