Question

Consider a compound slab consisting of two different materials having equal thickness and thermal conductivities $K$ and $2K$ respectively. The equivalent thermal conductivity of the slab is

• A. $3K$
• B. $\frac{4}{3}K$
• C. $\frac{2}{3}K$
• D. $\sqrt{2}K$

Answer 1

Answer: (B)

The quantity of heat following through a slab in time t,
$Q=\frac{KA\Delta \theta }{l}$
For same heat flow through each slab and composite slab, we have
$\frac{K_{1}A\left(\Delta {\theta }_1\right)}{l}=\frac{K_{2}A\left(\Delta {\theta }_2\right)}{l}$
$=\frac{K^{\prime }A\left(\Delta {\theta }_1+\Delta {\theta }_2\right)}{2l}$
or $K_1\Delta {\theta }_1=K_2\Delta {\theta }_2$
$=\frac{K^{\prime }}{2}\left(\Delta {\theta }_1+\Delta \theta -2\right)=C$ (say)
So, $\Delta {\theta }_1=\frac{C}{K_1},\Delta {\theta }_2=\frac{C}{K_2}$
and $\left(\Delta {\theta }_1+\Delta {\theta }_2\right)=\frac{2C}{K^{\prime }}$
or $\frac{C}{K_1}+\frac{C}{K_2}=\frac{2C}{K^{\prime }}$
or $C\left(\frac{K_1+K_2}{K_1{K}_2}\right)=\frac{2C}{K^{\prime }}$
$\therefore K^{\prime }=\frac{2K_1{K}_2}{K_1+K_2}$
Given $K_1=K,K_2=2K$
So, $K^{\prime }=\frac{2K\times 2K}{K+2K}=\frac{4}{3}$