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 $A,K_A=3K_B$. 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)

$\frac{Q}{At}=\frac{K\left({\theta }_1-{\theta }_2\right)}{d}=$ 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, $\theta -{\theta }_2=15^{\circ }C$
$\therefore {\theta }_1-\theta ={\theta }_1-{\theta }_2+{\theta }_2-\theta$
$=\left({\theta }_1-{\theta }_2\right)-\left(\theta -{\theta }_2\right)$
$=20^{\circ }C-15^{\circ }C=5^{\circ }C$