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The Carnot Cycle

The theoretical limit of efficiency for any heat engine.

By chemca Team • Updated Jan 2026

Proposed by Sadi Carnot in 1824, the Carnot Cycle is a theoretical thermodynamic cycle that provides the maximum possible efficiency that any heat engine can achieve. It consists of four reversible processes operating between a Source (High Temp, $T_1$) and a Sink (Low Temp, $T_2$).

1. The Four Steps of the Cycle

Imagine an ideal gas inside a cylinder with a frictionless piston.

Step 1: Reversible Isothermal Expansion

The system absorbs heat $Q_1$ from the Source at temperature $T_1$. The gas expands, doing work on the surroundings.

$\Delta U = 0$, $W_1 = -Q_1$ (Work output)

Step 2: Reversible Adiabatic Expansion

The system is insulated ($Q=0$). The gas continues to expand, doing work at the cost of internal energy. Temperature falls from $T_1$ to $T_2$.

$\Delta U < 0$, $W_2$ (Work output)

Step 3: Reversible Isothermal Compression

The system rejects heat $Q_2$ to the Sink at temperature $T_2$. Surroundings do work on the gas.

$\Delta U = 0$, $W_3 = -Q_2$ (Work input)

Step 4: Reversible Adiabatic Compression

The system is insulated ($Q=0$). Surroundings do work on the gas, raising internal energy. Temperature rises from $T_2$ back to $T_1$.

$\Delta U > 0$, $W_4$ (Work input)

2. Carnot Efficiency ($\eta$)

The efficiency of a heat engine is defined as the net work output divided by the heat absorbed.

$$ \eta = \frac{W_{net}}{Q_{absorbed}} = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1} $$

For a reversible cycle involving an ideal gas, the ratio of heat exchanged equals the ratio of absolute temperatures:

$$ \eta = 1 - \frac{T_2}{T_1} $$

Where $T_1$ is the Source Temperature and $T_2$ is the Sink Temperature (in Kelvin).

3. Carnot's Theorem

• No heat engine operating between two given heat reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs.
• All Carnot engines operating between the same two temperatures have the same efficiency, regardless of the working substance.

Why is 100% Efficiency Impossible?

From the formula $\eta = 1 - T_2/T_1$, to achieve $\eta = 1$ (100%), $T_2$ must be 0 K (Absolute Zero). Since Absolute Zero is practically unattainable (Third Law of Thermodynamics), 100% efficiency is impossible.

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