Question

The glowing filament in a lamp is radiating energy at a rate of 60 W. At the filament’s temperature of 1500∘C, the emissivity is 0.23.

Part A

What is the surface area of the filament?

Answer 1

Using the eq the power of radiation is,

$$ \begin{aligned} H &=A e \sigma T^{4} \\ A &=\frac{H}{e \sigma T^{4}} \\ &=\frac{60 \mathrm{~W}}{(0.23)\left(5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}^{4}\right)\left(1773^{\circ} \mathrm{K}\right)^{4}} \\ &=4.65 \times 10^{-4} \mathrm{~m}^{2} \end{aligned} $$