Q. 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 $36℃$. Then the difference of temperature at the two surfaces of $A$ will be
A) $6℃$
B) $12℃$
C) $18℃$
D) $24℃$
Explanation
(b) Suppose thickness of each wall is $x$ then
$\left( \frac{Q}{t} \right)_{combination} = \left( \frac{Q}{t} \right)_{A} \Rightarrow \frac{K_{S}A(\theta_{1} - \theta_{2})}{2x} = \frac{2KA(\theta_{1} - \theta)}{x}$
$\because K_{S} = \frac{2 \times 2K \times K}{(2K + K)} = \frac{4}{3}K$ and $\left( \theta_{1} - \theta_{2} \right) = 36{^\circ}$
$\Rightarrow \frac{\frac{4}{3}KA \times 36}{2x} = \frac{2KA(\theta_{1} - \theta)}{x}$
Hence temperature difference across will $A$ is
$\left( \theta_{1} - \theta \right) = 12℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.