Two plates of same area are placed in contact. Their thickness as well as their thermal conductivities are in the ratio $2: 3$. The outer surface of one plate is maintained at $10^{\circ} \mathrm{C}$ and that of the other at $0^{\circ} \mathrm{C}$. Temperature at the common surface is

Question

Aicharya · Accepted Answer

Given that, two plates are having same surface area $(A)$. Let $K_1, K_2$ and $t_1, t_2$ be their conductivities and thickness respectively,

According to question, $ \frac{K_1}{K_2}=\frac{t_1}{t_2}=\frac{2}{3}...(i) $ Since, the rate of flow of heat through both plates be same, then $ \begin{aligned} & \frac{\Delta Q}{\Delta t}=\frac{K_1 A \Delta T_1}{t_1}=\frac{K_2 A \Delta T_2}{t_2} \\ & \Rightarrow \frac{K_1}{K_2} \times \frac{t_2}{t_1} \times\left(T_1-T\right)=\left(T-T_2\right) \end{aligned} $ [where $T$ be the temperature of common surface] Substituting the values, we get $ \begin{aligned} \frac{2}{3} \times \frac{3}{2}(10-T) & =(T-0) \\ T & =5^{\circ} \mathrm{C} \end{aligned} $

Two plates of same area are placed in contact. Their thickness as well as their thermal conductivities are…

Solution