Question

A furnace has a two layered wall as shown. Each layer has the same area of cross -section. The temperature $\theta$ at the interface of two layers can be reduced by (keeping other parameters constant)

• A. increasing the thermal conductivity of outer layer
• B. decreasing the thermal conductivity of inner layer
• C. by increasing the thickness of the inner layer
• D. by decreasing the thickness of the outer layer

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A increasing the thermal conductivity of outer layer
B decreasing the thermal conductivity of inner layer
C by increasing the thickness of the inner layer
D by decreasing the thickness of the outer layer

Using thermal analogy of Ohm's law
Rate of heat flow $H=\frac{\Delta \theta }{R}$
As two layers are in series, heat flow is equal in both of them. So,
$H=\frac{800-80}{\frac{l_1}{k_{1}A}+\frac{l_0}{k_{0}A}}=\frac{800-\theta }{\frac{l_1}{k_{1}A}}$
$\Rightarrow \theta =800-\frac{720}{1+\left(\frac{k_1}{k_0}\right)\left(\frac{l_0}{l_1}\right)}$
$\theta$ can be reduced by reducing $\left(\frac{k_1}{k_0}\frac{l_0}{l_1}\right)$ i.e. by reducing thickness of outer layer and thermal conductivity of inner layer, by increasing thickness of the inner layer and thermal conductivity of outer layer.