Question

The temperature of the outer surface 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 the temperature of common interface be $TK$ rate of heat flow $H=\frac{Q}{t}=\frac{KA\Delta T}{l}$
$\therefore H_1={\left(\frac{Q}{t}\right)}_1=\frac{2KA(T-T_1)}{4x}$

and $H_2={\left(\frac{Q}{t}\right)}_2=\frac{KA(T_2-T)}{x}$

In steady-state, the rate of heat flow should be the same in the whole system i.e.,
$H_1=H_2$

or $\frac{2KA(T-T_1}{4x}=\frac{KA(T_2-T)}{x}$

$\frac{T-T_1}{2}=T_2-T$ or $T-T_1=2T_2-2T$

$T=\frac{2T_2+T_1}{3}$ ........(i)

Hence heat flow from composite slab is
$H=\left[\frac{KA(T_2-T)}{x}\right]=\frac{KA}{x}(T_2-\frac{2T_2+T_1}{3})$ [from eqn.(i)]

$=\frac{KA}{3x}(T_2-T_1)$ ..........(ii)

According to question,$H=\left[\frac{A(T_2-T_1)K}{x}\right]f$ ...........(iii)
by comparing eqs.(ii) and (iii) we get $f=\frac{1}{3}$