- #1

teggenspiller

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## Homework Statement

the filament of a light bulb has a temperature of 2580Cel. and radiates 60W of power. The emissivity of the filament is 0.36. What is the surface area of the filament?

## Homework Equations

P= [tex]\sigma[/tex]*A*emissivity*T^4

A=?

[tex]\sigma[/tex]=[tex]\sigma\ =\ 5.670400(40)\ \times\ 10^{-8}\ W\ ^{-2}\ K^{-4}[/tex]

T=2580C and for this equation we need to use the absolute K temp, which is 2580+273, so it is 2583K

P=60, so set the equation to 60

and e= .36

## The Attempt at a Solution

60W=A*[tex]\sigma\ =\ 5.670400(40)\ \times\ 10^{-8}\ W\ ^{-2}\ K^{-4}[/tex]*(.36)*(2583 ^4)

and when I solve for A i get hugely long decimals (obviously), by dividing 60 by this huge number. the multiple choice answers are

a)44mm^2

b) 62

c) 74

d) 91

3) 144

all in mm^2