Q. Two slabs are of the thickness $d_{1}$ and $d_{2}$. Their thermal conductivities are $K_{1}$ and $K_{2}$ respectively. They are in series. The free ends of the combination of these two slabs are kept at temperature $\theta_{1}$ and $\ \theta_{2}$. Assume $\theta_{1\ } > \ \theta_{2}$. The temperature $\theta$ of their common junction is
A) $\frac{K_{1}\theta_{1} + K_{2}\theta_{2}}{\theta_{1} + \theta_{2}}$
B) $\frac{K_{1}\theta_{1}d_{1} + K_{2}\theta_{2}d_{2}}{K_{1}d_{2} + K_{2}d_{1}}$
C) $\frac{K_{1}\theta_{1}d_{2} + K_{2}\theta_{2}d_{1}}{K_{1}d_{2} + K_{2}d_{1}}$
D) $\frac{K_{1}\theta_{1} + K_{2}\theta_{2}}{K_{1} + K_{2}}$
Explanation
(c) For first slab,
Heat current, $H_{1} = \frac{K_{1}\left( \theta_{1} - \theta \right)A}{d_{1}}$
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For second slab,
Heat current, $H_{2} = \frac{K_{2}\left( \theta - \theta_{2} \right)A}{d_{2}}$
As slabs are in series
$H_{1} = H_{2}$
$\therefore$ $\frac{K_{1}\left( \theta_{1} - \theta \right)A}{d_{1}} = \frac{K_{2}\left( \theta - \theta_{2} \right)A}{d_{2}}$
$\Rightarrow$ $\theta = \frac{K_{1}\theta_{1}d_{2} + K_{2}\theta_{2}d_{1}}{K_{2}d_{1} + K_{1}d_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.