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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
efficiency = 1 – Tsink/Tsource +1 = Tsink/TsourceAs Tsource is constant(2 +1)/(1 +1) = Tsink2/Tsink11.6/1.3 = x/90x = 110.76%increase in teperature of the source = 20.76*100/90 = 23%Hope it clears.....................................
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