Question

The temperatures of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively, are $T_2$ and $T_1$ ($T_2>T_1$). The rate of heat transfer through the slab, in a steady state is $\left(\frac{A(T_2-T_1)K}{x}\right)f$, with f equal to

• A. 1
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $\frac{1}{3}$

Answer 1

The correct option is D

$\frac{1}{3}$

Let A be the area of each slab. In the steady state, the rate of heat flow through the composite slab is given by
$\frac{Q}{t}=\frac{T_2-T_1}{\frac{l_1}{K_{1}A}+\frac{l_2}{K_{2}A}}=\frac{A(T_2-T_1)}{\frac{l_1}{K_1}+\frac{l_2}{K_2}}$ .....(1)
Given that, $l_1=x,l_2=4x,K_1=K$ and $K_2=2K$
Using these values in equation (1), $\frac{Q}{t}=\frac{A(T_2-T_1)}{\frac{x}{K}+\frac{4x}{2K}}=\left[\frac{A(T_2-T_1)K}{x}\right]\times \frac{1}{3}$
Comparing this with the given rate of heat transfer, we get $f=\frac{1}{3}$.
Hence, the correct choice is (d).