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A carnot engine whose heat sink is at 27 degree celsius has an efficiency of 40%.If it is desired to increase the efficiency by 20%. By how much degree the temperature of heat sink has to be changed ?
Dear student Vhange is same in celsius and Kelvin scale as it ia difference. 1 - 300/T1 = 40/100 T1 = 500K = 0.4 + 0.04 = 0.44 T1' = 300 /0.56 = 535.7 K Hence change is = 535.7 - 500 = 35.7
Dear student T2 = 27 = 300K n = 40%n = 1-(T2/T1)T1 = 500K n'= 40% + 10% of 40% = 44%T2' be temp of sink apply formula n' = 1-(T2'/T1)T1 = 500K , n' = 0.44T2' = 280KTemp of sink must be decreased by 300-280 = 20K
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