1.

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)

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`


Discussion

No Comment Found

Related InterviewSolutions