Carnot's Theorem and Engine Efficiency

A power plant burns coal at 800 K and rejects waste heat to a river at 300 K. No matter how cleverly it is engineered, no one will ever build one that converts more than 62.5% of the heat into work. This is not an engineering limitation. It is a law of nature, and Carnot's theorem is the proof. Carnot's theorem answers a deceptively simple question: how efficient can a heat engine possibly be? The answer matters because it sets an absolute ceiling on the performance of every engine, refrigerator, and heat pump ever conceived. It also hands us, almost as a byproduct, the definition of a temperature scale that does not depend on the properties of any particular substance. This article proves Carnot's theorem from the Kelvin and Clausius statements of the second law. You will see why reversibility is the key to maximum efficiency. Then you will see why all reversible engines sharing two heat reservoirs must be equally efficient. Finally you will see how this result opens the door to the thermodynamic temperature scale and the Clausius inequality.

Key takeaways

Carnot's theorem states that no engine operating between two fixed temperatures can exceed the efficiency of a reversible engine operating between the same two temperatures.
The proof is a proof by contradiction: assume a super-efficient engine exists, use it to drive a reversible engine in reverse, and show this violates the Clausius statement of the second law.
All reversible engines operating between the same two reservoirs have identical efficiency, regardless of the working substance or cycle details.
This efficiency depends only on the two reservoir temperatures, which is the seed of the thermodynamic (Kelvin) temperature scale.
Real engines are always irreversible, so their efficiency is always strictly less than the Carnot limit.
Key idea: Efficiency is capped by temperature ratios alone, because any engine that beat this cap could power a heat pump that moves heat uphill with no external work, violating the second law.

The core insight

If an engine could ever exceed the Carnot efficiency between two reservoirs, it could be coupled to a reversible refrigerator to produce a net flow of heat from cold to hot with no other effect anywhere in the universe — impossible by the Clausius statement — so no such engine can exist, and reversible engines between the same two temperatures are the unbeatable, unique-in-value benchmark.

The physical picture

Picture two heat reservoirs, a hot one at temperature $T_h$ and a cold one at temperature $T_c$ , with $T_h > T_c$ . A heat engine is any device that operates in a cycle, absorbs heat $q_h$ from the hot reservoir, delivers work $w$ to the surroundings, and dumps the leftover heat $q_c$ into the cold reservoir. Because the engine returns to its initial state each cycle, energy conservation requires $q_h = w + q_c$ . The efficiency is defined as the fraction of absorbed heat converted to useful work:

\varepsilon = \frac{w}{q_h} = 1 - \frac{q_c}{q_h}

Carnot's theorem asserts something remarkable: among all possible engines operating between $T_h$ and $T_c$ — steam engines, gas turbines, fuel cells reconfigured as heat engines, anything obeying the laws of thermodynamics — the reversible ones achieve the maximum possible efficiency, and all reversible ones achieve exactly the same efficiency. The chemical identity of the working fluid, the shape of the cycle on a pressure–volume diagram, none of it matters once reversibility is assumed. Only the two temperatures matter. The proof strategy is elegant: suppose a "super engine" existed that beat the reversible one. Run the reversible engine backward, using the super engine's work output to drive it as a refrigerator. The combination pumps heat from cold to hot with zero net work input from outside — a direct violation of the Clausius statement of the second law, which says heat never flows spontaneously from a colder to a hotter body without external work. Since nature forbids that, the super engine cannot exist.

The core relations

\varepsilon_{\text{rev}} = 1 - \frac{T_c}{T_h}

where: $\varepsilon_{\text{rev}}$ is the efficiency of any reversible engine, $T_c$ and $T_h$ are the cold and hot reservoir temperatures on the thermodynamic scale.

\varepsilon_{\text{irrev}} < \varepsilon_{\text{rev}} = 1 - \frac{T_c}{T_h}

where: $\varepsilon_{\text{irrev}}$ is the efficiency of any real (irreversible) engine operating between the same two reservoirs.

\frac{q_c}{q_h} = \frac{T_c}{T_h} \quad \text{(reversible engine)}

where: this ratio, being independent of the working substance, is what defines the thermodynamic temperature scale.

Derivation AProof of Carnot's Theorem

Goal. Show that no engine operating between reservoirs at $T_h$ and $T_c$ can have efficiency greater than that of a reversible engine operating between the same two reservoirs, and that all reversible engines between them share one efficiency value.

Step 1: Set up the comparison. Let engine R be reversible, operating between $T_h$ and $T_c$ , absorbing $q_h$ from the hot reservoir, delivering work $w_R$ , and rejecting $q_c = q_h - w_R$ to the cold reservoir. Suppose, for contradiction, an engine X exists (reversible or not) operating between the same two reservoirs with efficiency $\varepsilon_X > \varepsilon_R$ . Let engine X also absorb the same $q_h$ from the hot reservoir and produce work $w_X > w_R$ .

Step 2: Couple X to a reversed R. Since R is reversible, it can be run backward as a refrigerator: consuming work $w_R$ , it pulls $q_c$ from the cold reservoir and dumps $q_h$ into the hot reservoir. Use part of the work output of X, namely $w_R$ , to drive R backward. This is always possible because $w_X > w_R$ , so X produces more than enough work to run R in reverse.

Step 3: Track the net heat flows over one combined cycle. The hot reservoir receives $q_h$ back from reversed R but loses $q_h$ to X — net zero exchange with the hot reservoir. The cold reservoir loses $q_c$ to reversed R but this heat came from nowhere else in the combination; meanwhile X, absorbing the same $q_h$ , rejects only

q_{c,X} = q_h - w_X < q_h - w_R = q_c

to the cold reservoir. So the cold reservoir's net heat change is

q_{c,X} - q_c < 0

meaning the cold reservoir loses heat overall, and by hot-reservoir bookkeeping this heat appears as leftover work,

w_X - w_R > 0,

available with no other net effect on either reservoir.

Step 4: Identify the violation. The combined device (X plus reversed R) has, over a full cycle, exchanged zero net heat with the hot reservoir, extracted a positive amount of heat from the cold reservoir, and delivered that exact amount as work, with nothing else changed anywhere in the universe. Equivalently, viewed as a single unit exchanging heat only with the cold reservoir, it converts heat completely into work in a cyclic process, with the hot reservoir acting merely as a heat-recycling intermediary that returns to its initial state.

Step 5: Apply the second law. This is precisely what the Kelvin statement forbids: no cyclic device can convert heat entirely into work while exchanging heat with only one reservoir, since that would allow a perpetual extraction of work from a single heat bath. Equivalently, this net heat transfer from cold to hot with zero external work violates the Clausius statement. Either way, the assumed engine X cannot exist.

Step 6: Conclude the inequality. Therefore no engine, however constructed, can have

\varepsilon_X > \varepsilon_R

so for any engine operating between $T_h$ and $T_c$ ,

\varepsilon \le \varepsilon_R.

Step 7: Show all reversible engines tie. Now let X itself be reversible. Run the argument again with the roles of X and R swapped: use R's excess work (if $\varepsilon_R > \varepsilon_X$ ) to drive X in reverse. The identical contradiction forbids $\varepsilon_R > \varepsilon_X$ too. Combining both inequalities,

\varepsilon_R = \varepsilon_X

for any two reversible engines sharing the same two reservoirs. Because a reversible engine's efficiency cannot depend on the working substance, it can only depend on the two temperatures, giving a universal function $\varepsilon_{\text{rev}} = f(T_h, T_c)$ . Analysis of two reversible engines run in series (not repeated in full here) shows this function must take the multiplicative form $q_c/q_h = T_c/T_h$ , so

\boxed{\varepsilon_{\text{rev}} = 1 - \frac{T_c}{T_h}}

which is the maximum efficiency attainable by any engine, and the exact efficiency of every reversible engine, operating between $T_h$ and $T_c$ .

Worked examples

Example 1 — Foundational: the ceiling on a simple engine. A geothermal engine draws heat from a reservoir at $T_h = 450\text{ K}$ and rejects it to surface air at $T_c = 300\text{ K}$ . Step 1. Compute the Carnot (maximum possible) efficiency:

\varepsilon_{\text{rev}} = 1 - \frac{300\text{ K}}{450\text{ K}} = 1 - 0.6667 = 0.333

So no real engine between these two reservoirs can exceed 33.3%. Step 2. Suppose the actual plant, being irreversible (friction, finite-rate heat transfer, turbulence), delivers 220 kJ of work for every 800 kJ of heat drawn from the hot reservoir. Its actual efficiency is

\varepsilon_{\text{actual}} = \frac{220\text{ kJ}}{800\text{ kJ}} = 0.275 = 27.5\%

Step 3. Sanity check against Carnot's theorem: $27.5\% < 33.3\%$ , consistent with the theorem. The plant operates at

\frac{0.275}{0.333} \times 100\% = 82.6\%

of the theoretical maximum — a realistic "second-law efficiency" for real hardware. Step 4. Heat rejected to the cold reservoir: $q_c = q_h - w = 800\text{ kJ} - 220\text{ kJ} = 580\text{ kJ}$ , all physically reasonable and positive, confirming energy conservation.

Example 2 — Applied: comparing two working fluids and exposing a false claim. An inventor claims to have built two candidate engines for a 900 K industrial exhaust stream rejecting heat to a 350 K cooling loop. Engine A uses steam and reportedly converts 620 kJ of heat into 380 kJ of work per cycle. Engine B uses a supercritical CO₂ cycle and reportedly converts 500 kJ of heat into 320 kJ of work per cycle. The inventor claims Engine B is "fundamentally better since it's more efficient." Step 1. Compute the Carnot ceiling for these reservoirs:

\varepsilon_{\text{rev}} = 1 - \frac{350}{900} = 1 - 0.3889 = 0.6111 = 61.1\%

Step 2. Compute Engine A's efficiency:

\varepsilon_A = \frac{380\text{ kJ}}{620\text{ kJ}} = 0.6129 = 61.3\%

Step 3. Compute Engine B's efficiency:

\varepsilon_B = \frac{320\text{ kJ}}{500\text{ kJ}} = 0.640 = 64.0\%

Step 4. Apply Carnot's theorem as a consistency check. Both claimed efficiencies must not exceed 61.1%. Engine A, at 61.3%, already exceeds the Carnot limit by a small margin — physically impossible, indicating a measurement error, a rounding error in reported heats, or (more likely, given how close it is) an unaccounted extra heat source. Engine B, at 64.0%, exceeds the limit by nearly three percentage points — a clear violation. Step 5. Conclusion: Carnot's theorem tells us immediately, without knowing anything about steam or supercritical CO₂ chemistry, that both reported numbers cannot be trusted. The inventor's claim that Engine B is "better" is moot; both readings are thermodynamically impossible as reported, most likely because the systems are not exchanging heat with only two reservoirs, or some work/heat is being double-counted. This demonstrates the theorem's real power: it is a diagnostic tool, not just an abstract ceiling.

Limiting cases & common traps

Situation	What happens	Why
$T_c \to 0\text{ K}$	$\varepsilon_{\text{rev}} \to 1$	Complete conversion of heat to work becomes theoretically possible, but $T_c = 0$ is unreachable (Third Law)
$T_c \to T_h$	$\varepsilon_{\text{rev}} \to 0$	No temperature difference means no work can be extracted at all
Irreversible engine, same reservoirs	$\varepsilon < \varepsilon_{\text{rev}}$ strictly	Friction, finite-rate heat flow, turbulence all generate entropy, wasting potential work
Reported $\varepsilon > \varepsilon_{\text{rev}}$	Physically impossible	Signals measurement error or an unaccounted extra reservoir, not a real breakthrough

The chemist behind it

Sadi Carnot published his single, slim memoir "Réflexions sur la puissance motrice du feu" in 1824, analyzing idealized heat engines using the (later discredited) caloric theory of heat, yet arriving at conclusions that survived the theory's demise intact. He reasoned that the motive power of heat depends only on the temperatures between which it "falls," in analogy with water falling between heights to drive a mill wheel. Carnot died of cholera in 1832 at age 36, and his work went largely unnoticed until Émile Clapeyron reformulated it graphically, and later Rudolf Clausius and William Thomson (Lord Kelvin) placed it on rigorous energy- and entropy-based footing. What is remarkable is that Carnot's central conclusion — that reversible engines set an unbeatable efficiency ceiling depending only on temperature — required no correct microscopic theory of heat at all, only careful reasoning about impossible perpetual processes.

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

Nicolas Léonard Sadi Carnot

1796-1832 · French military engineer whose 1824 analysis of idealized heat engines founded the science of thermodynamics before the first law was even formulated

Ask the tutor: How does this two-reservoir result generalize into the Clausius inequality and the definition of entropy?

Does Carnot's theorem mean building a more efficient engine is impossible

No. It means efficiency is bounded above by $1 - T_c/T_h$ for any engine operating between those two temperatures. Real engines are always irreversible and sit below this ceiling, so there is usually plenty of room for engineering improvement. Carnot's theorem tells you where the improvement must stop, not that improvement is futile. Raising $T_h$ or lowering $T_c$ raises the ceiling itself, which is why combined-cycle power plants push combustion temperatures as high as materials allow.

Why does the proof use a reversible engine run backward as a refrigerator

Reversibility is exactly the property that lets a cycle run in either direction with the heat and work simply reversed in sign. An irreversible engine cannot be "run backward" using the same heat and work amounts; forcing it in reverse would require different inputs due to internal dissipation. The proof depends on being able to exactly undo the reversible engine's cycle, which only a reversible process guarantees.

Is Carnot's theorem specific to steam engines or heat engines generally

It applies to any device operating in a closed thermodynamic cycle that exchanges heat with exactly two reservoirs at fixed temperatures and produces or consumes work. This includes gas turbines, Stirling engines, and even refrigerators and heat pumps run in reverse, where the analogous statement bounds the maximum coefficient of performance rather than efficiency.

How does Carnot's theorem lead to a temperature scale

Because all reversible engines between two reservoirs share one efficiency depending only on temperature, the ratio $q_c/q_h$ is a universal function of $T_h$ and $T_c$ alone. Defining temperature so that this function equals $T_c/T_h$ gives a scale that requires no reference substance like mercury or an ideal gas. This thermodynamic temperature scale turns out to coincide numerically with the ideal-gas scale, and is the subject of the next topic.

What is the connection to the Clausius inequality

The Clausius inequality generalizes the reasoning used here beyond two-reservoir engines to arbitrary cyclic processes exchanging heat with any number of reservoirs at varying temperatures. It states that the cyclic integral of $dq_{\text{rev}}/T$ around any reversible loop is zero, while for irreversible cycles the analogous integral of $dq/T$ is strictly negative. Carnot's theorem is the two-reservoir special case of this broader statement, and it is the historical stepping stone toward defining entropy as a state function.