The thickness of a filament wire is 0.15 mm and length 10 cm. The filament ts heated to incandescence at 2000K. Calculate the energy radiated per second. If the power rating of the bulb is 40W, then calculate the relative emittance or surface emissivity.

Explanation

T`=2xx10^(3)K` thickness `=0.15xx10^(-3)m`
length=0.10m
Surface area of the filament `=2pirL=2xx3.142xx0.075xx10^(-3)xx0.10=4.173xx10^(-5)m^2`
luminous power `=AsigmaT^4`
`=4.713xx10^(-1)=42.756W`
Since `esigmaT^(4)A=P`
`e=P/(sigmaT^4A)=(40)/(42.576)=0.939" "e~~0.94`