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Grade: 11
        
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
   
8 months ago

Answers : (2)

Khimraj
3008 Points
							
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.....................................
 
8 months ago
Aastha Singh
askIITians Faculty
204 Points
							
DEAR STUDENT ,
KINDLY REFER TO THE IMAGE BELOW
579-2489_IMG_20190207_180100.jpg
7 months ago
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  • Discussion Forum
  • Previous Year Exam Questions


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