Radiation..........................?

Question

The filament of a light bulb has a temperature of 3.00 103 °C and radiates 51 watts of power. The emissivity of the filament is 0.35. Find the surface area of the filament.

only one equation for radiation that i know off and it doesn't work for this type of problem

Answer 1

EDIT: Note corrections.

You can find the watt/m^2 from this calculator http://www.spectralcalc.com/spectralcalc.php
(use the blackbody calculator).
I assume 3.00 103ºC means 3.00 * 10^3 ºC, which is 3.273*10^3 ºK

The result is 2.28*10^6 W/m^2, so the area is 51/2.28 * 10^-6 m^2 or 22.4 mm^2

You can find Plank's formula for black body radiation here:
http://en.wikipedia.org/wiki/Black_body_radiation
but you would have to integrate that to get total radiated power. I'm sure there is some reference that has that already done.