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Two black bodies at temperautre 400K and 500K are placed in an evacuated enclosure whose wall are at 300K . Find the ratio of their rates of cooling. |
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Answer» Solution :HEAT LOST per SECOND by radiation E = `bar(SIGMA)(T^(4)-T_(0)^(4))` Ratio of the rates of cooling= `E_(1)/E_(2)` `E_(1)/E_(2)=((T_(1)^(4)-T_(0)^(4)))/((T_(2)^(4)-T_(0)^(4))` `T_(1)=400K, T_(2) = 500K, T_(0)= 300K` `E_(1)/E_(2)=(400^(4)-300^(4))/(500^(4)-300^(4))` `E_(1):E_(2)=175:544` |
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