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# The filament of an evacuated light bulb has a length 10 cm, diameter 0.2 mm and emissivity 0.2. Calculate the power itradiatesat2000K.(cr=5.6710^{-}^{8}W/m^{2}K^{4})

- 21.5w
- 15.5w
- 8.9w
- 11.4w

## The correct answer is: 11.4w

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