Q. Two slabs A and B of equal surface area are placed one over the other such that their surfaces are completely in contact. The thickness of slab A is twice that of B. The coefficient of thermal conductivity of slab A is twice that of B. The first surface of slab A is maintained at 100$℃$, while the second surface of slab B is maintained at 25$℃$. The temperature at contact of their surfaces is
A) 62.5$℃$
B) 45$℃$
C) 55$℃$
D) 85$℃$
Explanation
(a) The temperature at the contact of the surface
$= \frac{K_{1}d_{2}\theta_{1} + K_{2}d_{1}\theta_{2}}{K_{1}d_{2} + K_{2}d_{1}}$
$\ = \ \frac{2K_{2}d_{2} \times 100 + 2d_{2} \times K_{2} \times 25}{2K_{2}d_{2} + K_{2}2d_{2}}$
$= \frac{200 + 50}{4} = 62.6℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.