Carnot Cycle

The Carnot cycle is an ideal heat-engine cycle that shows the highest possible efficiency for any engine operating between two reservoir temperatures. For a reversible engine between a hot reservoir at $T_H$ and a cold reservoir at $T_C$, the maximum efficiency is

$$\eta = 1 - \frac{T_C}{T_H}$$

with temperatures in Kelvin. Real engines do not reach this limit, but the Carnot cycle tells you what the limit is.

What the Carnot cycle means

The cycle has four stages: two isothermal stages, where heat is exchanged at constant temperature, and two reversible adiabatic stages, where no heat is exchanged and the temperature changes.

Its importance is simple: it gives a benchmark. If two engines work between the same hot and cold reservoirs, no engine can be more efficient than a reversible Carnot engine.

The four stages in order

1. Isothermal expansion at $T_H$. The gas absorbs heat $Q_H$ from the hot reservoir and does work while staying at the hot temperature.
2. Reversible adiabatic expansion. No heat enters or leaves. The gas keeps expanding, does work, and its temperature falls from $T_H$ to $T_C$.
3. Isothermal compression at $T_C$. The surroundings do work on the gas while the gas rejects heat $Q_C$ to the cold reservoir at constant cold temperature.
4. Reversible adiabatic compression. No heat is exchanged. The gas is compressed until its temperature rises from $T_C$ back to $T_H$.

After the fourth stage, the system returns to its starting state, so the process can repeat as a cycle.

When you can use the Carnot efficiency formula

Use

$$\eta = 1 - \frac{T_C}{T_H}$$

only when the engine is reversible and operates between two reservoirs at fixed absolute temperatures.

Why does it work? In a reversible Carnot cycle, the entropy gained from the hot reservoir matches the entropy delivered to the cold reservoir in magnitude, so

$$\frac{Q_H}{T_H} = \frac{Q_C}{T_C}$$

which gives

$$\frac{Q_C}{Q_H} = \frac{T_C}{T_H}$$

and then

$$\eta = \frac{W}{Q_H} = 1 - \frac{Q_C}{Q_H} = 1 - \frac{T_C}{T_H}$$

Do not apply this unchanged to a real engine with friction, turbulence, or heat transfer across a finite temperature difference. In that case, the Carnot value is still an upper bound, not the actual efficiency.

Worked example: maximum efficiency between two temperatures

Suppose an ideal Carnot engine operates between $T_H = 600\ \mathrm{K}$ and $T_C = 300\ \mathrm{K}$ and absorbs $Q_H = 900\ \mathrm{J}$ from the hot reservoir each cycle.

Its efficiency is

$$\eta = 1 - \frac{300}{600} = 0.50$$

So the maximum possible efficiency is $50\%$.

The work done per cycle is

$$W = \eta Q_H = 0.50 \times 900 = 450\ \mathrm{J}$$

The remaining heat must be rejected to the cold reservoir:

$$Q_C = Q_H - W = 900 - 450 = 450\ \mathrm{J}$$

This example shows the main idea clearly: once the reservoir temperatures are fixed, the maximum efficiency is fixed too. Increasing engineering quality can help a real engine get closer to that limit, but not exceed it.

Common mistakes in Carnot cycle problems

One common mistake is using Celsius in the efficiency formula. The ratio $T_C/T_H$ must use Kelvin.

Another mistake is treating the Carnot cycle as a realistic model of an everyday engine. It is an ideal reversible benchmark, not a claim about what normal engines actually do.

A third mistake is memorizing the four stages without tracking where heat enters and leaves. Heat enters during the hot isothermal expansion and leaves during the cold isothermal compression. The adiabatic stages have $Q = 0$.

It is also easy to overread the efficiency formula. It does not say an engine becomes efficient just because $T_H$ is large. Materials limits, irreversibility, and design constraints still matter in real machines.

Where the Carnot cycle is used

The Carnot cycle appears in thermodynamics because it connects entropy, reversibility, and engine efficiency in one clean model. It is used to set upper efficiency limits, to compare real engines with ideal ones, and to build intuition for refrigerators and heat pumps as well as heat engines.

If you already know the second law of thermodynamics, the Carnot cycle is one of the clearest ways to see that law turn into a quantitative limit.

Try a similar problem

Try your own version with $T_H = 500\ \mathrm{K}$ and $T_C = 350\ \mathrm{K}$. Compute the Carnot efficiency first, then choose a value of $Q_H$ and find the work and rejected heat. If you want to go one step further, compare that ideal answer with a real engine that runs at a lower efficiency and explain why the gap appears.