The Second Law, Entropy and Carnot

Heat engines, the Kelvin–Planck and Clausius statements, entropy, the Carnot limit and isentropic efficiency.

On this page

  1. Heat engines and the second law
  2. The Carnot limit
  3. Entropy
  4. Irreversibility and lost work
  5. Isentropic efficiency
  6. Reading a T–s diagram
  7. FAQ

Heat engines and the second law

The first law says energy is conserved; it does not say which way a process runs. Heat flows from hot to cold by itself, never the reverse; a stirred cup does not unstir; a flywheel slows by friction and never speeds up. The second law captures this. Two classic statements:

  • Kelvin–Planck: no device operating in a cycle can take heat from one reservoir and produce a net amount of work with no other effect. Some heat must be rejected to a colder reservoir.
  • Clausius: no device operating in a cycle can move heat from a cold body to a hot body without work input.

A heat engine takes heat QH from a hot source, produces work W and rejects QC. Its thermal efficiency is η = W ÷ QH = 1 − QC ÷ QH and is always below 100 percent. A refrigerator or heat pump is the same machine run backward, measured by its coefficient of performance (see Refrigeration and Heat Pumps).

The Carnot limit

Between two reservoirs at absolute temperatures TH and TC, no engine is more efficient than a reversible one, and all reversible engines have the same efficiency, the Carnot efficiency:

ηCarnot = 1 − TC ÷ TH

Carnot efficiency and real engines3006009001200150018000.00.20.40.60.81.0hot reservoir temperature (K), cold reservoir 300 Kthermal efficiencyGeothermal ORC 12%Steam power plant 38%Diesel engine 42%Combined cycle 60%Petrol engine 30%Carnot limit: 1 − T cold ÷ T hotTemperatures are the peak cycle temperatures (the diesel and petrol points are plotted at the axis limit). No heatengine can exceed the Carnot line for its reservoirs.
Carnot efficiency for a 300 K sink. Real plants reach about half to two thirds of the limit for their temperatures.

Worked example: how good is a 38 percent power plant?

A plant takes 100 MW of heat from steam at 800 K (527 °C) and rejects heat to a river at 300 K.

  • Carnot limit: η = 1 − 300 ÷ 800 = 62.5 percent, so at most 62.5 MW of work.
  • The real plant makes 38 MW (38 percent), i.e. 61 percent of the Carnot value, and rejects 62 MW to the river.
  • The missing work is entropy generated by friction, heat transfer across finite temperature differences and the like: Sgen = QC ÷ TC − QH ÷ TH = 207 − 125 = 82 kW/K, which is positive as the second law demands.

Entropy

Entropy s (kJ/kg·K) is a property that measures how spread out the energy is. For heat transferred reversibly at temperature T, dS = δQ ÷ T. The second law in one line: for any process, the entropy of the system plus its surroundings never decreases, ΔStotal ≥ 0, with equality only for a reversible process.

For an ideal gas with constant specific heats: s₂ − s₁ = cp ln(T₂ ÷ T₁) − R ln(p₂ ÷ p₁). An adiabatic reversible process is therefore isentropic (constant entropy), which is why p Vk = constant. For water and steam, s comes from tables.

Worked example: heat flowing from hot to cold

Let 1 000 kJ pass from a body at 400 K to one at 300 K.

  • Hot body: ΔS = −1 000 ÷ 400 = -2.500 kJ/K. Cold body: ΔS = +1 000 ÷ 300 = 3.333 kJ/K.
  • Total ΔS = +0.833 kJ/K: positive, so the process is possible and irreversible. The reverse (cold to hot) would give −0.833, forbidden.
  • Constant-pressure heating of 1 kg of air from 300 K to 600 K: Δs = 1.005 × ln 2 = 0.697 kJ/kg·K.

Irreversibility and lost work

Every real process generates entropy, and each unit of entropy generated wastes work: Wlost = T0 Sgen, where T0 is the environment temperature. In the heat-flow example above, 1 000 kJ at 400 K could have driven an engine to produce work; letting it simply conduct to 300 K destroys 250.0 kJ of that opportunity (T0 × Sgen = 300 × 0.833). This is the thermodynamic reason to avoid large temperature differences, throttling and mixing wherever you can. The maximum useful work from a stream relative to the environment is its exergy.

Isentropic efficiency

Real turbines, compressors and pumps are not reversible. Their performance is compared with the ideal isentropic device between the same pressures:

ηturbine = (h₁ − h₂actual) ÷ (h₁ − h₂s)    ηcompressor = (h₂s − h₁) ÷ (h₂actual − h₁)

Worked example: a real steam turbine and a real compressor

  • The ideal turbine of Steam Power Cycles: Rankine drops steam from h₁ = 3 213.6 to h₂s = 2 143.8 kJ/kg, a work of 1 069.8 kJ/kg. With η = 85 percent the actual work is 909.3 kJ/kg, so h₂ = 2 304.3 and the exit quality is 0.883 against 0.816 for the ideal turbine. A less efficient turbine leaves drier exhaust because the lost work reappears as heat in the steam.
  • A compressor with a pressure ratio of 6 and η = 80 percent: the ideal exit is 500.6 K, the actual 550.7 K, and the work rises from 201.6 to 251.9 kJ/kg.

Reading a T–s diagram

On a temperature–entropy diagram, the area under a path is the heat transferred (Q = ∫ T dS) and an isentropic process is a vertical line. The dome of liquid and vapor for water and the cycle’s outline show at a glance where heat enters and leaves. Steam cycles and refrigeration cycles are usually drawn this way (see Steam, Water and Property Tables). A Carnot cycle is a rectangle: two isotherms and two isentropes.

A heat pump shows the same law at work: it can deliver more heat than the work it uses, because it moves heat rather than making it. The Carnot limit for lifting heat 30 K from 5 °C to 35 °C is COP = 308 ÷ 30 = 10.3, as Refrigeration and Heat Pumps explains.

Related

FAQ

Why can’t an engine be 100 percent efficient?

Because the cycle must reject heat to a cold reservoir; otherwise the heat engine would be a perpetual motion machine of the second kind. The only reservoir you can use is the environment, so the Carnot limit is fixed by the temperatures.

Is entropy disorder?

It is a useful picture, but the engineering meaning is simpler: entropy is the property whose generation tells you how much work potential was lost.

What does isentropic mean?

Constant entropy. A reversible adiabatic process is isentropic. It is the ideal reference for a turbine, a compressor or a nozzle.

Do I always need entropy?

For first-law problems (energy balances) no. You need it for cycle limits, isentropic efficiencies and for finding exit states of turbines and compressors.

References

  • Çengel, Y. A. and Boles, M. A., Thermodynamics: An Engineering Approach, McGraw-Hill (second law, entropy, isentropic efficiencies).
  • Moran, M. J., et al., Fundamentals of Engineering Thermodynamics, Wiley.
  • Bejan, A., Advanced Engineering Thermodynamics, Wiley (exergy and entropy generation).

Last reviewed: October 2026. Efficiencies are typical.