Question

A wall has two layers $A$ and $B$ each made of different material. The layer $A$ is $10$ cm thick and $B$ is $20$ cm thick. The thermal conductivity of $A$ is thrice that of $B$. Under thermal equilibrium temperature difference across the wall is ${35}^{\circ }C$. The difference of temperature across the layer $A$ is

• A. ${30}^{\circ }C$
• B. ${14}^{\circ }C$
• C. ${8.75}^{\circ }C$
• D. $5^{\circ }C$

Answer 1

The correct option is D $5^{\circ }C$

Given: $K_A=3K_B$

$L_A=10$ cm $L_B=20$ cm

Let the area of the layers be $A$.

Using: $\frac{d{Q}_A}{dt}=\frac{d{Q}_B}{dt}$

$\therefore$ $\frac{K_{A}A(T+35-T^{\prime })}{L_A}=\frac{K_{B}A(T^{\prime }-T)}{L_B}$

$\frac{\left(3K_B\right)A(T+35-T^{\prime })}{10}=\frac{K_{B}A(T^{\prime }-T)}{20}$

$7T+210=7T^{\prime }$ $⟹T-T^{\prime }=-{30}^{o}C$

Temperature difference across layer A,

$\Delta T_A=T+35-T^{\prime }=-30+35=5^{o}C$