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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 constant, the source temperature must increased by |
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Answer» SOLUTION :`T_(2)/T_(1)=1-eta=1-40/100=3/5` `rArrT_(1)=5/3T_(2) rArrT_(1)=5/3xx300=500K` NEW EFFICIENCY `eta=60%` `T_(2)/T_(1)=1-eta^(.)=1-60/100=2/5` `T_(1)^(.)=5/2xx300=750K, DeltaT=750-500=250K` |
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