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An electric heater of surface area 200cm...

An electric heater of surface area `200cm^2` emits radiant energy of 60 kJ at time interval of 1 min. Determine its emissive power. If its emissivity be 0.45, what would be the radiant energy emitted by a black body in one hour, identical to the electrical heater in all respects?

Text Solution

Verified by Experts

Given `DeltaQ=60kJ`, `A=200xx10^-4m^2` and `Deltat=60s`
using, emissive power `E=((DeltaQ)/(Deltat))(1)/(A)`, we have
`E=(60kJ)/(200xx10^-4m^2xx60s)=50 kW//m^(2)`
`:.` Emissivity (of a suface)
`e=("emissive power (of that surface)")/("emissive power(of a black body)")`
`:.` Emissive power of a block body
`=(50)/(0.45)kW//m^(2)=111.11 kW//m^(2)`
`:.` Heat radiated `DeltaQ=EADeltat`
`=(111.11(kW)/(m^2))(200xx10^-4m^2)(60xx60s)=8000kJ`
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