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Efficiency of a heat engine whose sink is at temperature of 300 K is 40% . To increase the efficiency to 60 % keeping the sink temperature same, the source temperature must be increased by |
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Answer» Solution :`(T_(2))/(T_(1))=1-eta=1-(40)/(100)=(3)/(5)` `RARR T_(1)=(5)/(3)T_(2) rArr T_(1)=(5)/(3)xx300=500K` New efficiency `eta=60%` `(T_(2))/(T_(1))=1-eta=1-(60)/(100)=(2)/(5)""T_(1)=(5)/(2)xx300=750K`, `DeltaT=750-500=250K` |
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