Question

Two materials having coefficients of thermal conductivity ′3K′ and ′K′ and thickness 'd' and ′3d′, respectively, are joined to form a slab as shown in the figure. The temperature of the outer surfaces are $\prime {\theta }_2\prime$ ​ and $\prime {\theta }_1\prime$ respectively, $\left({\theta }_2>{\theta }_1\right).$ . The temperature at the interface is?

A) ${\displaystyle \frac{{\theta }_1+{\theta }_2}{2}}$
B) ${\displaystyle \frac{{\theta }_1}{10}}+{\displaystyle \frac{9{\theta }_2}{10}}$
C) ${\displaystyle \frac{{\theta }_1}{3}}+{\displaystyle \frac{2{\theta }_2}{3}}$
D) ${\displaystyle \frac{{\theta }_1}{6}}+{\displaystyle \frac{5{\theta }_2}{6}}$

Answer 1

Hint : In this solution, we will assume that the system has been in a steady-state for a long time. Then both the slabs will have the same rate of heat transfer and we will use this to determine the solution of the interface.

Formula used: In this solution, we will use the following formula:
Rate of heat transfer in a slab with different temperatures: $Q={\displaystyle \frac{kA\mathrm{\Delta }T}{L}}$ where $A$ is the area of the surface, $\mathrm{\Delta }T$ is the temperature difference, and $L$ is the length of the slab.

Complete step by step answer
In the system given to us, we will assume that the system has been left in this situation for a long time. This implies that the heat transfer will have occurred over a long time and the system can be said to have achieved a steady-state.
In this steady-state situation, we can say that the heat transfer in both the slabs will be the same. Let us assume that the temperature of the surface is $\theta$ . Then since the heat flow rate is constant for both the surfaces will be the same and we can write
 ${\displaystyle \frac{3kA({\theta }_2-\theta )}{d}}={\displaystyle \frac{kA(\theta -{\theta }_1)}{3d}}$
Dividing both sides by $kA/d$ and cross multiplying the denominators, we get
 $9{\theta }_2-9\theta =\theta -{\theta }_1$
Solving for $\theta$ , we get
$\Rightarrow \theta ={\displaystyle \frac{9{\theta }_2}{10}}+{\displaystyle \frac{{\theta }_1}{10}}$ which corresponds to option (B).

Note
Unless mentioned otherwise, we must take such systems to be in a steady-state i.e. the system has been in this state for a long time. Only with this assumption can we assume that both the slabs will have the same rate of heat transfer.