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A blackbody of sarface 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. |
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Answer» 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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