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A black body initially at 27^(@)C is heated to 327^(@)C.How many times is the total radiation emitted at the higher temperature than that emitted at the lower temperature ? What is the wavelength of the maximum energy radiation at the higher temperature ? Wien's constant = 2.898xx10^(-3) mK. |
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Answer» Solution :`T = 27+273 = 300 K, T_(2)=327+273=600` Let `E_(1)` and `E_(2)` be the radiation emitted at temperatures `T_(1)K` and `T_(2)K` `E_(1)/E_(2)=(sigmaT_(1)^(4))/(sigmaT_(2)^(4))=300^(4)/600^(4)= 1/16` `E_(2)=16 E_(1)` The radiation emitted at `327^(@)`C is 16 times that at `27^(@)`C `lamda_(m)=b/T=(2.989xx10^(-3))/600=4.83xx10^(-6)` m |
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