Determine the surface area of the filament of a 100W incandescent lamp radiating out its labelled power at 3000K. Given `sigam = 5.7 xx 10^(-8)Wm^(-2)K^(-4)`, and emissivity `epsilon` of the material of the filament = 0.3.

Explanation

Here, `A = ?, P=100W`
`T = 3000 K, epsilon = 0.3` ,
`sigma = 5.7 xx 10^(-8) Wm^(-2)K^(-4)`
`P = EA = (sigma epsilon T^(4))A`
or `A = (P)/(sigma epsilon T^(4)) = (100)/((5.7xx 10^(-8)) xx 0.3 xx (3000)^(4))`
`=7.25 xx 10^(-5)m^(2)`.