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A wall has two layer $A$ and $B$ each made of different material, both the layers have the same thickness. The thermal conductivity of the material $A$ is twice that of $B$. Under thermal equilibrium the temperature difference across the wall $B$ is $36^o C$. The temperature difference across the wall $A$ is ....... $^oC$
$6$
$18$
$12$
$72$
Solution
Consider two walls '$A$' and '$B$' of thickness '$t$' each, and thermal conductivity of' $2 K ^{\prime}$ and ' $K$ ' respectively.
The temperatures at the left of $A=T_a$
The temperatures at the right of $B=T_b$
The temperature at the junction $= T$
Heat flow is a constant in steady state:
$\dot{ Q }= KA \frac{ dT }{ dx }=$ constant
Equating heat flow for both walls we get:
$2 KA \frac{ T – T _{ a }}{ t -0}= KA \frac{ T _{ b }- T }{2 t – t }$
Which simplifies to: $3 T = T _b+2 T _a$
Also given that the temperature difference between the walls is $36^{\circ} C$
$T_b-T_2=36$
Combining the two equation in $T, T_a, T_b$ and eliminating $T_b$
We get $T – T _{ a }=12^{\circ} C$
