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 temperatures at the outer surfaces are $T_1$ and $T_2$ , the temperature at the common wall is
(a) $\frac{k_1 T_1 d_2 + k_2 T_2 d_1}{k_1 d_2 + k_2 d_1}$ (b) $\frac{k_1 T_1 + k_2 d_2}{d_1 + d_2}$ (c) $\left(\frac{k_1 d_1 + k_2 d_2}{T_1 + T_2}\right) T_1 T_2$ (d) $\frac{k_1 d_1 T_1 + k_2 d_2 T_2}{k_1 d_1 + k_2 d_2}$
Text Solution
Verified by ExpertsA
In series both walls have same rate of heat flow. Therefore
$\frac{\mathrm{d}Q}{\mathrm{d}t} = \frac{K_1 A (T_1 - \theta)}{d_1} = \frac{K_2 A (\theta - T_2)}{d_2} \\ \Rightarrow K_1 d_2 (T_1 - \theta) = K_2 d_1 (\theta - T_2) \\ \Rightarrow \theta = \frac{K_1 d_2 T_1 + K_2 d_1 T_2}{K_1 d_2 + K_2 d_1}$
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