Question

A wall has two layers $A$ and $B$, each made of different material. Both the layers have the same thickness. The thermal conductivity of material of $A$ is twice that of $B$. Under thermal equilibrium,the temperature difference across the wall is $42$${}^o$C . The temperature difference across the layer $A$ is :

• A. $8^{o}C$
• B. ${14}^{o}C$
• C. ${18}^{o}C$
• D. ${24}^{o}C$

Answer 1

The correct option is B ${14}^{o}C$

Here we have to find temperature across A, so 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

$K_1={2K}_2$

$T_1-T_2={42}^{0C}$

$T_2=T_1-{42}^{0C}$

$\frac{K_1(T_1-T)}{l_1}=\frac{K_2(T-T_2)}{l_2}$

So, on substituting the values in above equation we find:

${2K}_2(T_1-T)=K_2(T-T_1+42)$

$T_1-T={14}^{0C}$

NOTE:$T_1=A$,$T_2=B$