Question

The walls of a closed cubical box of edge $60cm$ are made of material of thickness $1mm$ and thermal conductivity $4\times 10^{-4}cal{s}^{-1}c{m}^{-1}{}^{\circ }{C}^{-1}$. The interior of the box is maintained $1000{}^{\circ }C$ above the outside temperature by a heater placed inside the box and connected across $400VDC$ supply. The resistance of the heater is

• A. 4.41 $\Omega$
• B. 44.1 $\Omega$
• C. 0.441 $\Omega$
• D. 441 $\Omega$

Answer 1

Answer: (C)

Heat transfer through conduction wall is given by mathematical expression,
$\frac{dQ}{dt}=kA\frac{\left(T_1-T_0\right)}{x}$
where, $\frac{dQ}{dt}=$ power transferred through the wall and
$x=$ thickness of the wall
Here, side of cube, $a=60cm$.
Hence, area $A=6a_{\text{, }}$
$\because$ Total area $=6$ (area of one side of the cube)
thickness $x=0.1cm$, thermal conductivity,
$k=4\times 10^{-4}cal{s}^{-1}c{m}^{-1}{}^{\circ }{C}^{-1}$,
temperature difference, $\left(T_1-T_0\right)=1000^{\circ }C$
and potential of DC source $V=400V$
Hence, the power
$P=\frac{kA\times 4.184\times \left(T_1-T_0\right)}{x}$
$P=\frac{k6a^2\times 4.184\times 10^3}{x}$
$P=\frac{4\times 10^{-4}\times 6\times (60{)}^2\times 10^3\times 4.184}{0.1}J$
$P=361.49W$
Given, voltage supply of $DC=400V$
Hence, power generated in a resistance,
$P=\frac{V^2}{R}=361.49$
$\Rightarrow R=\frac{V^2}{P}$
$R=\frac{400\times 400}{361.49\times 10^3}=0.4426\Omega$