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
${\left(\frac{Q}{t}\right)}_{\text{combination }}={\left(\frac{Q}{t}\right)}_A$
$\Rightarrow \frac{K_{S}A\left({\theta }_1-{\theta }_2\right)}{2x}=\frac{2KA\left({\theta }_1-\theta \right)}{x}$
$\because K_S=\frac{2\times 2K\times K}{(2K+K)}=\frac{4}{3}K$ and $\left({\theta }_1-{\theta }_2\right)=36^{\circ }$
$\Rightarrow \frac{\frac{4}{3}KA\times 36}{2x}=\frac{2KA\left({\theta }_1-\theta \right)}{x}$

Hence temperature difference across wall $A$ is
$\left({\theta }_1-\theta \right)=12^{\circ }C$