## Homework Statement

The filament of a light bulb has a temperature of 3.2*10^3°C and radiates 59 watts of power. The emissivity of the filament is 0.38. Find the surface area of the filament.

## Homework Equations

$$\Delta$$Q/$$\Delta$$T=e$$\sigma$$AT^4

$$\sigma$$=5.67*10^-8

## The Attempt at a Solution

59=(.38)(5.67*10^-8)(A)(3.2*10^3)^4

Use needed Algebra to solve for A and I get 2.61e-5, but it's not the right answer.
Can someone please tell me what I am doing wrong?

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rock.freak667
Homework Helper
Your temperature 3.2*10^3°C is in Celsius, you need to convert it to Kelvin (K).

gneill
Mentor
What temperature scale is implied by the Stefan-Boltzmann constant?