Question

A partition Wall has two layers $A$ and $B$, in contact, each made of a different material. They have the same thickness but the thermal conductivity of layer $A$ is twice that of layer $B$. If the steady state temperature difference across the wall is $60K$, then the corresponding difference across the layer $A$ is

• A. 10 K
• B. 20 K
• C. 30 K
• D. 40 K

Answer 1

Answer: (B)

20 K

At steady state total rate of heat flow for A and B is
$\frac{dQ}{dt}=\frac{d{Q}_A}{dt}=\frac{d{Q}_B}{dt}$

$\frac{k^{\prime }A(T_2-T_1)}{l_A+l_B}=\frac{K_{A}A(T-T_1)}{l_A}=\frac{K_{B}A(T_2-T)}{l_B}$

Given that $K_A=2k;K_B=k;l_A=l_B=l;(T_2-T_1)=60K$

By solving for $T$

$\frac{K_{A}A(T-T_1)}{l_A}=\frac{K_{B}A(T_2-T)}{l_B}\frac{2kA(T-T_1)}{l}=\frac{kA(T_2-T)}{l}T=\frac{T_2+2T_1}{3}$

then substitute $T$ in above equation

$\frac{k^{\prime }A(T_2-T_1)}{l_A+l_B}=\frac{2kA(T-T_1)}{l_A}\frac{k^{\prime }A(T_2-T_1)}{2l}=\frac{2kA(\frac{2T_1+T_2}{3}-T_1)}{l}\frac{k^{\prime }A(T_2-T_1)}{2l}=\frac{2kA(T_2-T_1)}{3l}{k}^{\prime }=\frac{4k}{3}$
Then the temperature difference between the ends of A is

$\frac{4kA(T_2-T_1)}{6l}=\frac{2kA(T-T_1)}{l}\frac{2\times 60}{6}=(T-T_1)(T-T_1)=20K$