Question

A partition wall has two layers of different materials $A$ and $B$ in contact with each other. They have the same thickness but the thermal conductivity of layer $A$ is twice that of layer $B$. At steady state if the temperature difference across the layer $B$ is $50$ $K$, then the corresponding difference across the layer $A$ is

• A. $50K$
• B. $12.5K$
• C. $25K$
• D. $60K$
• E. $6.25K$

Answer 1

The correct option is C $25K$

From fouriers law of conduction

Heat transfer will be same for both walls in contact

so $Q_1=Q_2$

so $\frac{2KA\Delta T_a}{l}$ =$\frac{KA\Delta T_b}{l}$

here thickness is same and cross sectional area is also same so will get cancelled

$\therefore 2T_a=T_b$

and given that $T_b=50$

so $T_a=25K$