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A carnot engine operating between temperature T_(1) andT_(2) has efficiency (1)/(6). When T_(2) is lowered by 62 K, its efficiency increases to (1)/(3). Then find the values of T_(1) and T_(2). |
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Answer» Solution :`"Efficiency"eta=1-(T_(2))/(T_(1))` `"In the first case :"1-(T_(2))/(T_(1))=(1)/(6) rArr (T_(2))/(T_(1))=(5)/(6)` `T_(1)=(6)/(5)T_(2)` `"In the SECOND case :"1-((T_(2)-62))/(T_(1))=(1)/(3),(T_(2)-62)/(T_(1))=(2)/(3)` `(T_(2)-62)/(((6)/(5))T_(2))=(2)/(3)` `T_(2)-62=(2)/(3)XX(6)/(5)T_(2)=(4)/(5)T_(2)rArr T_(2)=310K` `T_(1)=(6)/(5)xx310=372K` `T_(1)=372K and T_(2)=210K.` |
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