Thermodynamicsη = 1 - T_C/T_H

Carnot Engine Simulator

Visualize the Carnot cycle on a PV diagram. Adjust hot and cold reservoir temperatures to see efficiency and work output.

— A→B Isothermal (T_H)— B→C Adiabatic— C→D Isothermal (T_C)— D→A Adiabatic

Reservoir Temperatures

Hot reservoir T_H

Cold reservoir T_C

Cycle Summary (W = 1000 J)

Efficiency η62.50%

Q_H absorbed1600.0 J

Q_C rejected600.0 J

COP (refrig.)0.600

T_H800 K

T_C300 K

η62.50 %

Q_H1600.0 J

W1000 J

Q_C600.0 J

COP0.600

About the Carnot Engine

The Carnot engine is a theoretical heat engine operating between two reservoirs at temperatures T_H (hot) and T_C (cold). It represents the upper bound on efficiency for any heat engine operating between those temperatures — a consequence of the Second Law of Thermodynamics.

ℹNo real engine can be more efficient than a Carnot engine operating between the same two temperature reservoirs. This is the Carnot theorem.

The Four Stages of the Carnot Cycle

Stage	Process	Description	Heat Exchange
A → B	Isothermal expansion	Gas absorbs heat Q_H from hot reservoir at T_H	Q_H absorbed
B → C	Adiabatic expansion	Gas expands, cools from T_H to T_C with no heat exchange	Q = 0
C → D	Isothermal compression	Gas rejects heat Q_C to cold reservoir at T_C	Q_C released
D → A	Adiabatic compression	Gas is compressed, warms from T_C to T_H with no heat exchange	Q = 0

Carnot Efficiency

The efficiency η of the Carnot engine is the fraction of heat absorbed from the hot reservoir that is converted to work:

η = 1 - \frac{T _{C}}{T _{H}} = \frac{W}{Q _{H}}

Since T_C and T_H are absolute temperatures (Kelvin), efficiency is always less than 1 (100%). Efficiency approaches 1 only as T_C → 0 K, which is unattainable.

Energy Balance

Q_{H} = W + Q_{C} ⟹ W = Q_{H} - Q_{C}

Worked Example

A Carnot engine operates between T_H = 800 K and T_C = 300 K. Find efficiency and heat flows for W = 1000 J of net work output:

η = 1 - \frac{300}{800} = 1 - 0.375 = 0.625 (62.5%)

Q_{H} = \frac{W}{η} = \frac{1000}{0.625} = 1600 J, Q_{C} = Q_{H} - W = 600 J

Carnot Refrigerator (Reverse Cycle)

Run the cycle in reverse and it becomes a refrigerator (or heat pump). The Coefficient of Performance (COP) for a Carnot refrigerator is:

COP_{refrigerator} = \frac{Q _{C}}{W} = \frac{T _{C}}{T _{H} - T _{C}}

💡A Carnot refrigerator with T_H = 300 K and T_C = 250 K has COP = 250/50 = 5: for every 1 J of work input, 5 J of heat is extracted from the cold reservoir.

Carnot Engine Formulas

Core Efficiency Formula

η_{Carnot} = 1 - \frac{T _{C}}{T _{H}}

Heat and Work Relations

Q_{H} = \frac{W}{η}, Q_{C} = Q_{H} - W = Q_{H} \cdot \frac{T _{C}}{T _{H}}

Coefficient of Performance

COP_{ref} = \frac{T _{C}}{T _{H} - T _{C}}, COP_{hp} = \frac{T _{H}}{T _{H} - T _{C}}

Quantity	Symbol	Formula
Efficiency	η	1 − T_C/T_H
Heat from hot reservoir	Q_H	W/η
Heat to cold reservoir	Q_C	Q_H − W
COP (refrigerator)	COP_r	T_C/(T_H−T_C)
COP (heat pump)	COP_hp	T_H/(T_H−T_C)

⚠All temperatures must be in Kelvin (absolute). Using Celsius will give incorrect efficiency values.

Frequently Asked Questions

Why can't real engines achieve Carnot efficiency?

Carnot efficiency requires all processes to be perfectly reversible, meaning they must be infinitely slow (quasi-static) with no friction, heat leaks, or other irreversibilities. Real engines operate at finite speeds, generating entropy, which always reduces efficiency below the Carnot limit.

What happens to efficiency as T_C approaches T_H?

As T_C → T_H, efficiency → 0. You cannot extract any net work when both reservoirs are at the same temperature — this is consistent with the Second Law, which requires a temperature difference to drive a heat engine.

Can efficiency ever equal 100%?

Only if T_C = 0 K (absolute zero). Since the Third Law of Thermodynamics states absolute zero is unattainable, 100% efficiency is impossible in practice.

What is the difference between a heat engine and a refrigerator?

A heat engine converts heat to work; a refrigerator uses work to transfer heat from cold to hot. The Carnot cycle can run in either direction — forward for engine operation, reverse for refrigeration. The COP of a refrigerator replaces efficiency as the performance metric.

Why is the Carnot cycle shown on a PV diagram?

The PV diagram shows the state of the working gas (usually an ideal gas) at each point in the cycle. The area enclosed by the cycle curve equals the net work output per cycle. Isothermal processes appear as hyperbolas; adiabatic processes as steeper hyperbolas.