The Carnot Cycle

Imagine a steam engine that could never be beaten, no matter how clever the engineering. That is exactly what Sadi Carnot imagined in 1824, decades before anyone had a proper concept of entropy. He asked a purely theoretical question. What is the best possible efficiency for turning heat into work? The answer turns out to depend on nothing but temperature. This is one of the most surprising results in classical physics. It applies equally to steam, gas, or any working substance, provided every step is reversible. This article builds the Carnot cycle step by step. You will see all four stages on a pressure-volume diagram. Then we derive the efficiency formula from first principles using the ideal gas law. Finally, we connect the result to the thermodynamic temperature scale and Carnot's theorem.

Key takeaways

The Carnot cycle is an idealized four-step engine cycle: isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression.
Only two reservoirs are used, a hot one at $T_h$ and a cold one at $T_c$ , and every step is reversible.
The efficiency depends only on the two temperatures.

\eta = 1 - \frac{T_c}{T_h}

Efficiency is always less than 1 because $T_c > 0$ can never be reached and $T_h$ is finite.
No engine operating between the same two reservoirs can exceed this efficiency (Carnot's theorem, covered next).
Key idea: because entropy change is zero over the full cycle, the heat exchanged at each reservoir must scale exactly with that reservoir's temperature, forcing $\eta$ to depend only on $T_h$ and $T_c$ .

The core insight

The Carnot cycle is a theoretical construction, not a real engine design, whose entire purpose is to show that reversibility plus two fixed reservoir temperatures fixes the maximum possible fraction of absorbed heat that can be converted into work, and that this maximum, $\eta = 1 - T_c/T_h$ , is independent of the working substance.

The physical picture

Picture a cylinder of ideal gas fitted with a frictionless, weightless piston, in thermal contact with either a hot reservoir at temperature $T_h$ or a cold reservoir at temperature $T_c$ , or thermally isolated from both. The gas undergoes four reversible steps in sequence, returning exactly to its starting state so the cycle can repeat indefinitely.

Step 1 — Isothermal expansion at $T_h$ . The gas sits in contact with the hot reservoir and expands slowly from volume $V_1$ to $V_2$ . Because the expansion is reversible and isothermal, the internal energy of the ideal gas does not change, so all the heat absorbed, $q_h$ , is converted directly into work done by the gas.

Step 2 — Adiabatic expansion. The gas is isolated from any reservoir and allowed to expand further, from $V_2$ to $V_3$ . No heat enters or leaves. The gas does work at the expense of its own internal energy, so its temperature falls from $T_h$ to $T_c$ .

Step 3 — Isothermal compression at $T_c$ . The gas, now in contact with the cold reservoir, is compressed from $V_3$ to $V_4$ . Work is done on the gas, and it releases heat $q_c$ to the cold reservoir to keep its temperature constant.

Step 4 — Adiabatic compression. The gas is isolated again and compressed from $V_4$ back to $V_1$ , raising its temperature from $T_c$ back to $T_h$ , closing the cycle exactly. The net effect over one full cycle: heat $q_h$ is absorbed at high temperature, heat $|q_c|$ is rejected at low temperature, and the difference emerges as net work done on the surroundings. The area enclosed by the four curves on a pressure-volume diagram equals that net work.

Figure

The Carnot cycle on a pressure-volume diagram: two isotherms bridged by two adiabats

Figure

Energy flow: heat in at T_h, heat out at T_c, work delivered

The core relations

\eta = \frac{w_{\text{net}}}{q_h} = 1 - \frac{|q_c|}{q_h}

where $w_{\text{net}}$ is the net work delivered per cycle, $q_h$ the heat absorbed from the hot reservoir, and $|q_c|$ the heat rejected to the cold reservoir.

\frac{|q_c|}{q_h} = \frac{T_c}{T_h}

where $T_h$ and $T_c$ are the absolute (thermodynamic) temperatures of the hot and cold reservoirs.

\eta_{\text{Carnot}} = 1 - \frac{T_c}{T_h}

where this is the maximum possible efficiency for any engine operating reversibly between these two temperatures.

w_{\text{max}} = q_h\left(1 - \frac{T_c}{T_h}\right)

where $w_{\text{max}}$ is the greatest work extractable from heat $q_h$ drawn from a reservoir at $T_h$ , when waste heat is dumped at $T_c$ .

Derivation AFrom the ideal-gas cycle to η = 1 − T_c/T_h

Goal. Compute the work and heat exchanged in each of the four steps for one mole of ideal gas, and combine them into the efficiency, showing it reduces to a function of temperature alone.

Step 1. Isothermal expansion at $T_h$ , from $V_1$ to $V_2$ . For an ideal gas, $\Delta U = 0$ on an isotherm, so $q_h = -w_1$ , and reversible work is

w_1 = -nRT_h\ln\frac{V_2}{V_1}, \qquad q_h = nRT_h\ln\frac{V_2}{V_1}.

The gas absorbs heat exactly equal to the work it performs, since $\ln(V_2/V_1) > 0$ .

Step 2. Adiabatic expansion, from $V_2, T_h$ to $V_3, T_c$ . No heat is exchanged, $q=0$ , and for a reversible adiabat with heat capacity ratio consistent with $C_{V,m}$ ,

T_h V_2^{\gamma-1} = T_c V_3^{\gamma-1}.

This fixes the relationship between $V_2$ and $V_3$ once $T_h$ and $T_c$ are chosen.

Step 3. Isothermal compression at $T_c$ , from $V_3$ to $V_4$ . By the same logic as Step 1, with $V_4 < V_3$ ,

q_c = nRT_c\ln\frac{V_4}{V_3}, \qquad w_3 = -q_c.

Because $V_4 < V_3$ , the logarithm is negative, so $q_c < 0$ : heat is released, not absorbed.

Step 4. Adiabatic compression, from $V_4, T_c$ back to $V_1, T_h$ . Again $q = 0$ , and the adiabat condition gives

T_c V_4^{\gamma-1} = T_h V_1^{\gamma-1}.

Step 5. Combine the two adiabat relations. Dividing the Step 2 relation by the Step 4 relation eliminates the temperatures:

\frac{V_2^{\gamma-1}}{V_1^{\gamma-1}} = \frac{V_3^{\gamma-1}}{V_4^{\gamma-1}} \quad\Longrightarrow\quad \frac{V_2}{V_1} = \frac{V_3}{V_4}.

This is the crucial geometric fact: the volume ratio of the hot isotherm equals the volume ratio of the cold isotherm.

Step 6. Form the ratio of heats. Using the Step 1 and Step 3 results together with $V_2/V_1 = V_3/V_4$ ,

\frac{q_c}{q_h} = \frac{nRT_c\ln(V_4/V_3)}{nRT_h\ln(V_2/V_1)} = \frac{-T_c\ln(V_3/V_4)}{T_h\ln(V_2/V_1)} = -\frac{T_c}{T_h}.

Taking magnitudes, $|q_c|/q_h = T_c/T_h$ exactly — every reference to $V$ has cancelled.

Step 7. Assemble the efficiency. Since the internal energy returns to its initial value after a full cycle, $\Delta U_{\text{cycle}} = 0$ , so by the first law

w_{\text{net}} = q_h + q_c = q_h - |q_c|.

Dividing by $q_h$ and substituting the Step 6 result,

\eta = \frac{w_{\text{net}}}{q_h} = 1 - \frac{|q_c|}{q_h} = 1 - \frac{T_c}{T_h}.

\boxed{\eta_{\text{Carnot}} = 1 - \frac{T_c}{T_h}}

Worked examples

Example 1 — A textbook Carnot engine. A reversible engine operates between a hot reservoir at $T_h = 500\text{ K}$ and a cold reservoir at $T_c = 300\text{ K}$ . On each cycle it draws $q_h = 1200\text{ J}$ from the hot reservoir. Step 1: efficiency.

\eta = 1 - \frac{300\text{ K}}{500\text{ K}} = 1 - 0.600 = 0.400.

Step 2: net work per cycle.

w_{\text{net}} = \eta q_h = 0.400 \times 1200\text{ J} = 480\text{ J}.

Step 3: heat rejected to the cold reservoir.

|q_c| = q_h - w_{\text{net}} = 1200\text{ J} - 480\text{ J} = 720\text{ J}.

Check: $|q_c|/q_h = 720/1200 = 0.600 = T_c/T_h$ , exactly as the derivation requires. The engine converts 40% of the heat drawn from the hot side into usable work; the rest, 720 J, is unavoidably dumped into the cold reservoir. No amount of clever engineering with any real gas or working substance can push this beyond 40% while operating between these same two temperatures.

Example 2 — Sizing a power plant against a river-cooled condenser. A proposed power plant burns fuel to maintain a boiler at $T_h = 823\text{ K}$ ( $550\,^\circ\text{C}$ , typical of superheated steam) and rejects waste heat to river water that keeps the condenser at $T_c = 313\text{ K}$ ( $40\,^\circ\text{C}$ ). Suppose the plant must deliver $w_{\text{net}} = 750\text{ MW}$ of electrical output, and operates at 85% of the Carnot limit due to real irreversibilities (friction, finite-rate heat transfer, generator losses). Step 1: the Carnot (ideal) efficiency ceiling.

\eta_{\text{Carnot}} = 1 - \frac{313\text{ K}}{823\text{ K}} = 1 - 0.3803 = 0.6197.

Step 2: the realistic efficiency at 85% of that ceiling.

\eta_{\text{real}} = 0.85 \times 0.6197 = 0.5267.

Step 3: required heat input rate from fuel.

\dot{q}_h = \frac{w_{\text{net}}}{\eta_{\text{real}}} = \frac{750\text{ MW}}{0.5267} = 1424\text{ MW}.

Step 4: waste heat rejected to the river per unit time.

\dot{q}_c = \dot{q}_h - w_{\text{net}} = 1424\text{ MW} - 750\text{ MW} = 674\text{ MW}.

Sanity check: $\eta_{\text{real}} = 750/1424 = 0.527$ , consistent with Step 2. Notice that even the theoretical Carnot ceiling here is only 62%, meaning at least 38% of the fuel's heat is fundamentally unrecoverable as work — a consequence of finite $T_c$ , not of engineering imperfection. Real plants, burdened further by irreversibility, typically land near 40–45% thermal efficiency, consistent with the 52.7% figure above being on the optimistic side for a single-stage cycle.

Limiting cases & common traps

Situation	What happens	Why
$T_c \to 0\text{ K}$	$\eta \to 1$	Third law forbids reaching absolute zero, so this limit is never physically realized
$T_h \to T_c$	$\eta \to 0$	No temperature difference means no work can be extracted
Using Celsius instead of kelvin	Wrong, often nonsensical, answer	The formula requires absolute thermodynamic temperature; a ratio of Celsius values has no physical meaning
Treating $\eta$ as achievable by real engines	Overestimate of real performance	Friction, finite-rate heat transfer, and non-quasistatic steps always reduce real efficiency below $\eta_{\text{Carnot}}$
Forgetting $q_c$ is negative by sign convention	Sign errors in the first-law bookkeeping	$q_c$ is heat leaving the gas, so it is negative; $\|q_c\|$ is the magnitude rejected to the reservoir

The chemist behind it

Sadi Carnot published his single major work, Réflexions sur la puissance motrice du feu (Reflections on the Motive Power of Fire), in 1824, at a time when the caloric theory of heat, treating heat as a conserved fluid, still dominated scientific thinking. Remarkably, Carnot's central result about maximum efficiency survives intact even though his underlying caloric picture was later overturned. His genius was to reason about the engine abstractly, asking what any heat engine could achieve in principle, rather than analyzing one particular design. Rudolf Clausius and William Thomson (Lord Kelvin) later reworked Carnot's ideas using the energy-conservation framework of the first law, and Kelvin used the Carnot cycle to define an absolute thermodynamic temperature scale independent of any particular thermometric substance — the bridge to the next topic in this sequence.

Origins · 1 minNicolas Léonard Sadi Carnot and how this was discovered

Nicolas Léonard Sadi Carnot

1796-1832 · French military engineer whose sole publication founded the theoretical study of heat engines and set the stage for the second law of thermodynamics

Ask the tutor: How does Carnot's theorem prove that no engine can beat this efficiency, even in principle?

Can a real engine ever reach the Carnot efficiency

No. The Carnot cycle requires every step to be reversible, meaning infinitely slow and free of friction, turbulence, or finite-rate heat transfer across a temperature gap. Real engines operate at finite speed to deliver finite power, which inherently introduces irreversibility. Consequently every real heat engine has efficiency strictly less than $\eta_{\text{Carnot}} = 1 - T_c/T_h$ for the same reservoir temperatures. The Carnot value is a theoretical ceiling, useful for judging how close real designs come to the best possible performance, not a target that can be met exactly.

Does the working substance matter for the Carnot efficiency formula

No, and that is the deep surprise of Carnot's result. Although the derivation above uses an ideal gas for concrete algebra, Carnot's theorem (the next topic) proves using the second law that any reversible engine, regardless of working substance, whether gas, steam, or something more exotic, operating between the same two reservoirs must have exactly the same efficiency. If it did not, you could couple a more efficient engine to a less efficient one running in reverse and build a perpetual motion machine of the second kind, violating the second law.

Why does the Carnot cycle use exactly two reservoirs rather than a continuous range

Restricting the cycle to two constant-temperature reservoirs, connected by adiabatic steps in between, is what makes the heat exchanges cleanly proportional to a single temperature at each stage. If heat were exchanged over a continuously varying temperature range, computing the reversible heat transfer would require integrating over the full temperature profile. The two-reservoir design isolates the essential physics, and generalizes cleanly into the definition of entropy, since $q_h/T_h + q_c/T_c = 0$ for the Carnot cycle.

How does the Carnot cycle relate to entropy

Every step in the Carnot cycle is reversible, so the total entropy change of the universe over one full cycle is zero. Working through the heats found above, $q_h/T_h = nR\ln(V_2/V_1)$ and $q_c/T_c = nR\ln(V_4/V_3)$ , and because $V_2/V_1 = V_3/V_4$ , these two terms exactly cancel. This gives $q_h/T_h + q_c/T_c = 0$ , which is precisely the special case of the Clausius inequality that defines entropy as a state function. It previews the formal statement of the second law.

What if the cycle ran in reverse

Running the Carnot cycle backward turns it into an idealized refrigerator or heat pump: work is now supplied to the system, heat is absorbed from the cold reservoir, and a larger quantity of heat is delivered to the hot reservoir. The coefficient of performance for refrigeration becomes $T_c/(T_h - T_c)$ , and for a heat pump $T_h/(T_h - T_c)$ , both derived from the same reversible heat and work relations used for the forward engine. This reversibility is exactly why Carnot's cycle, uniquely among cycles, sets a two-way bound on both engines and refrigerators.