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Temperatures of the source and sink between which an ideal heat engine is working are x and y respectively (Kelvin)
Its efficiency is n.
What is its new efficiency if the source temperature is tripled and sink temperature is doubled?
a)(2n-1)/3
b)(2n+1)/3
c)(3n+1) /2
d)(n+2)/3
Plz explain..
We know that n = 1 - y/x => n-1 = -y/x ---(1)
similarly, let n' be the new efficiency, and n' = 1 - y'/x' whare y' = 2y and x' = 3x => n' = 1- 2y/3x => n'-1 = -2y/3x ---(2)
(1) / (2) => n-1/n'-1 = 3/2 => n' = (2n + 1)/3 ,therefore option is (b)
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