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A bulb made of tungsten filament of surface are a`0.5cm^(2)` is heated to a temperature `3000 K` when operatred at `220V`. The emissivity of the filament is `e=0.35` and take `sigma=5.7xx10^(-8)mks units`. Then calculate the wattage of the bulb (in watt) |
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Answer» The emissive power `"watt"//m^(2)` is `E=esT^(4)` Therefore the power of the bulb is `P=exx`area(Watts) `:. P=e A sigma T^(4)` `:. P=0.35xx0.5xx10^(-4)xx5`, `7xx10^(-8)xx(3000)^(4)` `implies P=80.8 W` |
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