Q. A composite slab consists of two parts of equal thickness. The thermal conductivity of one is twice that of the other. What will be the ratio of temperature difference across the two layers in the state of equilibrium ?

Thermal Properties of Matter Report Error

Solution:

We use
$\frac{k_{1} A\left(\Delta T_{1}\right)}{l} =\frac{k_{2} A\left(\Delta T_{2}\right)}{l}$
$\frac{\Delta T_{1}}{\Delta T_{2}} =\frac{k_{2}}{k_{1}}$
$=\frac{2 k_{1}}{k_{1}}=2$