Carnot Cycle — The Thermodynamic Speed Limit

Four reversible processes around a P–V loop. The Carnot cycle is the theoretical maximum efficiency any heat engine can achieve between two reservoirs.

Visualization

Input

T_H

Hot reservoir temperature

\mathrm{K}

T_C

Cold reservoir temperature

\mathrm{K}

V_1

Starting volume (state 1)

\mathrm{L}

V_2

Volume after isothermal expansion (state 2)

\mathrm{L}

n

Moles of gas

\mathrm{mol}

\gamma

Heat capacity ratio (γ = C_p/C_v)

Show the formal P–V diagram

The thermodynamic speed limit

Pick any two heat reservoirs — a hot one at $T_H$ and a cold one at $T_C$ . No heat engine you can build, no matter how clever, can convert more than this fraction of the heat it takes from the hot side into useful work:

\eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H}

Source: OpenStax University Physics §4.5; Wikipedia — Carnot cycle. This is the second law of thermodynamics as a budget — the temperatures alone tell you the ceiling. A power plant burning fuel at 750 K and dumping waste heat at 300 K cannot exceed 60% efficiency, ever. The only way up is to raise $T_H$ or lower $T_C$ .

The cycle that achieves it

The Carnot cycle is the only cycle that hits the ceiling above. It does so by being entirely reversible — every heat transfer happens at vanishing temperature difference, no friction, no leaks. It's the thermodynamic equivalent of frictionless surfaces in mechanics: a useful idealisation that real machines approach but never reach.

Four stages take the gas around the closed loop in the P–V diagram:

Isothermal expansion at $T_H$ (1→2). The gas absorbs heat $Q_H$ from the hot reservoir while expanding; its temperature stays put because the heat exactly matches the work done.
Adiabatic expansion (2→3). The gas is isolated; expansion continues but now the work comes out of internal energy, so the temperature falls to $T_C$ .
Isothermal compression at $T_C$ (3→4). The gas rejects heat $Q_C$ to the cold reservoir while being compressed.
Adiabatic compression (4→1). The gas is isolated again and compressed back to its starting state, with the temperature climbing back to $T_H$ .

The shaded area inside the P–V loop is the net work done per cycle:

W = Q_H - Q_C

For an ideal gas the heats follow from integrating $p\,dV$ along the isotherms:

Q_H = n R T_H \ln\!\frac{V_2}{V_1}, \qquad Q_C = n R T_C \ln\!\frac{V_3}{V_4}

The Carnot constraint forces $V_3/V_4 = V_2/V_1$ , so $Q_C/Q_H = T_C/T_H$ , which is exactly the efficiency formula again — the cycle is defined to satisfy the bound.

Why the temperatures matter

Because $\eta = 1 - T_C/T_H$ , the only way to push the limit higher is to widen the temperature gap. That's why jet engines run combustion at thousands of kelvin and dump exhaust to ambient air, and why geothermal plants struggle when the source rock is only modestly hotter than the surface. Slide $T_H$ up and watch the loop fatten; slide $T_C$ down and watch the same effect from the other side.

Show your work

Volume after adiabatic expansion (state 3): $V_3 \;=\; V_2 \cdot \left(\dfrac{T_H}{T_C}\right)^{1/(\gamma - 1)} \;=\; \textcolor{#dc4124}{\left(0.00223\right)} \cdot \left(\dfrac{T_H}{T_C}\right)^{1/(\gamma - 1)} \;=\; 0.0088088\;\mathrm{m^{3}}$
Volume after isothermal compression (state 4): $V_4 \;=\; V_1 \cdot \left(\dfrac{T_H}{T_C}\right)^{1/(\gamma - 1)} \;=\; \textcolor{#dc4124}{\left(0.001\right)} \cdot \left(\dfrac{T_H}{T_C}\right)^{1/(\gamma - 1)} \;=\; 0.0039501\;\mathrm{m^{3}}$
Pressure at state 1: $P_1 \;=\; \dfrac{n R T_H}{V_1} \;=\; \dfrac{n \textcolor{#dc4124}{\left(8.314\right)} T_H}{\textcolor{#dc4124}{\left(0.001\right)}} \;=\; 6.236 \times 10^{5}\;\mathrm{Pa}$
Pressure at state 2: $P_2 \;=\; \dfrac{n R T_H}{V_2} \;=\; \dfrac{n \textcolor{#dc4124}{\left(8.314\right)} T_H}{\textcolor{#dc4124}{\left(0.00223\right)}} \;=\; 2.796 \times 10^{5}\;\mathrm{Pa}$
Pressure at state 3: $P_3 \;=\; \dfrac{n R T_C}{V_3} \;=\; \dfrac{n \textcolor{#dc4124}{\left(8.314\right)} T_C}{\textcolor{#dc4124}{\left(0.0088088\right)}} \;=\; 28315\;\mathrm{Pa}$
Pressure at state 4: $P_4 \;=\; \dfrac{n R T_C}{V_4} \;=\; \dfrac{n \textcolor{#dc4124}{\left(8.314\right)} T_C}{\textcolor{#dc4124}{\left(0.0039501\right)}} \;=\; 63142\;\mathrm{Pa}$
Heat absorbed from hot reservoir (isothermal expansion 1→2): $Q_H \;=\; n R T_H \ln\!\left(\dfrac{V_2}{V_1}\right) \;=\; n \textcolor{#dc4124}{\left(8.314\right)} T_H \ln\!\left(\dfrac{\textcolor{#dc4124}{\left(0.00223\right)}}{\textcolor{#dc4124}{\left(0.001\right)}}\right) \;=\; 500.1\;\mathrm{J}$
Heat rejected to cold reservoir (isothermal compression 3→4): $Q_C \;=\; n R T_C \ln\!\left(\dfrac{V_3}{V_4}\right) \;=\; n \textcolor{#dc4124}{\left(8.314\right)} T_C \ln\!\left(\dfrac{\textcolor{#dc4124}{\left(0.0088088\right)}}{\textcolor{#dc4124}{\left(0.0039501\right)}}\right) \;=\; 200\;\mathrm{J}$
Net work per cycle: $W \;=\; Q_H - Q_C \;=\; 300.1\;\mathrm{J}$
Carnot efficiency (from temperatures): $\eta \;=\; 1 - \dfrac{T_C}{T_H} \;=\; 0.6$

Worked Example

OpenStax University Physics Vol. 2 §4.5 Example 4.2: a Carnot engine between T_H = 750 K and T_C = 300 K has efficiency η = 1 − 300/750 = 0.60 (60%). With Q_H = 500 J the engine produces W = 300 J of work and rejects Q_C = 200 J. Efficiency depends only on temperatures, so it must reproduce exactly regardless of n, V, γ.

Quantity	Expected	Computed	Δ
eta	0.6	0.6	0
eta_pct	60	60	0

Source: OpenStax, University Physics Vol. 2 §4.5 Example 4.2 — Carnot efficiency

Worked Example

Same OpenStax Example 4.2: with V₁ = 1.0 L, V₂ = 2.23 L, n = 0.1 mol so that nRT_H ln(V₂/V₁) ≈ 500 J, the model should reproduce Q_H ≈ 500, Q_C ≈ 200, W ≈ 300 J. (V₂/V₁ = 2.23 chosen so exp(500/(0.1·8.314·750)) ≈ 2.23.)

Quantity	Expected	Computed	Δ
Q_h	500	500.1	0.08809
Q_c	200	200	0.03524
W_net	300	300.1	0.05285

Source: OpenStax, University Physics Vol. 2 §4.5 Example 4.2 — energy budget

Sources

[1] OpenStax. University Physics Volume 2 — §4.5 The Carnot Cycle, eq. e = 1 − T_C/T_H (Eq. 4.5); four reversible stages: isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression (openstax.org)
[2] Wikipedia — Carnot cycle, eq. η = 1 − T_C/T_H; Q_H/T_H = −Q_C/T_C; W = Q_H − Q_C (en.wikipedia.org)

Assumptions

Working substance is an ideal gas (OpenStax §4.5).
All four processes are reversible (Wikipedia: Carnot cycle).
No friction, no heat loss to surroundings outside the prescribed reservoir contacts (Wikipedia: Carnot cycle).
Heat reservoirs are large enough that their temperatures stay constant during the cycle.
γ = C_p/C_v is constant for the gas across the cycle (idealisation).