Question

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

• A. ${12}^{o}C$
• B. ${18}^{o}C$
• C. $6^{o}C$
• D. ${24}^{o}C$

Answer 1

The correct option is A ${12}^{o}C$

Here, $K_a=2K_B,T_A-T_B={36}^{o}C$
Let T is the temperature of the junction.
As ${\left(\frac{△T}{△t}\right)}_A={\left(\frac{△T}{△t}\right)}_B$

$\therefore \frac{K_{A}A(T_A-T)}{x}=\frac{K_{B}A(T-T_B)}{x}$

$2K_B(T_A-T)=K_B(T-T_B)$
$2(T_A-T)=T-T_B$
Add $(T_A-T)$ on both sides,we get
$3(T_A-T)=T_A-T+T-T_B$
$3(T_A-T)=T_A-T_B$
$T_A-T=\frac{T_A-T_B}{3}=\frac{36}{3}={12}^{o}C$
$\therefore$ temperature difference across the layer
$A=T_A-T={12}^{o}C$