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How do you find the surface area of a filament with an amissivity of .36? The bulb temp is 2580C and radiates 60W of power
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P/A = esT^4 P = power radiated, A = area of radiator, e= emissivity of radiator, s = 5.67x10^-8 W/m^2-K^4, and T = temperature in Kelvin.
Found it on yahoo, justa plug and chug.
so solve for A and you get Area=\[P/(esT^4)=(60W)/((.36)(5.6*10^-8)(2853.15K))=[4.4911*10^-5 m^2]\]
Don't forget to compute T^4
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