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Grade: 12

                        

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 ?

5 months ago

Answers : (2)

Arun
24737 Points
							
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
5 months ago
Vikas TU
12119 Points
							
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
5 months ago
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