Question

One end of a thermally insulated rod is kept at a temperature $T_1$ and the other at $T_2$. The rod is composed of two sections of lengths $l_1$ and $l_2$ and thermal conductivities $k_1$ and $k_2$ respectively as shown in the figure. The temperature at the interface of the two sections is :

Answer 1

Step 1. Given data

1. One end of a thermally insulated rod is kept at a temperature $T_1$
2. Another end of a thermally insulated rod is kept at a temperature $T_2$
3. The rod is composed of two sections of lengths $l_1$ and $l_2$
4. The thermal conductivity is $k_1$ and $k_2$

Step 2. Formula used

$\frac{Q_{\circ }}{t}=\frac{\left(T_{\circ }-T\right)k_{\circ }A}{L}$

Where $k_{\circ }$ is thermal conductivity, $Q_{\circ }$ is the amount of heat transferred, $A$ area of surface, $L$ distance between the two isothermal planes, $T_{\circ }-T$ is the difference in temperature and t is the time.

Step 3. Find the temperature at the interface of the two sections is

Let $Q_1$ and $Q_2$ be the amount of heat flow from the left and the right side respectively.

$$\begin{gathered} \therefore \frac{Q_1}{t}=\frac{\left(T_1-T\right)k_{1}A}{L_1} \\ \frac{Q_2}{t}=\frac{\left(T_2-T\right)k_{2}A}{L_2} \\ \frac{Q_1}{t}=\frac{Q_2}{t} \\ \Rightarrow \frac{\left(T_1-T\right)k_{1}A}{L_1}=\frac{\left(T_2-T\right)k_{2}A}{L_2} \\ \Rightarrow \left(T_1-T\right)k_1{L}_2=\left(T_2-T\right)k_2{L}_1 \\ \Rightarrow T=\frac{k_1{L}_2{T}_1+k_2{L}_1{T}_2}{k_1{L}_2+k_2{L}_1} \end{gathered}$$

Hence, the temperature at the interface is $\frac{k_1{L}_2{T}_1+k_2{L}_1{T}_2}{k_1{L}_2+k_2{L}_1}$.