Need help on this Light bulb Filament Temperature with emissivity and δ given

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SUMMARY

The discussion focuses on calculating the temperature of a tungsten filament in a 100 W light bulb, which radiates 2.85 W of light. The filament's surface area is 0.400 mm², and its emissivity is 0.952. The relevant formula used is P = eδAT⁴, where δ is 5.6696 x 10^-8 W/m²·K⁴. Participants are seeking clarification on applying this formula correctly to derive the filament's temperature in Kelvin.

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Hi, I am stuck on this question, any help would be appreciated. Thanks!

  1. The tungsten filament of a certain 100 W light bulb radiates 2.85 W of light. (The other 97.15 is carried away by convection and conduction) The filament has surface area of 0.400 mm^2 and an emissivity of 0.952. Find the filament's temperature in kelvin. (melting point of tungsten is 3683 K) Take δ 5.6696 X 10^-8 W/m^2.K^4

I tried to use the formula: P=eδAT^4 but I still can't get the right answer. I tried to do a search on internet, come across a lot of problem related to the above mentioned but the resistance and ohm is known. If you could point me to the right way, providing the right formula would be very much appreciated.
 
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What temperature did you get?
 

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