The temperature drop through a two layer furnace wall is `900^(@)C`. Each layer is of equal area of cross section. Which of the following actions will result in lowering temperature `theta` of the interface ?

Options

A. By increasing the thermal conductivity of outer layer.
B. By increasing the thermal conductivity of inner layer.
C. By increasing thickness of outer layer.
D. By increasing thickness of inner layer.

Answer

A, D

Explanation

Rate of heat flow `(K_(i)A(1000 - theta))/(l_(1)) = (K_(0)A (theta - 100))/(l_(0)) implies (theta-100)/(900) = (1)/(1 + (K_(0)l_(i))/(K_(i)l_(0)))`
Now, we can see that `theta` can be decrease by increasing thermal conductivity of outer layer `(K_(0))` and thickness of inner layer ` (l_(1))`.