The Carnot cycle is an ideal cycle in thermodynamics consisting of two reversible isothermal processes and two reversible adiabatic processes. When operated as a heat engine, it absorbs heat from a hot reservoir, converts part of that input into work, and rejects the remainder to a cold reservoir. Its efficiency is the greatest possible for an engine exchanging heat only with those two reservoirs. Named after Sadi Carnot, who proposed the underlying model in 1824, it provides a theoretical benchmark rather than a practical machine design. (openstax.org)
Physical assumptions
The cycle operates between reservoirs at constant absolute temperatures (T_H) and (T_C), with (T_H>T_C>0). A reservoir can supply or receive heat without appreciably changing temperature. The working substance returns to its initial state after every complete cycle, allowing the operation to be repeated. (openstax.org)
Each stage is a reversible process: the system and its surroundings can, in principle, be restored to their initial states without residual changes. This excludes friction, uncontrolled expansion, and heat transfer across finite temperature differences. During reversible heat exchange, the working substance and reservoir differ in temperature only infinitesimally. The idealization therefore describes a limiting process, not ordinary finite-rate operation. (ocw.mit.edu)
The four stages
A gas enclosed by a movable piston provides a standard illustration. Starting at the hot temperature, the engine follows four stages:
- Isothermal expansion. The gas remains in contact with the hot reservoir at (T_H), absorbs heat (Q_H), and expands while doing work. This is an isothermal process.
- Adiabatic expansion. Thermal contact is removed, and the gas continues expanding without heat exchange. Work is supplied at the expense of its internal energy, lowering its temperature to (T_C). This is an adiabatic process.
- Isothermal compression. At (T_C), external work compresses the gas while it rejects heat (Q_C) to the cold reservoir.
- Adiabatic compression. With the gas thermally isolated again, further compression raises its temperature to (T_H) and restores the initial state. (openstax.org)
For an ideal gas, internal energy depends only on temperature, so its change vanishes during either isothermal stage. The adiabatic stages instead change internal energy without transferring heat. These gas-specific properties illustrate the cycle; the Carnot efficiency itself does not require an ideal-gas working substance. (ocw.mit.edu)
Work and efficiency
Let (Q_H) and (Q_C) denote positive magnitudes of heat absorbed and rejected per cycle. Because internal energy is a state function, its net change over a complete cycle is zero. The first law of thermodynamics therefore gives
[ W_{\mathrm{net}}=Q_H-Q_C, \qquad \eta=\frac{W_{\mathrm{net}}}{Q_H} =1-\frac{Q_C}{Q_H}. ]
For the reversible Carnot cycle,
[ \frac{Q_C}{Q_H}=\frac{T_C}{T_H}, \qquad \boxed{\eta_C=1-\frac{T_C}{T_H}}. ]
Temperatures must be expressed on an absolute scale, ordinarily in kelvins, rather than as Celsius or Fahrenheit readings. Efficiency depends on the temperature ratio, not simply their difference. Raising (T_H) while holding (T_C) fixed, or lowering (T_C) while holding (T_H) fixed, increases the theoretical limit. (ocw.mit.edu)
For example, reservoirs at 1,000 K and 300 K give a Carnot efficiency of (0.70), or 70%. Even this ideal engine must reject 30% of its heat input rather than convert it into work. (ocw.mit.edu)
Entropy and graphical representation
Entropy provides a working-substance-independent derivation. For reversible heat transfer, (dS=\delta Q_{\mathrm{rev}}/T). The hot isotherm increases the working substance’s entropy by (Q_H/T_H), while the cold isotherm decreases it by (Q_C/T_C). Both reversible adiabatic stages have constant entropy. Since the substance returns to its initial state,
[ \frac{Q_H}{T_H}-\frac{Q_C}{T_C}=0. ]
This directly establishes the heat-to-temperature ratio used in the efficiency formula. (openstax.org)
A gas cycle appears as a clockwise loop on a pressure–volume diagram, with the enclosed area representing net work output. On a temperature–entropy diagram it forms a rectangle: the isotherms are horizontal and the constant-entropy paths vertical. If their entropy separation is (\Delta S), its area is ((T_H-T_C)\Delta S=W_{\mathrm{net}}). (openstax.org)
Carnot’s theorem and irreversibility
Carnot’s theorem states that no engine operating between two fixed-temperature reservoirs can exceed the efficiency of a reversible engine, and that all reversible engines between those reservoirs have equal efficiency. It is a consequence of the second law of thermodynamics; exceeding the limit would permit a combined engine-and-refrigerator arrangement that violates that law. (openstax.org)
The working substance has zero net entropy change in any complete cycle, including an irreversible one. What distinguishes reversible operation is zero total entropy generation in the system and surroundings. For an irreversible two-reservoir engine,
[ S_{\mathrm{gen}}=\frac{Q_C}{T_C}-\frac{Q_H}{T_H}>0, ]
which implies (\eta<\eta_C). Thus, heat rejection is necessary even without friction; irreversibility imposes an additional efficiency penalty. (openstax.org)
Reversed operation
Running the cycle backward requires work input and transfers heat from the cold reservoir to the hot one. It then operates as a refrigerator or heat pump, depending on whether cooling or heating is the intended output. Their maximum coefficients of performance are
[ \mathrm{COP}{R}=\frac{T_C}{T_H-T_C}, \qquad \mathrm{COP}{HP}=\frac{T_H}{T_H-T_C}. ]
A coefficient greater than one does not violate conservation of energy: the device moves heat as well as supplying energy through work. The hot-side heat delivery equals the cold-side heat extraction plus work input. (openstax.org)