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
$H=\frac{Q}{t}=\frac{K\,A\,\Delta T}{l}$
$\therefore H_{1}=\left(\frac{Q}{t}\right)_{1}=\frac{2K\,A\left(T-T_{1}\right)}{4x}$
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.,
$H_{1}=H_{2}$
$\Rightarrow \frac{2KA\left(T-T_{1}\right)}{4x}=\frac{KA\left(T_{2}-T\right)}{x}$
$\Rightarrow \frac{T-T_{1}}{2}=T_{2}-T$
$\Rightarrow T-T_{1}=2T_{2}-2T$
$\Rightarrow T=\frac{2T_{2}+T_{1}}{3}\quad\ldots\left(i\right)$
Hence, heat flow from composite slab is
$H=\frac{KA\left(T_{2}-T\right)}{x}$
$=\frac{KA}{x}\left(T_{2}-\frac{2T_{2}+T_{1}}{3}\right)=\frac{KA}{3x}\left(T_{2}-T_{1}\right)\quad\ldots\left(ii\right)$
[from Eq. $(i)$]
Accordingly,
$H=\left[\frac{A\left(T_{2}-T_{1}\right)K}{x}\right]f\quad\ldots\left(iii\right)$
By comparing Eqs. $\left(ii\right)$ and $\left(iii\right)$,we get
$\Rightarrow f=\frac{1}{3}$