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 ?

Question

Arun · Answer

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

Vikas TU · Answer

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' = 0.44
T2' = 280K
Temp of sink must be decreased by 300-280 = 20K

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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 ?

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 ?

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