The Carnot Engine and Maximum Efficiency

Designing a reversible cycle between two reservoirs

Suppose a hot reservoir is at TH and a cold reservoir at TC, with TH > TC > 0 in kelvin. A reversible engine must avoid friction and exchange heat across an infinitesimal temperature difference. Therefore it absorbs heat while its working substance is at TH and releases heat while at TC. These two heat-exchange stages must be isothermal in the reversible limit.

The working substance must also travel between TH and TC. If a non-adiabatic temperature-changing stage were made reversible by contacting reservoirs at progressively different temperatures, the engine would no longer use only the stated two reservoirs. Instead, reversible adiabatic expansion lowers the gas temperature from TH to TC without heat exchange, and reversible adiabatic compression raises it back. This constructs the Carnot cycle: two isotherms joined by two adiabats. Carnot studied the ideal efficiency question in 1824, before the modern heat-and-energy framework was complete.

Use one fixed ideal-gas sample as the working substance for a calculable model. States 1 and 2 lie on the hot isotherm, and states 3 and 4 on the cold isotherm. Volumes increase on 1→2 and 2→3, then decrease on 3→4 and 4→1. QH and QC below are positive heat magnitudes; every Wᵢⱼ remains signed work done by the gas. Keeping this convention throughout avoids changing a compression work symbol into an unannounced positive magnitude.

Hot isothermal expansion: 1 → 2

Put the gas in reversible thermal contact with the hot reservoir at TH and slowly reduce the external load. It expands from V₁ to V₂, absorbing heat while maintaining constant temperature. Ideal-gas U is unchanged. Consequently every joule entering as heat is transferred outward as work on this leg.

Adiabatic expansion and cooling: 2 → 3

Thermally insulate the gas and let it continue expanding quasi-statically until its temperature falls to TC. No heat enters. Positive work now reduces internal energy; the cooling occurs because the gas spends its own internal energy on work rather than receiving replacement energy from a reservoir.

Cold isothermal compression: 3 → 4

Bring the gas into reversible thermal contact with the cold reservoir. Increase the external load infinitesimally so the gas compresses from V₃ to V₄ at TC. Work is done on the gas. To prevent its temperature from rising, the same energy leaves as heat. Both signed gas work and signed heat input are negative; the magnitude QC is positive.

Adiabatic compression and warming: 4 → 1

Finally insulate the gas and continue compressing quasi-statically. The work supplied to the gas raises its U and T from TC to TH. The endpoint is chosen so that the gas returns to its starting P₁, V₁ and TH. The magnitude of this compression work equals that of the earlier adiabatic expansion work when n and Cᵥ are the same.

Why the isothermal volume ratios are equal

On the expansion adiabat, TH V₂^(γ−1) = TC V₃^(γ−1). Rearranging gives (V₃/V₂)^(γ−1) = TH/TC. On the compression adiabat, TC V₄^(γ−1) = TH V₁^(γ−1), so (V₄/V₁)^(γ−1) = TH/TC. Both positive ratios therefore have the same value.

Deriving Carnot efficiency

Adding the signed work of all four stages cancels the adiabatic terms. Thus Wnet = nRTH ln(V₂/V₁) − nRTC ln(V₃/V₄) = QH − QC. The equal volume ratios make the logarithms identical, so QC/QH = TC/TH. Divide net work by heat absorbed to obtain efficiency.

The efficiency rises if TH increases at fixed TC, or if TC decreases at fixed TH. If TH and TC approach each other, ηC approaches zero: there is little temperature difference available to drive the engine. For finite TH and positive TC, ηC is below 1. Setting TC = 0 K or TH infinitely large is not an attainable way to construct a perfect practical engine. The limit is a ceiling on efficiency, not a prediction of the power of a real machine.

Why no engine between the same reservoirs can do better

Let R be a reversible engine and imagine a rival engine I operating between the same hot and cold reservoirs with a greater efficiency. Scale their operations so each, when used as an engine, would absorb the same heat QH from the hot reservoir. If R would deliver work W, the supposedly better rival delivers W′ > W. Run the rival forward and the reversible device backward as a refrigerator, using part of the rival’s work to drive it.

The rival removes QH from the hot reservoir and rejects QH − W′ to the cold reservoir. The reversed R returns exactly QH to the hot reservoir and extracts QH − W from the cold reservoir. The hot reservoir therefore has zero net change. The cold reservoir loses (QH − W) − (QH − W′) = W′ − W. The combined device delivers exactly W′ − W as net work. Its sole result would be to remove heat from one reservoir and convert all of it to work, violating Kelvin–Planck. Therefore W′ > W is impossible: no engine can exceed the reversible efficiency.

Now suppose two reversible engines with different working substances had different efficiencies. The more efficient one could run forward and power the less efficient one backward, producing the same forbidden result. Interchanging their roles rules out either strict inequality. All reversible engines operating only between the same two reservoirs must therefore have equal efficiency, independent of working substance. The ideal-gas calculation supplies a convenient derivation of a universal result, not a restriction of Carnot’s theorem to ideal gases.

A temperature scale independent of the working substance

For a reversible two-reservoir engine, QC/QH = TC/TH holds regardless of which substance performs the cycle. Heat-transfer ratios can therefore define ratios on a universal thermodynamic temperature scale. Fixing a reference gives numerical temperatures. For an ideal gas this agrees with the absolute gas-temperature scale. Celsius ratios cannot be inserted into Carnot efficiency because Celsius has an arbitrary shifted zero; ratios of kelvin temperatures have the required physical meaning.

Every ideal Carnot stage can be reversed. Running the cycle anticlockwise takes heat QC from the cold reservoir, receives work Win = QH − QC and releases QH to the hot reservoir. It is a reversible refrigerator. Combining α = QC/(QH−QC) with QC/QH = TC/TH gives αrev = TC/(TH−TC). This short consequence connects the engine derivation to the performance concept; it does not remove the need for work when TH > TC.

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