The filament of typical light bulb is at a temperature of about 3,000 K. For a 100 W bulb, what is the surface area of the filament. You can assume that the entire 100 W goes into heating the filament and the filament has an emissivity of 1.
The equation is P= e(sigma)(A)(T^4)
P is power
e is the emissivity
sigma is the constant 5.67 x 10 ^-5
The Attempt at a Solution
I have gotten to the point where i got 2.2 x 10^-5 m.
The answer in the back of the book is 22 mm^2.
I can't figure out how this was gotten.