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## Homework Statement

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.

## Homework 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

## 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.