Question

The walls of a closed cubical box of edge $60\, cm$ are made of material of thickness $1\, mm$ and thermal conductivity $4 \times 10^{-4} \; cal \; s^{-1} \; cm^{-1} {^{\circ}C^{-1}}$. The interior of the box is maintained $1000 \; {^{\circ}C}$ above the outside temperature by a heater placed inside the box and connected across $400\, V\, DC$ 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,
$\frac{d Q}{d t}=k A \frac{\left(T_{1}-T_{0}\right)}{x}$
where, $\frac{d Q}{d t}=$ power transferred through the wall and
$x=$ thickness of the wall
Here, side of cube, $a=60 \,cm$.
Hence, area $A=6 a_{\text {, }}$
$\because$ Total area $=6$ (area of one side of the cube)
thickness $x=0.1 \,cm$, thermal conductivity,
$k=4 \times 10^{-4} cal s ^{-1} cm ^{-1}{ }^{\circ} C ^{-1}$,
temperature difference, $\left(T_{1}-T_{0}\right)=1000^{\circ} C$
and potential of DC source $V=400 \,V$
Hence, the power
$P =\frac{k A \times 4.184 \times\left(T_{1}-T_{0}\right)}{x}$
$P =\frac{k 6 a^{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 $
$P =361.49 \,W$
Given, voltage supply of $DC =400 \,V$
Hence, power generated in a resistance,
$P=\frac{V^{2}}{R}=361.49$
$ \Rightarrow R=\frac{V^{2}}{P} $
$R=\frac{400 \times 400}{361.49 \times 10^{3}}=0.4426\, \Omega$