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 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) $\frac{k_{1}d_{1}T_{1} + k_{2}d_{2}T_{2}}{k_{1}d_{1} + k_{2}d_{2}}$
Explanation
(a) In series both walls have same rate of heat flow. Therefore
$\frac{dQ}{dt} = \frac{K_{1}A(T_{1} - \theta)}{d_{1}} = \frac{K_{2}A(\theta - T_{2})}{d_{2}}$
$\Rightarrow K_{1}d_{2}\left( T_{1} - \theta \right) = K_{2}d_{1}(\theta - T_{2})$
$\Rightarrow \ \theta = \frac{K_{1}d_{2}T_{1} + K_{2}d_{1}T_{2}}{K_{1}d_{2} + K_{2}d_{1}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.