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 the material of A is twice that of B. Under thermal equilibrium, the temperature difference across the wall is ${36}^o$C. The temperature difference across the layer A is?

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

Answer 1

Answer: (B)

${12}^o$C

Given :A wall has two layers A and B, each made of different material. Both the layers have the same thickness. The thermal conductivity of the material of A is twice that of B. Under thermal equilibrium, the temperature difference across the wall is 36${}^o$C The temperature difference across the layer A is?

Solution :

${\theta }_A-{\theta }_B={36}^{o}C$

$K_A=2K_B$

${\theta }_C=\frac{\frac{K_A{\theta }_A}{l}+\frac{K_B{\theta }_B}{l}}{\frac{K_A}{l}+\frac{K_B}{l}}$

${\theta }_C=\frac{2{\theta }_A+{\theta }_A-36}{3}$

on solving we get ${\theta }_A-{\theta }_C={12}^{o}C$

The Correct Opt = B