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 of wall is ${36}^{o}C$. Then the difference of temperature at the two surfaces of A will be

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Suppose thickness of each wall is x then
$(\frac{Q}{t}{)}_{combination}\Rightarrow (\frac{Q}{t}{)}_A\Rightarrow \frac{K_{s}A({\theta }_1-{\theta }_2)}{2x}=\frac{2KA({\theta }_1-\theta )}{x}$
$\because K_s=\frac{2\times 2K\times K}{(2K+K)}=\frac{4}{3}Kand({\theta }_1-{\theta }_2)={36}^o$
$\Rightarrow \frac{\frac{4}{3}KA\times 36}{2x}=\frac{2KA({\theta }_1-\theta )}{x}$

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