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a carnot engine whose sink is at 90 k has an efficiency of 30%. by how much the teperature of the source be increased to have its efficiency equal to 50%, keeping sink teperature constant

a carnot engine whose sink is at 90 k has an efficiency of 30%. by how much the teperature of the source be increased to have its efficiency equal to 50%, keeping sink teperature constant
   

Grade:11

2 Answers

Khimraj
3007 Points
5 years ago
efficiency
\eta = 1 – Tsink/Tsource
\eta +1 = Tsink/Tsource
As Tsource is constant
(\eta2 +1)/(\eta1​ +1) = Tsink2/Tsink1
1.6/1.3 = x/90
x = 110.76
%increase in teperature of the source = 20.76*100/90 = 23%
Hope it clears.....................................
 
Aastha Singh
askIITians Faculty 212 Points
5 years ago
DEAR STUDENT ,
KINDLY REFER TO THE IMAGE BELOW
579-2489_IMG_20190207_180100.jpg

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