Boltzman Constant, Emissivity and Surface Area of a filament

AI Thread Summary
The discussion revolves around calculating the surface area of a tungsten filament heated to 2.30 x 10^3 K with an emissivity of 0.31, given that the lamp delivers 35.0 W of power. The relevant equation used is Stefan's Law, which relates power, emissivity, area, and temperature. The calculation shows that the surface area A is derived from the equation 35.0 = 0.31(5.670x10^-8)(2.30*10^3)^4A, leading to A being approximately 7.12 x 10^-5 m². The user confirms the solution is correct and expresses gratitude for the assistance received. The problem has been resolved successfully.
kavamo
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Homework Statement



A tungsten filament in a lamp is heated to a temperature of 2.30 x 10^3 K by an electric current. The tungsten has an emissivity of 0.31. What is the surface area of the filament if the lamp delivers 35.0 W of power?

Homework Equations

Stefan's Law

(a greek letter that looks like P?) P(net)=ekAT^4

The Attempt at a Solution



35.0=0.31(5.670x10^-8 w/m^2*K^4)(2.30*10^3)^4A

35.0 = 491876.5257A

35.0/491876.5257=A

7.1156069x10^-5


is this correct?

Thanks in advance for your help.
 
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