Question

The temperature 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 $\left(\frac{A\left(T_2-T_1\right)K}{x}\right)f$ , with $f$ equals to

• A. $1$
• B. $1/2$
• C. $2/3$
• D. $1/3$

Answer 1

Answer: (D)

Let the temperature of common interface be $T^{\circ }C$
Rate of heat flow
$H=\frac{Q}{t}=\frac{KA\Delta T}{l}$
$\therefore H_1={\left(\frac{Q}{t}\right)}_1=\frac{2KA\left(T-T_1\right)}{4x}$
and $H_2={\left(\frac{Q}{t}\right)}_2=\frac{KA\left(T_2-T\right)}{x}$
In steady state, the rate of heat flow should be same in whole system i.e.,
$H_1=H_2$
$\Rightarrow \frac{2KA\left(T-T_1\right)}{4x}=\frac{KA\left(T_2-T\right)}{x}$
$\Rightarrow \frac{T-T_1}{2}=T_2-T$
$\Rightarrow T-T_1=2T_2-2T$
$\Rightarrow T=\frac{2T_2+T_1}{3}\dots \left(i\right)$
Hence, heat flow from composite slab is
$H=\frac{KA\left(T_2-T\right)}{x}$
$=\frac{KA}{x}\left(T_2-\frac{2T_2+T_1}{3}\right)=\frac{KA}{3x}\left(T_2-T_1\right)\dots \left(ii\right)$
[from Eq. $(i)$]
Accordingly,
$H=\left[\frac{A\left(T_2-T_1\right)K}{x}\right]f\dots \left(iii\right)$
By comparing Eqs. $\left(ii\right)$ and $\left(iii\right)$,we get
$\Rightarrow f=\frac{1}{3}$