Explain why the Carnot cycle is described as a reversible cycle and discuss what this implies about the total entropy change of the universe during one complete Carnot cycle.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
The Carnot cycle consists of two isothermal processes and two adiabatic processes, all carried out quasi-statically.
Model answer (4 marks)
A reversible process is carried out quasi‑statically, so the system remains in thermodynamic equilibrium and there are no dissipative effects such as friction or turbulence.
In the Carnot cycle the two isothermal steps involve heat exchange with a reservoir at the same temperature as the working substance; therefore the heat transfer is reversible and no temperature gradient drives an irreversible flow.
Because every step of the cycle is reversible, the entropy change of the universe (system plus surroundings) for each step is zero. Hence the total entropy change of the universe over one complete Carnot cycle is zero.
A zero total entropy change means the Carnot cycle reaches the maximum possible efficiency for a heat engine operating between the two temperatures; any real cycle generates entropy in the universe and therefore has a lower efficiency.
In the Carnot cycle the two isothermal steps involve heat exchange with a reservoir at the same temperature as the working substance; therefore the heat transfer is reversible and no temperature gradient drives an irreversible flow.
Because every step of the cycle is reversible, the entropy change of the universe (system plus surroundings) for each step is zero. Hence the total entropy change of the universe over one complete Carnot cycle is zero.
A zero total entropy change means the Carnot cycle reaches the maximum possible efficiency for a heat engine operating between the two temperatures; any real cycle generates entropy in the universe and therefore has a lower efficiency.
Examiner tips
- Use the word ‘reversible’ and link it to quasi‑static, equilibrium and no dissipation. Mention that heat transfer is at equal temperatures. State that ΔS_universe = 0 for each step and for the whole cycle. Explain the implication for maximum efficiency.
Common mistakes
- Confusing reversible with ideal; forgetting to mention quasi‑static. Saying entropy of the system is zero instead of entropy of the universe. Claiming the Carnot cycle is irreversible because it has two different temperatures.
Mark scheme (4 marks)
- A reversible process is one carried out quasi-statically (infinitely slowly) so the system is always in thermodynamic equilibrium, with no dissipative effects such as friction or turbulence.
- During each isothermal process, heat is exchanged reversibly with a reservoir at the same temperature as the working substance, so there is no temperature difference driving heat transfer irreversibly.
- The entropy change of the universe (system plus surroundings) during each step is zero because every process is reversible, so the total entropy change of the universe over one complete Carnot cycle is zero.
- This zero total entropy change means the Carnot cycle achieves the maximum possible efficiency for a heat engine operating between the same two temperatures; any real cycle produces entropy in the universe and therefore must have a lower efficiency.
Key terms in this question
Related
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