Question

A wall has two layers A and B each made of a different materials. Both layers have the same thickness. The thermal conductivity of the material of A is twice that of B. Under thermal equilibrium, the temperature difference across the wall is 36$^{\circ}$C. The temperature difference across layer A is

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

Answer 1

Answer: (B)

12$^{\circ}$C

Answer 2

Let $\theta$ be the temperature of interface. As
$\, \, \, \, \, \, \, \, \bigg(\frac{\Delta Q}{\Delta t}\bigg)_A =\bigg(\frac{\Delta Q}{\Delta t}\bigg)_B$
$\therefore \, \, \, \, \, \, \, \, \, \, \, \, K_1A \times \frac{\Delta T_1}{\Delta_x} =K_2A \frac{\Delta T_2}{\Delta x}$
$\, \, \, \, \, \, \, (2 K_2) \frac{(36-\theta)}{\Delta x} =K_2 \frac{(\theta -0)}{\Delta x}$
$\therefore \, \, \, \, \, \, \, \, \, \, \, 72-2 \theta \Rightarrow \, \theta=24^{\circ}C$
$\therefore \, \,$ Temperature difference across layer A
\, \hspace30mm =36-24=12^{\circ}C