1. The problem statement, all variables and given/known data 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. 2. Relevant equations 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 3. 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.