- #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