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A blackbody of surface area 10cm^(2) is...

A blackbody of surface area `10cm^(2)` is heated to `127^(@)C` and is suspended in a room at temperature `27^(@)C` calculate the initial rate of loss of heat from the body to the room.

Text Solution

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For a blackbody at temperature T, the rate of emission is `u=sigmaAT^(4)` . When it is kept in a room at temperature `T_(0)`, the rate of absorption is `u_(0)=sigmaA(T^(4)-(t^(4)0)` .Here `A=10xx10^(4)m^(2)` , `T=400K` and `T_(0)=300K` . Thus, `u-u_(0)`
`=(5.67xx10^(-8)Wm^(-2)K^(-4)(10xx10^(-4)m^(2))(400^(4)-(300^(4))K^(4)`
`=0.99W`.
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