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A carnot engine whose efficiency is 40 % recives heat at 500 K . If the efficiency is to be 50 %. The source temperature for the sameexhaust temperature is

Answer»

900 K
600 K
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Solution :Efficiency of Carnot engine, `b= 1- T_(2)/T_(1)`
Where ` T_(1) andT_(2) ` be the TEMPERATURES of source and sinkrespectively .
` T_(2)/T_(1) = 1 -n=1- 40/100=60/100 = 3/5 `
` T_(2) = 3/5 T_(1) = 3/5 xx 500 K = 300 K `
LET ` T_(1)`be the TEMPERATUREOF the source for the same sinktemperature.
` thereforeT_(2)/T_(1) = 1-n =1 - 50/100 = 1/2`
` T_(1) = 2T_(2) = 2xx 300 K = 600 K `


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