Question

A wall is made of equally thick layers $A$ and $B$ of different materials. Thermal conductivity of $A$ is twice that of $B$. In the steady state, the temperature difference across the wall is $36^{\circ}C$. The temperature difference across the layer $A$ is

• A. $12\,{}^{\circ}C$
• B. $18\,{}^{\circ}C$
• C. $6\,{}^{\circ}C$
• D. $24\,{}^{\circ}C$

Answer 1

Answer: (A)

Here, $K_A = 2K_B$, $T_A - T_B = 36^{\circ}C$
Let $T$ is the temperature of the junction.
As $\left(\frac{\Delta T}{\Delta t}\right)_{A}=\left(\frac{\Delta T}{\Delta t}\right)_{B}$
$\therefore \frac{K_{A}A\left(T_{A}-T\right)}{x}=\frac{K_{B}A\left(T-T_{B}\right)}{x}$
$2K_{B}\left(T_{A}-T\right)=K_{B}\left(T-T_{B}\right)$
$2\left(T_{A}-T\right)=T-T_{B}$
Add $\left(T_{A}-T\right)$ on both sides, we get
$3\left(T_{A}-T\right)=T_{A}-T+T-T_{B}$
$3\left(T_{A}-T\right)=T_{A}-T_{B}$
$T_{A}-T=\frac{T_{A}-T_{B}}{3}$
$=\frac{36}{3}=12\,{}^{\circ}C$
$\therefore $ Temperature difference across the layer
$A=T_{A}-T=12\,{}^{\circ}C$