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

  $\frac{K_1 \theta_1+K_2 \theta_2}{\theta_1+\theta_2}$
• B

  $\frac{K_1 \theta_1 d_1+K_2 \theta_2 d_2}{K_1 d_2+K_2 d_1}$
• Correct answer
  C

  $\frac{K_1 \theta_1 d_2+K_2 \theta_2 d_1}{K_1 d_2+K_2 d_1}$
• D

  $\frac{K_1 \theta_1+K_2 \theta_2}{K_1+K_2}$

Solution

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, $H_1=H_2$ $\begin{array}{ll} \therefore & \frac{K_1\left(\theta_1-\theta\right) A}{d_1}=\frac{K_2\left(\theta-\theta_2\right) A}{d_2} \\ \Rightarrow \quad \theta & \quad \theta=\frac{K_1 \theta_1 d_2+K_2 \theta_2 d_1}{K_2 d_1+K_1 d_2} \end{array}$

Hence, the correct answer is option: 'C'.