Q.26 A Carnot engine receives 4000 J of heat from a reservoir at 500 K, does some work, rejects 2000 J heat to reservoir at 200 K.
(a) How much work does it do?
(b) What is its efficiency?
Best answer
Answer: (a) 2000 J (b) 50% from the given heats (the Carnot formula with 500 K and 200 K would give 60%)
Heat absorbed \(Q_H = 4000\) J at \(T_H = 500\) K; heat rejected \(Q_C = 2000\) J at \(T_C = 200\) K.
(a) For a complete cycle \(\Delta U = 0\), so by the first law the work done is
\[W = Q_H - Q_C = 4000 - 2000 = 2000\ \text{J}\](b) Efficiency
\[\eta = \frac{W}{Q_H} = 1 - \frac{Q_C}{Q_H} = 1 - \frac{2000}{4000} = 0.5 = 50\%\]Note: the question calls it a Carnot engine. For an ideal (reversible) Carnot engine \(\dfrac{Q_C}{Q_H} = \dfrac{T_C}{T_H}\), so \(\eta = 1 - \dfrac{T_C}{T_H} = 1 - \dfrac{200}{500} = 0.6 = 60\%\), and it would reject only \(4000\times\dfrac{200}{500} = 1600\) J. The printed heat values (2000 J rejected) do not match the reservoir temperatures, so the data are not self-consistent. The efficiency from the given heats is 50%; 60% is the maximum (Carnot) efficiency possible between 500 K and 200 K. Writing both, with this explanation, is the safe exam answer.
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