Question

A 80 watt filament of a lamp loses all its energy by radiation from its surface. If the emissivity of the surface of filament is 0.4 and the area of the surface is $5\times {10}^{-5}{m}^2$, calculate the temperature of the filament.
[$\sigma =5.67\times {10}^{-8}W/m^2{K}^4$. Neglect temperature of surroundings]

Answer 1

Given : $P=80W$ $e=0.4$ $A=5\times {10}^{-5}{m}^2$

Using $P=\sigma Ae{T}^4$

Or $80=5.67\times {10}^{-8}\times 5\times {10}^{-5}\times 0.4\times T^4$

$⟹T=3000K$