Question

A wall has two layers $A$ and $B$ each made of different materials. The thickness of both the layers is the same. The thermal conductivity of $A, K_{A}=3 K_{B}$. The temperature difference across the wall is $20^{\circ} C$ in thermal equilibrium.

• A. The temperature difference across $A$ is $15^{\circ} C$.
• B. Rate of heat transfer across $A$ is more than across $B$.
• C. Rate of heat transfer across both is same.
• D. Temperature difference across $A$ is $5^{\circ} C$.

Answer 1

Answer: (D)

$\frac{Q}{A t}=\frac{K\left(\theta_{1}-\theta_{2}\right)}{d}=$ constant
$\therefore K_{A}\left(\frac{\theta_{1}-\theta}{d}\right)$
$=K_{B}\left(\frac{\theta-\theta_{2}}{d}\right)$
$\frac{K_{A}}{K_{B}}=\frac{\theta-\theta_{2}}{\theta_{1}-\theta}$
or $3=\frac{\theta-\theta_{2}}{\theta_{1}-\theta}$
or $3 \theta_{1} +\theta_{2}=4 \theta\,\,\, (1)$
Given $\theta_{1}-\theta_{2}=4 \theta\,\,\, (2)$
Solving (1) and (2) we have, $\theta-\theta_{2}=15^{\circ} C$
$\therefore \theta_{1}-\theta=\theta_{1}-\theta_{2}+\theta_{2}-\theta$
$=\left(\theta_{1}-\theta_{2}\right)-\left(\theta-\theta_{2}\right)$
$=20^{\circ} C -15^{\circ} C =5^{\circ} C$