Question

The emissivity and surface area of tungsten filament of an electric bulb are $0.35$ and $0.25 \times 10^{-4} metre ^{2}$ respectively. The operating temperature of filament is $3000\, K$. If $\sigma=5.67 \times 10^{-8}$ watt metre $^{-2} K ^{-4}$, then power of bulb is approximately:

• A. 40 watt
• B. 143 watt
• C. 3000 watt
• D. 1050 watt

Answer 1

Answer: (A)

Given that
$e=0.35$
$A=0.25 \times 10^{-4} m ^{2}$
$\sigma=5.67 \times 10^{-8} Wm ^{-2} K ^{-4}$
$T=3000\, K$
We use $P=A e \sigma T^{4}$
$\Rightarrow P=0.25 \times 10^{-4} \times 0.35 \times 5.67 \times 10^{-8} \times(3000)^{4}$
$\Rightarrow P=40.2\, W$
$\Rightarrow P \approx 40\, W$