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

Question

Khimraj · Answer

efficiency = 1 – T sink /T source +1 = T sink /T source As T source is constant ( 2 +1)/( 1 ​ +1) = T sink2 /T sink1 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 · Answer

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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

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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

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