Question

A wall is made up of two layers, \( A \) and \( B \). The thickness of the two layers is the same, but the materials are different. The thermal conductivity of \( A \) is double that of \( B \). If in thermal equilibrium, the temperature difference between the two ends is \( 36^{\circ} C \), then the difference in temperature between the two surfaces of \( A \) will be: 1. \( 6^{\circ} C \) 2. \( 12^{\circ} C \) 3. \( 18^{\circ} C \) 4. \( 24^{\circ} C \)

Answer 1

Correct option is (2) 12^∘C

Consider two walls 'A' and 'B' of thickness 't' each, and thermal conductivity of '2K' and 'K' respectively.

The temperatures at the left of A = T_a​

The temperatures at the right of B = T_b​

The temperature at the junction = T

Heat flow is a constant in steady state:

\(Q = KA \frac{dT}{dx}\) ​= constant

Equating heat flow for both walls we get:

\(2KA \frac{T-T_a}{t - 0} = KA \frac{T_b - T}{2t - t}\)

Which simplifies to: 3T = T_b​ + 2T_a​

Also given that the temperature difference between the walls is 36^∘C

T_b​ − T_a​ = 36

Combining the two equation in T, T_a​, T_b​ and eliminating T_b​

We get T − T_a ​= 12^∘C