Question

A wall has two layers $A$ and $B$, each of same thickness and same area of cross-section but made of different materials. The thermal conductivity of $A$ is three times that of $B$. The temperature difference across the wall is $20$${}^o$C . In thermal equilibrium :

a) the temperature difference across the layer $A$ is 15${}^o$C
b) the temperature difference across the layer $B$ is 15${}^o$C
c) the rate of flow of heat across $A$ is more than across $B$
d) the rate of flow of heat across $A$ and $B$ are same

• A. a, b correct
• B. b, c correct
• C. c, d correct
• D. b, d correct

Answer 1

The correct option is D b, d correct

Heat transfer from the wall is given as:
$Q=\frac{KA(T_1-T_2)}{L}$
Here, area and thickness of both the wall is equal, so for equilibrium condition equate both of them, also
$K_1={3K}_2$
$T_1-T_2={20}^{\circ }C$
$T_2=T_1-{20}^{\circ }C$
$T_1=T_2+{20}^{\circ }C$
Therefor temperature across 1:
$\frac{K_1(T_1-T)}{l_1}=\frac{K_2(T-T_2)}{l_2}$
So on substituting the values in above equation we find:
${3K}_2(T_1-T)=K_2(T-T_1+20)$
$T_1-T=5^{\circ }C$
For surface 2:
${3K}_2(20+T_2-T)=K_2(T-T_2)$
$T-T_2={15}^{\circ }C$
$Q=K(T_1-T_2)$
So for surface 1:

$Q\propto K(T_1-T)=3K\left(5\right)={15}^{\circ }C\left(K\right)$
For surface 2: $Q\propto K(T-T_2)=K\left(15\right)={15}^{\circ }C\left(K\right)$
Same for both as both of them have same length and area.
NOTE: $T_1=A$, $T_2=B$