Question

Two slabs are of the thicknesses $d_1$ 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 temperatures $\theta_1$ and $\theta_2$ . Assume $\theta_1 > \theta_2$ .The temperature $\theta$ of their common junction is.............

• A. $\frac{K_1\theta_1d_2+K_2\theta_2d_1}{K_1 d_2+K_2d_1}$
• B. $\frac{K_1\theta_1+K_2\theta_2}{K_1 +K_2}$
• C. $\frac{K_1\theta_1+K_2\theta_2}{\theta_1 +\theta_2}$
• D. $\frac{K_1\theta_1d_1+K_2\theta_2d_2}{K_1d_2 +K_2d_1}$

Answer 1

Answer: (A)

$\frac{K_1\theta_1d_2+K_2\theta_2d_1}{K_1 d_2+K_2d_1}$

Answer 2

For first slab

Heat current, $H_{1}=\frac{K_{1}\left(\theta_{1}-\theta\right) A}{d_{1}}$
For second slab,
Heat current, $H_{2}=\frac{K_{2}\left(\theta-\theta_{2}\right) A}{d_{2}}$
As slabs are in series
$H_{1}=H_{2}$
$\therefore \frac{K_{1}\left(\theta_{1}-\theta\right) A}{d_{1}}=\frac{K_{2}\left(\theta-\theta_{2}\right) A}{d_{2}}$
$\Rightarrow \theta=\frac{K_{1} \theta_{1} d_{2}+K_{2} \theta_{2} d_{1}}{K_{2} d_{1}+K_{1} d_{2}}$