Question

Two metal cubes A and B of same size are arranged as shown in figure. The extreme ends of the combination are maintained at the indicated temperatures. The arrangement is thermally insulated. The coefficients of thermal conductivity of A and B are 300 W/m ${}^{\circ }C$,and 200 W/m ${}^{\circ }C$, respectively. After steady state is reached the temperature T of the interface will be ….

• A. $T={60}^{\circ }C$
• B. $T={70}^{\circ }C$
• C. $T={80}^{\circ }C$
• D. $T={90}^{\circ }C$

Answer 1

The correct option is A $T={60}^{\circ }C$

Thermal resistance, $R=\frac{1}{KA}$.
i.e. R $\infty \frac{1}{K}$, I and A being the same for both the blocks
$\therefore \frac{R_A}{R_B}=\frac{K_B}{R_B}=\frac{200}{300}$
Rate of flow of hear (thermal current ) along both the cubes wil be equal (in series). Therefore, temperature difference across both the cubes will be in the ratio of their thermal resistances. Hence,
$\frac{(TD{)}_A}{(TD{)}_B}=\frac{R_A}{R_B}=\frac{2}{3}or\frac{{100}^{\circ }C-T}{T-0^{\circ }C}=\frac{2}{3}$
$\Rightarrow {300}^{\circ }C-3T=2T$
Or $T={60}^{\circ }C$