Carnot efficiency from the second law alone
Statement
No heat engine operating between two reservoirs can exceed a maximum efficiency that depends only on their temperatures — and not at all on the working substance or the engine design.
Why it matters
This is the rare physical result that constrains all possible future technology. It is also the cleanest example in physics of a bound derived without a model: no assumption whatsoever is made about what the engine contains or how it works.
Assumptions
Derivation
Result
Reading. Equality holds if and only if the cycle is reversible. Nothing about the working substance appears — steam, air, or a single atom, the bound is identical. It also shows that η = 1 requires TC = 0, which is the Kelvin statement of the third law's practical consequence.
Units check. A ratio of absolute temperatures; dimensionless. Note that the result is false with Celsius, a useful reminder that thermodynamic temperature is not an arbitrary scale.
Limiting cases
- TC → TH: the bound → 0. No temperature difference, no available work.
- TC → 0: the bound → 1. Perfect efficiency requires an unreachable absolute-zero sink.
- Reversible limit: the inequality becomes equality, defining the ideal Carnot cycle at zero power.
Breaks when
- The reservoirs are finite and their temperatures drift during operation.
- Non-thermal energy sources are involved — a fuel cell converts chemical energy directly and is not bounded by Carnot at all, a point widely misunderstood.
- The working system is small enough for fluctuations to matter — individual cycles of a microscopic engine can transiently exceed the bound, though the average cannot.
Failure modes
- Using Celsius. The single most common error in the topic and it produces nonsense including efficiencies above one.
- Believing Carnot efficiency is a design target. It is achieved only by an infinitely slow cycle, which produces zero power.
- Applying the bound to devices that are not heat engines. Heat pumps have coefficients of performance exceeding one and violate nothing.
The Curzon–Ahlborn bound for maximum power rather than maximum efficiency gives 1 − √(TC ⁄ TH), a different and generally tighter benchmark. A bound is only as meaningful as its stated conditions.
Worked number
A modern combined-cycle gas turbine runs at roughly TH = 1,700 K with TC = 300 K, giving a Carnot bound of 82%. Real plants achieve about 62%. The gap is not incompetence — it is irreversibility: finite-rate heat transfer, friction, and turbulence. The Curzon–Ahlborn power bound gives 1 − √(300/1700) = 58%, which real plants now exceed slightly.
Run the check →Discussion
The striking feature of this result is what is absent from it. The efficiency bound contains only two temperatures and no property of the engine at all — no heat capacity, no equation of state, no mention of pistons or turbines or working fluid. That silence is the whole point. The derivation used exactly one physical law, the second law in the form ΔStotal ≥ 0, and one bookkeeping fact, that a cyclic device returns to its own initial entropy. Everything specific to a particular machine cancels because it lives inside ΔSengine, which is forced to zero over a cycle. What survives is a statement purely about the two reservoirs.
The structure η ≤ 1 − TC/TH is best read as a constraint on waste, not on work. Step 4 bounds QC from below: an engine is obliged to dump at least QH TC/TH of heat into the cold reservoir simply to carry away the entropy it received from the hot one. Entropy enters the engine at rate QH/TH and must leave at rate at least that much; because it can only leave at the lower temperature TC, it drags a minimum quantity of energy out with it. The unavailable energy TC ΔS is not lost to error — it is the toll charged for exporting entropy to a finite-temperature sink.
Equality holds if and only if ΔStotal = 0, i.e. the cycle is reversible, and this is the hinge that connects the result to the rest of thermodynamics. Because the reversible bound is independent of substance, one can turn the logic around and define thermodynamic temperature by the reversible heat ratio, QC/QH = TC/TH — this is Kelvin's absolute scale, fixed by nothing but a reversible engine and one reference point. Running the same argument backwards drives the refrigerator and heat-pump bounds (COPref ≤ TC/(TH−TC)), and running it differentially over a varying reservoir reproduces the Clausius inequality ∮ dQ/T ≤ 0. Carnot's result is the second law wearing its most practical face.
Statistical mechanics sharpens rather than replaces this picture. The equality QC/TC = QH/TH is the macroscopic shadow of entropy being a state function of the microstate distribution; the second-law inequality is the statement that a spontaneously chosen path almost never decreases the number of accessible microstates. This is why the “breaks when” note about microscopic engines is not a loophole but a consequence: for a system of a few degrees of freedom the fluctuation theorems (Jarzynski, Crooks) make ΔStotal < 0 an event of small but nonzero probability, so a single cycle can transiently beat Carnot while the ensemble average ⟨ΔStotal⟩ ≥ 0 is preserved exactly. The bound is a statement about expectation values, and it becomes sharp precisely in the thermodynamic limit where relative fluctuations vanish.
Common misconceptions. (i) Carnot efficiency is not a design target — it is reached only by a reversible, therefore infinitely slow, cycle that delivers zero power; for maximum power the relevant figure is the Curzon–Ahlborn value 1−√(TC/TH). (ii) The bound applies only to heat engines cycling between reservoirs; a fuel cell converts chemical free energy directly and is not capped by Carnot, and a heat pump legitimately has a coefficient of performance far above one. (iii) The temperatures must be absolute — using Celsius does not merely shift the answer, it can produce efficiencies above unity, which is the clearest possible signal that thermodynamic temperature is a genuine physical scale and not a convention.
Worked examples
Example 1 — A reversible engine, heat and work per cycle. A Carnot engine works between superheated steam at TH = 550 K and a condenser at TC = 300 K, absorbing QH = 2000 J of heat per cycle. Find its efficiency, the work delivered, and the heat rejected, and verify the cycle is reversible.
Answer. The engine converts 909 J of the 2000 J intake into work each cycle and rejects 1091 J to the condenser; the equal entropy flows confirm reversibility.
Example 2 — An irreversible engine: entropy generated and lost work. A real engine absorbs QH = 1500 J from a reservoir at TH = 800 K and is measured to deliver W = 600 J per cycle, rejecting the rest to the environment at TC = 320 K. Find the entropy generated per cycle and the work lost to irreversibility.
Answer. The engine generates 0.94 J/K of entropy per cycle and thereby destroys 300 J of otherwise-available work — exactly the gap between its 600 J output and the 900 J a reversible engine would deliver from the same heat intake.
Problems
- A Carnot engine operates between reservoirs at 127 °C and 27 °C. What is its maximum efficiency? (Watch the temperature scale.)
Solution
Convert to absolute temperature first: TH = 127 + 273 = 400 K, TC = 27 + 273 = 300 K. Then ηmax = 1 − 300/400 = 0.25 = 25%. (Using Celsius directly would give the meaningless 1 − 27/127 = 79%, illustrating why absolute temperature is essential.)
- An inventor claims an engine that is 70% efficient while rejecting heat to the ambient environment at 300 K. What is the minimum hot-reservoir temperature this would require, and is the claim plausible?
Solution
Set the claimed efficiency equal to the Carnot ceiling (the most generous case): 0.70 = 1 − 300/TH, so TH = 300/(1 − 0.70) = 300/0.30 = 1000 K. The engine would need a source of at least 1000 K and would additionally have to be essentially reversible (zero power). Achieving 70% at any finite power from a 1000 K source is not plausible.
- A power station delivers 500 MW of electrical power using a source at 850 K and a cooling sink at 290 K, running at 65% of the Carnot efficiency. Find the rate of heat intake from the source and the rate of waste-heat rejection.
Solution
ηCarnot = 1 − 290/850 = 0.6588. Actual η = 0.65 × 0.6588 = 0.4282. Heat intake rate Q̇H = Ẇ/η = 500/0.4282 = 1168 MW. Waste rate Q̇C = Q̇H − Ẇ = 1168 − 500 = 668 MW. (Note the plant sheds more energy as waste heat than it delivers as electricity — typical of thermal generation.)
- An engineer reports an engine that absorbs 3000 J at 500 K and rejects 2000 J at 250 K each cycle. Is this consistent with the second law? If so, how much entropy does it generate per cycle and how much work is lost relative to a reversible engine?
Solution
Work per cycle W = 3000 − 2000 = 1000 J, so η = 33.3%. Carnot ceiling ηC = 1 − 250/500 = 50%, so the engine runs below the bound — permissible. Entropy generated: ΔStotal = −3000/500 + 2000/250 = −6 + 8 = +2 J/K > 0, consistent with the second law. A reversible engine would give Wmax = 0.50 × 3000 = 1500 J, so lost work = 1500 − 1000 = 500 J, equal to TC ΔStotal = 250 × 2 = 500 J. ✓
- A finite hot body of heat capacity C = 5000 J/K, initially at TH = 600 K, is used to drive a sequence of infinitesimal reversible engines rejecting to a large cold reservoir at TC = 300 K, until the body has cooled to TC. Derive the maximum work extractable and evaluate it. (This is the “breaks when” case of a drifting reservoir.)
Solution
Because every stage is reversible, total entropy is unchanged: the body's entropy fall must be matched by the cold reservoir's rise. Body: ΔSbody = ∫THTC C dT/T = C ln(TC/TH). Reversibility requires QC/TC = −ΔSbody, so QC = C TC ln(TH/TC). The heat leaving the body is QH = C(TH − TC). Hence
Wmax = QH − QC = C(TH − TC) − C TC ln(TH/TC)Numerically: C(TH−TC) = 5000 × 300 = 1.500 × 106 J; C TC ln(2) = 5000 × 300 × 0.6931 = 1.040 × 106 J. Therefore Wmax = 4.60 × 105 J ≈ 460 kJ. Note this is far below the 1.5 MJ of heat released, because the body's average temperature (and hence its instantaneous Carnot efficiency) falls continuously as it cools — the fixed-reservoir bound of the derivation no longer applies.