The Carnot Cycle

The Carnot cycle is an idealized thermodynamic cycle that describes the most efficient possible heat engine operating between two temperatures.

It was developed by French physicist Sadi Carnot in the 19th century.

A Carnot engine operates between:

• a hot reservoir at temperature T_h
• a cold reservoir at temperature T_c

and converts some of the thermal energy received from the hot reservoir into useful work.

Its importance comes from a fundamental result:

No heat engine operating between the same two temperatures can be more efficient than a Carnot engine.

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Why the Carnot Cycle Matters

Real engines contain unavoidable sources of irreversibility such as:

• friction
• turbulence
• electrical resistance
• heat transfer across finite temperature differences
• rapid expansion and compression
• combustion

The Carnot cycle assumes that all processes are:

perfectly reversible

Therefore, it produces no entropy through irreversibility.

This makes it an ideal benchmark for determining:

the maximum possible efficiency of a heat engine

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The Carnot Engine

A Carnot engine is an ideal heat engine containing a working substance, usually modeled as an ideal gas.

During one complete cycle, the gas:

1. expands isothermally
2. expands adiabatically
3. compresses isothermally
4. compresses adiabatically

After the fourth stage, the gas returns to:

its original thermodynamic state

The cycle can then repeat.

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The Four Stages

The Carnot cycle consists of:

isothermal expansion

↓

adiabatic expansion

↓

isothermal compression

↓

adiabatic compression

↓

original state

Two processes occur at constant temperature.

Two processes occur without heat transfer.

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Isothermal Processes

An isothermal process occurs at:

constant temperature

Therefore:

ΔT = 0

For an ideal gas, internal energy depends only on temperature.

So during an isothermal process:

ΔU = 0

From the First Law:

ΔU = Q − W

therefore:

Q = W

during an ideal-gas isothermal expansion.

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Adiabatic Processes

An adiabatic process occurs with:

no heat transfer

Therefore:

Q = 0

From:

ΔU = Q − W

we obtain:

ΔU = −W

during expansion.

Thus the gas performs work using its internal energy and its temperature decreases.

During adiabatic compression, work is done on the gas and its temperature increases.

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Stage 1: Isothermal Expansion

The gas begins in contact with the:

hot reservoir at T_h

It expands slowly while remaining at constant temperature.

During this stage:

• heat Q_h enters the gas
• gas expands
• gas performs work
• temperature remains T_h
• internal energy remains constant for an ideal gas

Energy relationship:

Q_h = W

for this stage.

The thermal energy entering the gas is converted into work while the temperature remains constant.

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Why Temperature Remains Constant

As the gas expands, it performs work.

Normally, this would reduce its internal energy and temperature.

However, the hot reservoir continuously supplies energy.

Therefore:

energy entering as heat = energy leaving as work

and:

ΔU = 0

So the temperature remains constant.

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Stage 2: Adiabatic Expansion

The gas is now thermally isolated.

It continues expanding.

During this stage:

Q = 0

The gas performs work on the surroundings.

Since no heat enters:

internal energy decreases

and therefore:

temperature decreases

The temperature falls from:

T_h → T_c

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Energy During Adiabatic Expansion

Since:

Q = 0

the First Law becomes:

ΔU = −W

Because the gas performs positive work:

W > 0

therefore:

ΔU < 0

The gas loses internal energy and cools.

---

Stage 3: Isothermal Compression

The gas is placed in thermal contact with the:

cold reservoir at T_c

The surroundings slowly compress the gas.

During this stage:

• work is done on the gas
• gas volume decreases
• thermal energy Q_c leaves the gas
• temperature remains constant at T_c

For an ideal gas:

ΔU = 0

Therefore, the work done on the gas is transferred as heat to the cold reservoir.

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Why Heat Leaves During Compression

Compression tends to increase the gas's internal energy and temperature.

However, the gas must remain at:

T_c

Therefore the energy supplied by compression is transferred to the cold reservoir.

The gas remains at constant temperature.

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Stage 4: Adiabatic Compression

The gas is thermally isolated again.

The surroundings continue compressing it.

During this stage:

Q = 0

Work is done on the gas.

Its internal energy increases.

Therefore its temperature rises:

T_c → T_h

At the end of the process, the gas returns to its original state.

The cycle is complete.

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The Complete Carnot Cycle

The four stages can be summarized as:

The working gas then returns to its initial state.

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Energy Flow Through the Carnot Engine

During the cycle:

hot reservoir

↓

Q_h

↓

Carnot engine

↙︎　　　　　　　　　↘︎

W　　　　　　　Q_c

　　　　　　　　　　　↓

　　　　　　　cold reservoir

Energy conservation requires:

Q_h = W + Q_c

Therefore:

W = Q_h − Q_c

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The P–V Diagram

The Carnot cycle is often shown on a:

pressure-volume (P–V) diagram

The cycle forms a closed loop.

The two isothermal curves correspond to:

T_h

and:

T_c

The two adiabatic curves connect them.

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Area Under a P–V Curve

For expansion or compression:

W = area under the P–V curve

During expansion:

work is done by the gas

During compression:

work is done on the gas

For a complete Carnot cycle:

net work = area enclosed by the cycle

This is an important graphical interpretation.

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Why the Cycle Produces Net Work

During the high-temperature expansion, the gas performs relatively large work.

During the lower-temperature compression, less work must be supplied to compress it.

Therefore:

work during expansion > work required for compression

The difference becomes:

net useful work

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Heat Engine Efficiency

Efficiency is:

η = W/Q_h

Since:

W = Q_h − Q_c

then:

η = (Q_h − Q_c)/Q_h

Therefore:

η = 1 − Q_c/Q_h

This equation applies to heat-engine efficiency in terms of heat transfers.

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Carnot Efficiency

For a reversible Carnot engine:

Q_c/Q_h = T_c/T_h

Therefore:

η_Carnot = 1 − T_c/T_h

where:

• η_Carnot = maximum theoretical efficiency
• T_c = cold reservoir temperature
• T_h = hot reservoir temperature

Temperatures must be measured in:

kelvin

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Worked Example: Carnot Efficiency

A Carnot engine operates between:

T_h = 600 K

and:

T_c = 300 K

Use:

η_Carnot = 1 − T_c/T_h

η = 1 − 300/600

η = 1 − 0.50

η = 0.50

Therefore:

η = 50%

The maximum possible efficiency is:

50%

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Worked Example: Higher Hot Temperature

Suppose:

T_h = 900 K

T_c = 300 K

η = 1 − 300/900

η = 1 − 0.333

η ≈ 0.667

Therefore:

η ≈ 66.7%

Increasing the hot-reservoir temperature increased the maximum possible efficiency.

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Worked Example: Lower Cold Temperature

Suppose:

T_h = 600 K

T_c = 200 K

η = 1 − 200/600

η = 1 − 0.333

η ≈ 66.7%

Lowering the cold-reservoir temperature also increases maximum efficiency.

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Temperature Difference and Efficiency

From:

η_Carnot = 1 − T_c/T_h

we can see:

Increasing T_h tends to increase efficiency.

Decreasing T_c tends to increase efficiency.

Therefore:

larger temperature difference → greater possible efficiency

However, the ratio T_c/T_h is what directly determines Carnot efficiency.

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Why Kelvin Must Be Used

Suppose:

T_h = 400°C

and:

T_c = 100°C

First convert:

T_h = 400 + 273

T_h = 673 K

T_c = 100 + 273

T_c = 373 K

Now:

η = 1 − 373/673

η ≈ 1 − 0.554

η ≈ 0.446

Therefore:

η ≈ 44.6%

Using Celsius directly would give an incorrect result.

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Finding the Cold-Reservoir Temperature

Suppose a Carnot engine has:

η = 60%

and:

T_h = 750 K

Use:

η = 1 − T_c/T_h

0.60 = 1 − T_c/750

T_c/750 = 0.40

T_c = 0.40 × 750

T_c = 300 K

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Finding the Hot-Reservoir Temperature

Suppose:

η = 40%

and:

T_c = 360 K

Use:

0.40 = 1 − 360/T_h

360/T_h = 0.60

T_h = 360/0.60

T_h = 600 K

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Calculating Work

Suppose a Carnot engine:

• receives Q_h = 5000 J
• operates at η = 60%

Use:

η = W/Q_h

Therefore:

W = ηQ_h

W = 0.60 × 5000

W = 3000 J

The engine produces:

3000 J of work

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Calculating Rejected Heat

From:

Q_h = W + Q_c

we have:

Q_c = Q_h − W

Q_c = 5000 − 3000

Q_c = 2000 J

Therefore:

• Q_h = 5000 J
• W = 3000 J
• Q_c = 2000 J

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Checking with Reservoir Temperatures

Suppose:

T_h = 750 K

T_c = 300 K

Then:

Q_c/Q_h = T_c/T_h

Q_c/5000 = 300/750

Q_c/5000 = 0.40

Q_c = 2000 J

This agrees with our previous calculation.

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Entropy During the Carnot Cycle

The Carnot cycle is completely reversible.

During isothermal expansion at T_h:

ΔS_hot = −Q_h/T_h

The working substance gains the corresponding entropy.

During isothermal compression at T_c, entropy is transferred to the cold reservoir.

For the complete reversible cycle:

ΔS_total = 0

No entropy is produced.

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The Working Substance Returns to Its Starting State

Entropy is a:

state function

After one complete cycle, the working substance returns to exactly the same state.

Therefore:

ΔS_working substance = 0

over the complete cycle.

Similarly:

ΔU_cycle = 0

because internal energy is also a state function.

This does not mean no energy has been transferred.

The engine has:

• absorbed Q_h
• rejected Q_c
• produced W

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Why the Carnot Engine Has Maximum Efficiency

Any real source of irreversibility produces entropy.

Examples include:

• friction
• turbulence
• heat transfer across finite temperature differences
• rapid expansion
• rapid compression
• electrical resistance

Entropy production reduces the maximum amount of energy that can be converted into useful work.

The Carnot engine avoids all these idealized losses.

Therefore:

no engine between the same T_h and T_c can be more efficient

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Carnot's Theorem

Carnot's theorem states that:

No heat engine operating between two reservoirs can be more efficient than a reversible engine operating between those same reservoirs.

It also implies that:

all reversible engines operating between the same two temperatures have the same efficiency

regardless of their particular design or working substance.

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The Working Substance Does Not Determine Maximum Efficiency

Imagine two ideal reversible engines.

Engine A uses:

helium

Engine B uses:

another ideal gas

If both operate between:

T_h = 700 K

and:

T_c = 350 K

then both have:

η = 1 − 350/700

η = 50%

Their maximum efficiency is determined by reservoir temperatures, not by the choice of working substance.

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Real Engines

Real engines cannot follow the Carnot cycle perfectly.

They contain:

• friction
• finite temperature differences
• rapid gas flow
• turbulence
• combustion
• heat leakage
• mechanical resistance

Therefore:

real engines are irreversible

and:

η_real < η_Carnot

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Comparing Carnot and Real Engines

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Example: Real vs Carnot Efficiency

A real engine operates between:

T_h = 800 K

and:

T_c = 300 K

Carnot limit:

η_Carnot = 1 − 300/800

η_Carnot = 1 − 0.375

η_Carnot = 62.5%

Suppose the real engine operates at:

40% efficiency

This is possible because:

40% < 62.5%

But an engine claiming:

70% efficiency

between the same temperatures would be impossible because:

70% > 62.5%

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Why Real Engines Cannot Be Carnot Engines

A perfectly reversible Carnot process must proceed through infinitesimally small differences in:

• temperature
• pressure

This means the processes must occur:

extremely slowly

But a practical engine must produce:

useful power

Power is:

P = W/t

If an engine took an infinitely long time to complete each cycle:

t → ∞

then its practical power output would approach:

P → 0

Thus perfect Carnot operation is not useful as a practical engine.

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Efficiency vs Power

This creates an important engineering trade-off.

To approach maximum reversible efficiency:

operate increasingly slowly

But to produce useful power:

operate at a finite rate

Real engineers therefore balance:

• efficiency
• power
• cost
• size
• reliability
• materials
• operating temperature

The most efficient theoretical engine is not necessarily the most practical engine.

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Carnot Cycle on a P–V Diagram

The Carnot P–V diagram contains four curves.

A → B

Isothermal expansion at T_h

Heat enters.

B → C

Adiabatic expansion

Temperature falls from T_h to T_c.

C → D

Isothermal compression at T_c

Heat leaves.

D → A

Adiabatic compression

Temperature rises from T_c to T_h.

The area enclosed by the loop represents:

net work produced per cycle

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Carnot Cycle on a Temperature-Entropy Diagram

The Carnot cycle can also be represented using a:

temperature-entropy (T–S) diagram

For an ideal Carnot cycle:

• isothermal processes appear horizontally
• reversible adiabatic processes occur at constant entropy and appear vertically

The result is a rectangular cycle on an idealized T–S diagram.

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Reversible Adiabatic Means Isentropic

During a reversible adiabatic process:

Q = 0

and no entropy is produced.

Therefore:

ΔS = 0

Such a process is called:

isentropic

Thus Stages 2 and 4 of the ideal Carnot cycle are:

reversible adiabatic = isentropic

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The Four Stages in Energy Terms

Stage 1

Isothermal expansion

Q_h enters.

Gas performs work.

Stage 2

Adiabatic expansion

No heat transfer.

Gas performs work and cools.

Stage 3

Isothermal compression

Work is done on gas.

Q_c leaves.

Stage 4

Adiabatic compression

No heat transfer.

Work is done on gas and it warms.

Then the cycle repeats.

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Why Heat Is Absorbed at T_h

During Stage 1, the working gas remains in thermal contact with the hot reservoir.

To make this transfer perfectly reversible, the temperature difference between reservoir and gas would need to be:

infinitesimally small

This allows heat transfer without entropy production in the ideal limit.

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Why Heat Is Rejected at T_c

During Stage 3, the working substance is in thermal contact with the cold reservoir.

Again, the temperature difference must approach:

zero

for perfectly reversible heat transfer.

The gas releases Q_c while remaining at T_c.

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Why 100% Carnot Efficiency Is Impossible

For:

η_Carnot = 1

we would require:

T_c/T_h = 0

Therefore either:

T_c = 0 K

or an unboundedly large T_h.

Neither provides a realizable finite-temperature heat engine.

Therefore:

η_Carnot < 100%

for engines operating between finite, physically realizable temperatures.

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Example: Very Hot Engine

Suppose:

T_h = 1500 K

and:

T_c = 300 K

η = 1 − 300/1500

η = 1 − 0.20

η = 80%

Even with an extremely hot reservoir, the maximum theoretical efficiency is still below 100%.

A real engine would achieve less than 80%.

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Engineering and High Temperatures

Increasing T_h can improve theoretical efficiency.

However, real materials place limits on operating temperatures.

At very high temperatures:

• metals weaken
• turbine blades can deform
• oxidation increases
• cooling becomes difficult
• components may melt or fail

Therefore engineers cannot simply increase T_h indefinitely.

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Engineering and Low Temperatures

Lowering T_c can also increase efficiency.

However, cooling the cold reservoir itself may require:

additional energy

and environmental conditions impose limits.

Therefore practical engines cannot arbitrarily lower T_c either.

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Steam Power Plants

Steam power plants do not operate on a perfect Carnot cycle.

They commonly use cycles related to the:

Rankine cycle

The system includes:

• boiler
• turbine
• condenser
• pump

The Carnot cycle remains useful because it establishes the maximum theoretical efficiency between given temperatures.

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Internal Combustion Engines

Gasoline engines are commonly modeled using idealized cycles such as the:

Otto cycle

Diesel engines can be modeled using the:

Diesel cycle

Gas turbines are often modeled using the:

Brayton cycle

These are not Carnot cycles.

But Carnot efficiency still provides a fundamental upper benchmark based on operating temperatures.

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Carnot Efficiency as a Benchmark

Suppose a real power plant has:

η_real = 45%

and its Carnot limit is:

η_Carnot = 60%

It may be useful to compare:

45/60 = 0.75

The real plant achieves about:

75% of the Carnot-limit efficiency

This does not mean the plant is 75% efficient overall.

Its actual efficiency remains:

45%

It means it achieves 75% of the maximum theoretical value available between those temperatures.

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Worked Example: Full Carnot Problem

A Carnot engine operates between:

T_h = 1000 K

and:

T_c = 400 K

It absorbs:

Q_h = 5000 J

Find the efficiency

η = 1 − 400/1000

η = 0.60

η = 60%

Find the work output

W = ηQ_h

W = 0.60 × 5000

W = 3000 J

Find Q_c

Q_c = Q_h − W

Q_c = 5000 − 3000

Q_c = 2000 J

Therefore:

Q_h = 5000 J

W = 3000 J

Q_c = 2000 J

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Worked Example: Testing an Engine Claim

An engineer claims an engine operating between:

700 K

and:

350 K

has an efficiency of:

60%

Find the Carnot limit:

η_Carnot = 1 − 350/700

η_Carnot = 0.50

η_Carnot = 50%

The claimed efficiency is:

60%

Since:

60% > 50%

the claim violates the Carnot limit.

Such an engine cannot operate as claimed between those two reservoirs.

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Worked Example: Comparing Two Engines

Engine A:

T_h = 600 K

T_c = 300 K

η_A = 1 − 300/600

η_A = 50%

Engine B:

T_h = 900 K

T_c = 300 K

η_B = 1 − 300/900

η_B ≈ 66.7%

Engine B has the higher theoretical maximum efficiency because it operates with the larger temperature ratio difference.

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Carnot Cycle and the Second Law

The Carnot cycle demonstrates several major consequences of the Second Law:

• heat engines require hot and cold reservoirs
• not all thermal energy can become work
• reversible processes establish maximum efficiency
• irreversible processes reduce efficiency
• entropy production reduces available useful work
• no engine can exceed the Carnot limit

The Carnot cycle therefore provides one of the clearest links between:

temperature, entropy, energy and efficiency

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Common Mistakes

Mistake: Saying the Carnot engine is a real engine design.
It is an idealized thermodynamic model.

Mistake: Saying the Carnot engine is 100% efficient.
Its efficiency is η = 1 − T_c/T_h and is below 100% for finite reservoir temperatures.

Mistake: Using Celsius in the Carnot equation.
Always use kelvin.

Mistake: Confusing the hot and cold temperatures.
T_h is the higher temperature and T_c is the lower temperature.

Mistake: Saying all four stages transfer heat.
The two adiabatic stages have Q = 0.

Mistake: Saying temperature changes during the isothermal stages.
Isothermal means constant temperature.

Mistake: Saying temperature stays constant during adiabatic expansion.
The gas cools because it performs work without receiving heat.

Mistake: Saying temperature stays constant during adiabatic compression.
The gas warms because work is done on it without heat leaving.

Mistake: Saying real engines can reach Carnot efficiency if friction is reduced enough.
Perfect Carnot efficiency requires completely reversible processes, an unattainable idealization.

Mistake: Saying a larger temperature difference always means a practical engine will automatically be more efficient.
Carnot efficiency sets the theoretical limit; real efficiency also depends on design and irreversibilities.

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Key Terms

Carnot cycle — ideal reversible thermodynamic cycle consisting of two isothermal and two adiabatic processes.

Carnot engine — ideal heat engine operating through a Carnot cycle.

Hot reservoir — high-temperature source supplying thermal energy Q_h.

Cold reservoir — lower-temperature region receiving Q_c.

Working substance — material undergoing expansion and compression during the cycle.

Isothermal process — process occurring at constant temperature.

Adiabatic process — process involving no heat transfer.

Isentropic process — process occurring at constant entropy; a reversible adiabatic process is isentropic.

Reversible process — ideal process producing no net entropy.

Carnot efficiency — maximum theoretical efficiency between two reservoir temperatures.

P–V diagram — graph of pressure against volume used to represent thermodynamic processes.

T–S diagram — graph of temperature against entropy.

Carnot's theorem — principle stating that no engine can exceed the efficiency of a reversible engine operating between the same temperatures.

Q_h — thermal energy absorbed from the hot reservoir.

Q_c — thermal energy rejected to the cold reservoir.

Net work (W) — useful work produced over one complete engine cycle.

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Key Takeaways

• The Carnot cycle represents an ideal reversible heat-engine cycle.
• It consists of four stages.
• Stage 1 is isothermal expansion at T_h.
• Stage 2 is adiabatic expansion from T_h to T_c.
• Stage 3 is isothermal compression at T_c.
• Stage 4 is adiabatic compression from T_c to T_h.
• Heat enters only during the high-temperature isothermal expansion.
• Heat leaves during the low-temperature isothermal compression.
• No heat is transferred during either adiabatic stage.
• The working substance returns to its initial state after each complete cycle.
• The net work produced equals Q_h − Q_c.
• On a P–V diagram, the enclosed area represents the net work per cycle.
• The Carnot engine is completely reversible.
• It therefore produces no net entropy.
• Carnot efficiency is η_Carnot = 1 − T_c/T_h.
• Temperatures in the Carnot equation must be measured in kelvin.
• The Carnot efficiency depends only on the two reservoir temperatures.
• Increasing T_h increases the theoretical maximum efficiency.
• Decreasing T_c increases the theoretical maximum efficiency.
• A Carnot engine operating between finite temperatures cannot be 100% efficient.
• No real engine operating between the same temperatures can exceed Carnot efficiency.
• Real engines have lower efficiency because they contain irreversible processes.
• Friction, turbulence and finite temperature differences are important sources of irreversibility.
• Real engines must operate at finite rates to produce useful power.
• A perfectly reversible Carnot engine would operate extremely slowly and is therefore an idealization.
• Steam turbines, gasoline engines, diesel engines and gas turbines use practical cycles rather than the Carnot cycle.
• The Carnot cycle provides a benchmark against which real engines can be evaluated.
• Carnot's theorem is a fundamental consequence of the Second Law of Thermodynamics.

Check Your Understanding

1. What is the Carnot cycle?
2. Why is the Carnot cycle important in thermodynamics?
3. Name the four stages of the Carnot cycle in order.
4. What happens during isothermal expansion?
5. Why does the temperature remain constant during isothermal expansion?
6. What happens to the temperature during adiabatic expansion?
7. Why does the gas cool during adiabatic expansion?
8. What happens during isothermal compression?
9. What happens during adiabatic compression?
10. During which two stages is Q = 0?
11. What does the area enclosed by a Carnot cycle on a P–V diagram represent?
12. Write the equation for Carnot efficiency.
13. Why must temperatures in the Carnot efficiency equation be measured in kelvin?
14. A Carnot engine operates between 800 K and 400 K. Calculate its maximum efficiency.
15. A Carnot engine operates between 900 K and 300 K and absorbs 6000 J. Calculate its work output.
16. For the engine in Question 15, calculate Q_c.
17. Explain why a Carnot engine cannot reach 100% efficiency.
18. Why are real engines less efficient than Carnot engines operating between the same temperatures?
19. Explain why a perfectly reversible Carnot engine would be impractical as a real engine.
20. A real engine operating between 700 K and 350 K claims an efficiency of 55%. Use the Carnot limit to evaluate whether this claim is physically possible.