Questions
An ideal heat engine is exhausting heat energy to a reservoir at . If the initial efficiency of 50% is decreased to 40% by changing the temperature of sink, the new temperature of sink is
detailed solution
Correct option is B
efficiency of heat engine=η=1-T2T1---(1)T1,T2 are temperatures of source and sinkgiven T2=27+273=300K;efficiency=50%=50100η=1−T2T1⇒50100=1−300T1⇒12=300T1⇒T1=600K---(2)new temperature of sink=T21; new efficiency=40%=40100;and eqn(2) substitute in following equation,η1=1−T12T140100=1−T21600⇒T21600=1−25=35 ⇒T21600=1−25T21600=35T21=35X600=360K.Talk to our academic expert!
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