Q: Two walls of thicknesses d_{1} and d_{2} and thermal conductivities K_{1} and K_{2} are in contact. In the steady state, if the temperature at the outer surfaces are T_{1} and T_{2} , the temperature at the common wall is

(a) $\displaystyle \frac{K_1 T_1 d_2 + K_2 T_2 d_1}{K_1 d_2 + K_2 d_1} $

(b) $\displaystyle \frac{K_1 T_1 + K_2 T_2 }{d_1 + d_2} $

(c) $\displaystyle (\frac{K_1 d_1 + K_2 d_2 }{T_1 + T_2})T_1 T_2 $

(d) $\displaystyle \frac{K_1 d_1 T_1 + K_2 d_2 T_2}{K_1 d_1 + K_2 d_2} $

**Click to See Answer : **

Since Heat current $\displaystyle i = \frac{\Delta T}{R}$ is same through both the walls .

$\displaystyle \frac{T_1 -T}{R_1} = \frac{T -T_2}{R_2}$

$\displaystyle \frac{T_1 -T}{d_1/K_1 A} = \frac{T -T_2}{d_2/K_2 A}$

$\displaystyle \frac{K_1 (T_1 -T)}{d_1} = \frac{K_2 (T -T_2)}{d_2} $

K_{1} d_{2} T_{1} – K_{1} d_{2} T = K_{2} d_{1} T – K_{2} d_{1} T_{2}

$\displaystyle T = \frac{K_1 T_1 d_2 + K_2 T_2 d_1}{K_1 d_2 + K_2 d_1} $