In a carnot engine with an efficiency of 40%, the heat rejected to the sink at 27 C is 60 kJ, then the heat supplied in kJ to the engine by the sources is
Answer / Ali Murtaza
First, let's find the heat rejected (Qr) and heat absorbed (Qa) using the efficiency formula: Efficiency = Qa / (Qa + Qr). Since efficiency is given as 40%, we can solve for Qa: Qa = (Efficiency x (Qa + Qr)) = (0.4 x (60 + Qa)) => Qa = 90 kJnThen, the heat supplied by sources (Qs) would be the sum of the heat absorbed and rejected: Qs = Qa + Qr = 90 + 60 = 150 kJ
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