Question

A wall is made up of two layers A and B. The thickness of the two layers is the same, but materials are different. The thermal conductivity of A is double than that of B. In thermal equilibrium the temperature difference between the two ends is ${36}^{\circ }C$. Then the difference of temperature at the two surfaces ofAwill be

• A. 6°C
• B. 12°C
• C. 18°C
• D. 24°C

Answer 1

The correct option is B

12°C

Suppose thickness of each wall is x then

${\left(\frac{Q}{T}\right)}_{combination}={\left(\frac{Q}{T}\right)}_A\Rightarrow \frac{K_{s}A({\theta }_1-{\theta }_2)}{2x}=\frac{2KA({\theta }_1-\theta )}{x}$

∵ $K_s=\frac{2\times 2K\times K}{(2K+K}=\frac{4}{3}Kand({\theta }_1-{\theta }_2)={36}^{\circ }$

$\Rightarrow \frac{\frac{4}{3}KA\times 36}{2x}=\frac{2KA({\theta }_1-{\theta }_2)}{x}$

Hence temperature difference across wall A is

$({\theta }_1-\theta )={12}^{\circ }C$