1.

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.

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)


Discussion

No Comment Found