Question

A partition wall has two layers of different materials $A$ and $B$ in contact with each other. They have the same thickness but the thermal conductivity of layer $A$ is twice that of $B$. At steady state, if the temperature difference across the layer $B$ is $50\text{K}$, then the corresponding temperature difference across the layer $A$ is
(Assume that the heat flow rate through both the layers is same)

• A. $50\text{K}$
• B. $12.5\text{K}$
• C. $25\text{K}$
• D. $60\text{K}$

Answer 1

The correct option is C $25\text{K}$

As we know
$q=-kA\frac{dT}{dx}$
Where, $k=$ Thermal conductivity

Let $T$ be the junction temperature.
Here, $k_A=2k_B,T-T_B=50\text{K}$
At the steady state, $q_A=q_B$
$\Rightarrow \frac{k_{A}A(T_A-T)}{\frac{L}{2}}=\frac{k_{B}A(T-T_B)}{\frac{L}{2}}$
$\Rightarrow 2k_B(T_A-T)=k_B(T-T_B)$
$\Rightarrow T_A-T=\frac{T-T_B}{2}=\frac{50}{2}=25\text{K}$