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)

$\frac{4}{3}K$

Answer 2

The quantity of heat following through a slab in time t,
$Q=\frac{K A \Delta \theta}{l}$
For same heat flow through each slab and composite slab, we have
$\frac{K_{1} A\left(\Delta \theta_{1}\right)}{l}=\frac{K_{2} A\left(\Delta \theta_{2}\right)}{l}$
$=\frac{K' A\left(\Delta \theta_{1}+\Delta \theta_{2}\right)}{2 l}$
or $K_{1} \Delta \theta_{1}=K_{2} \Delta \theta_{2}$
$=\frac{K'}{2}\left(\Delta \theta_{1}+\Delta \theta-2\right)=C$ (say)
So, $\Delta \theta_{1}=\frac{C}{K_{1}}, \Delta \theta_{2}=\frac{C}{K_{2}}$
and $\left(\Delta \theta_{1}+\Delta \theta_{2}\right)=\frac{2 C}{K^{\prime}}$
or $\frac{C}{K_{1}}+\frac{C}{K_{2}}=\frac{2 C}{K'}$
or $C\left(\frac{K_{1}+K_{2}}{K_{1} K_{2}}\right)=\frac{2 C}{K'}$
$\therefore K'=\frac{2 K_{1} K_{2}}{K_{1}+K_{2}}$
Given $K_{1}=K, K_{2}=2 K$
So, $K'=\frac{2 K \times 2 K}{K+2 K}=\frac{4}{3}$