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 of $A$ will be

• A. $6^{\circ} C$
• B. $12^{\circ} C$
• C. $18^{\circ} C$
• D. $24^{\circ} C$

Answer 1

Answer: (B)

Suppose thickness of each wall is $x$ then
$\left(\frac{Q}{t}\right)_{\text {combination }}=\left(\frac{Q}{t}\right)_{A}$
$\Rightarrow \frac{K_{S} A\left(\theta_{1}-\theta_{2}\right)}{2 x}=\frac{2 K A\left(\theta_{1}-\theta\right)}{x}$
$\because K_{S}=\frac{2 \times 2 K \times K}{(2 K+K)}=\frac{4}{3} K$ and $\left(\theta_{1}-\theta_{2}\right)=36^{\circ}$
$\Rightarrow \frac{\frac{4}{3} K A \times 36}{2 x}=\frac{2 K A\left(\theta_{1}-\theta\right)}{x}$

Hence temperature difference across wall $A$ is
$\left(\theta_{1}-\theta\right)=12^{\circ} C$