Explain why no real heat engine operating between two fixed temperature reservoirs can exceed the efficiency of a Carnot engine operating between the same two reservoirs.

IB DP Physics Higher Level (2023 syllabus) — B.4 Thermodynamics (HL only) · Explain · 4 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

Model answer (4 marks)

A Carnot engine is fully reversible, so all its processes are quasi‑static and no entropy is produced inside the engine. Any real engine has irreversible processes (friction, finite‑temperature heat transfer, turbulence, etc.) that generate entropy. By the second law the entropy of the universe cannot decrease, so the irreversible processes increase the total entropy. To keep the entropy balance, a real engine must reject more heat to the cold reservoir than a Carnot engine would, which means less work is produced from the same heat input. Consequently the efficiency of any real engine operating between the same two reservoirs is always lower than the Carnot efficiency.

Examiner tips

  • Use the term ‘reversible’ and ‘irreversible’ early; link irreversibility to entropy generation.
  • Show the entropy balance: ΔS_univ=ΔS_engine+ΔS_surroundings≥0, and explain how extra ΔS_engine forces Q_c larger.
  • Mention that efficiency η=1−Q_c/Q_h, so larger Q_c gives lower η.
  • Keep the answer concise – 4 marks allow a short paragraph with 3–4 sentences.

Common mistakes

  • Confusing the Carnot cycle with any idealised cycle; not stating that Carnot is reversible.
  • Failing to mention that the extra entropy must be compensated by rejecting more heat to the cold reservoir.
  • Using vague phrases like ‘real engines are less efficient’ without explaining the entropy reason.

Mark scheme (4 marks)

  1. A Carnot engine is fully reversible, meaning all processes are quasi-static and no entropy is generated within the engine itself.
  2. Any real engine contains irreversible processes (such as friction, heat losses across finite temperature differences, or turbulence) which generate entropy.
  3. By the second law of thermodynamics, entropy of an isolated system (or the universe) cannot decrease, so irreversible processes increase the total entropy.
  4. Because more heat must be rejected to the cold reservoir (to account for extra entropy produced), less work is extracted from the same heat input, so efficiency is lower than the Carnot efficiency.

Key terms in this question

Carnot engine

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