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Find the emissive power from "0.3 cm"^(2) surface of filament of bulb of tungsten at 3000 K temperature. Take e=0.4 for tungsten bulb. |
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Answer» Solution :From Stefan - Boltzmaan law, `H=AesigmaT^(4)` `=0.3xx10^(-4)xx0.4xx5.67xx10^(-8)xx(3000)^(4)` `=55.1124` `:.` H `~~` 60 W (Acc. to textbook VALUE) |
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