R=kσT⁴ = 0.92•5.67•10⁻⁸•3000⁴=… W/m²
R=P/A =>
P=AR= A•kσT⁴ =
=8•10⁻⁶•0.92•5.67•10⁻⁸•3000⁴= 33.8 W
Calculate the electric power which must be supplied to the filament of of a light bulb operating at 3000K. The total surface area of the filament is
8×10−6m2
and its emissivity is 0.92.
1 answer
1 answer