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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

Khimraj
3007 Points
2 years ago
efficiency
$\eta$ = 1 – Tsink/Tsource
$\eta$ +1 = Tsink/Tsource
As Tsource is constant
($\eta$2 +1)/($\eta$1​ +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