Carnot Cycle
Maximum Efficiency heat engine — Formula, PV Diagram, Derivation & Solved Problems
Last Updated: March 2026
📌 Key Takeaways
- The Carnot cycle is a theoretical ideal — it represents the maximum possible efficiency any heat engine can achieve between two given temperatures.
- Four reversible processes: Isothermal expansion → Adiabatic expansion → Isothermal compression → Adiabatic compression.
- Efficiency formula: ηCarnot = 1 − TL/TH (temperatures must be in Kelvin).
- Carnot efficiency depends only on reservoir temperatures — not on the working fluid, engine design, or cycle details.
- Carnot’s theorem: No heat engine operating between two reservoirs can be more efficient than a Carnot engine between the same reservoirs.
- All real engines operate below Carnot efficiency due to irreversibilities like friction and finite-rate heat transfer.
1. What is the Carnot Cycle?
The Carnot cycle is a theoretical thermodynamic cycle proposed by French engineer Sadi Carnot in 1824. It describes an idealised heat engine that operates with the highest possible efficiency between two thermal reservoirs — a hot source at temperature TH and a cold sink at temperature TL.
Carnot’s insight was revolutionary: he showed that the maximum efficiency of a heat engine depends only on the temperatures of the two reservoirs, not on the working substance (gas, steam, air) or the mechanical design of the engine. This result placed a fundamental physical limit on all heat engines — a limit imposed not by engineering constraints, but by the laws of nature themselves.
The Carnot cycle is composed entirely of reversible processes — processes that can be reversed without any change in the system or surroundings. Since all real processes involve some degree of irreversibility (friction, rapid changes, heat transfer across finite temperature differences), the Carnot cycle cannot be built in practice. It serves as a theoretical benchmark — the gold standard against which real engines are measured.
2. The Four Processes — Step by Step
The Carnot cycle consists of four sequential reversible processes. Imagine a gas enclosed in a piston-cylinder assembly:
Process 1→2: Reversible Isothermal Expansion
The gas is placed in contact with the hot reservoir at temperature TH. The gas expands slowly while absorbing heat QH from the reservoir. Because the temperature remains constant (isothermal), the internal energy of an ideal gas does not change, and all the absorbed heat is converted into work done by the expanding gas.
- Temperature: constant at TH
- Volume: increases (V&sub1; → V&sub2;)
- Pressure: decreases
- Heat: QH absorbed from hot reservoir
- Work: done BY the gas (positive)
Process 2→3: Reversible Adiabatic Expansion
The gas is now insulated (no heat transfer) and continues to expand. Since no heat enters, the expansion reduces the gas temperature from TH to TL. The gas does work at the expense of its internal energy.
- Temperature: decreases from TH to TL
- Volume: increases (V&sub2; → V&sub3;)
- Heat: Q = 0 (adiabatic)
- Work: done BY the gas (positive)
Process 3→4: Reversible Isothermal Compression
The gas is placed in contact with the cold reservoir at temperature TL. The gas is compressed slowly, and the heat generated by compression is rejected to the cold reservoir as QL. The temperature remains constant at TL.
- Temperature: constant at TL
- Volume: decreases (V&sub3; → V&sub4;)
- Heat: QL rejected to cold reservoir
- Work: done ON the gas (negative)
Process 4→1: Reversible Adiabatic Compression
The gas is insulated again and compressed further. The compression raises the gas temperature from TL back to TH, returning the system to its initial state and completing the cycle.
- Temperature: increases from TL to TH
- Volume: decreases (V&sub4; → V&sub1;)
- Heat: Q = 0 (adiabatic)
- Work: done ON the gas (negative)
| Process | Type | Temperature | Heat | Work |
|---|---|---|---|---|
| 1→2 | Isothermal Expansion | TH (constant) | QH absorbed | Done by gas (+) |
| 2→3 | Adiabatic Expansion | TH → TL | 0 | Done by gas (+) |
| 3→4 | Isothermal Compression | TL (constant) | QL rejected | Done on gas (−) |
| 4→1 | Adiabatic Compression | TL → TH | 0 | Done on gas (−) |
3. PV and TS Diagrams
PV Diagram (Pressure-Volume)
On a PV diagram, the Carnot cycle appears as a closed loop formed by two isotherms (horizontal curves at TH and TL) and two adiabats (steeper curves connecting them). The area enclosed by the loop represents the net work output of the cycle. The isothermal curves are hyperbolic (PV = constant for ideal gas at constant T), while the adiabatic curves are steeper (PVγ = constant).
TS Diagram (Temperature-entropy)
On a TS diagram, the Carnot cycle is a simple rectangle:
- The top horizontal line (1→2) represents isothermal heat addition at TH — entropy increases from S&sub1; to S&sub2;.
- The right vertical line (2→3) represents adiabatic expansion — entropy remains constant (isentropic), temperature drops from TH to TL.
- The bottom horizontal line (3→4) represents isothermal heat rejection at TL — entropy decreases from S&sub2; back to S&sub1;.
- The left vertical line (4→1) represents adiabatic compression — entropy remains constant, temperature rises from TL to TH.
The rectangular shape on the TS diagram makes efficiency calculations intuitive:
QH = TH × (S&sub2; − S&sub1;) — area under the top line
QL = TL × (S&sub2; − S&sub1;) — area under the bottom line
Wnet = QH − QL = (TH − TL) × (S&sub2; − S&sub1;) — area of the rectangle
Exam tip: The TS diagram is almost always easier to work with for Carnot cycle problems than the PV diagram. The rectangular shape makes it straightforward to identify heat input, heat rejection, and net work.
4. Carnot Efficiency — Derivation
The thermal efficiency of any heat engine is defined as the ratio of net work output to heat input:
General Efficiency
η = Wnet / QH = (QH − QL) / QH = 1 − QL/QH
For the Carnot cycle, using the TS diagram results:
QH = TH(S&sub2; − S&sub1;) and QL = TL(S&sub2; − S&sub1;)
Therefore: QL/QH = TL/TH
Carnot Efficiency
ηCarnot = 1 − TL / TH
Where:
- ηCarnot = maximum possible thermal efficiency (dimensionless, or × 100 for percentage)
- TL = absolute temperature of cold reservoir (K)
- TH = absolute temperature of hot reservoir (K)
⚠️ Temperatures MUST be in Kelvin. Using Celsius gives wrong results.
Key observations from the formula:
- Efficiency increases as TH increases (hotter source = more efficiency).
- Efficiency increases as TL decreases (colder sink = more efficiency).
- Efficiency = 100% only if TL = 0 K (absolute zero), which is physically unattainable.
- Efficiency = 0% if TH = TL (no temperature difference, no work possible).
5. Carnot’s Theorem
Carnot’s theorem makes two powerful statements:
Statement 1: No heat engine operating between two given thermal reservoirs can have a higher efficiency than a reversible (Carnot) engine operating between the same two reservoirs.
Statement 2: All reversible engines operating between the same two reservoirs have the same efficiency, regardless of the working fluid or cycle details.
These statements are direct consequences of the Second Law of Thermodynamics. They can be proved by contradiction — assuming a more-efficient engine exists leads to a violation of either the Kelvin-Planck or Clausius statement.
Practical implication: when an engineer designs a power plant operating between a combustion temperature of 800 K and an ambient temperature of 300 K, the Carnot efficiency is 1 − 300/800 = 62.5%. The actual plant might achieve 35–40% efficiency. The gap between actual and Carnot efficiency measures how much room exists for improvement through reducing irreversibilities.
6. Reversed Carnot Cycle — Refrigerator & Heat Pump
When the Carnot cycle is operated in reverse, it becomes the most efficient possible refrigerator or heat pump:
| Device | Purpose | Performance Measure | Carnot Value |
|---|---|---|---|
| Carnot Refrigerator | Remove heat from cold space | COPR = QL / W | COPR = TL / (TH − TL) |
| Carnot Heat Pump | Deliver heat to warm space | COPHP = QH / W | COPHP = TH / (TH − TL) |
Useful Relationship
COPHP = COPR + 1
The COP of a heat pump is always exactly 1 more than the COP of a refrigerator operating between the same temperatures.
Note that COP values can be (and often are) greater than 1 — a heat pump with COP = 4 delivers 4 kW of heating for every 1 kW of electrical work input. This does not violate energy conservation; the heat pump is moving energy from outside, not creating it.
7. Worked Numerical Examples
Example 1: Basic Carnot Efficiency
Problem: A Carnot engine operates between a hot reservoir at 527°C and a cold reservoir at 27°C. Calculate the maximum efficiency.
Solution
Convert to Kelvin: TH = 527 + 273 = 800 K, TL = 27 + 273 = 300 K
ηCarnot = 1 − TL/TH = 1 − 300/800 = 1 − 0.375
ηCarnot = 0.625 = 62.5%
No heat engine operating between 800 K and 300 K can exceed 62.5% efficiency.
Example 2: Finding Heat Rejected
Problem: A Carnot engine receives 1,000 kJ of heat from a source at 600 K and rejects heat to a sink at 300 K. Calculate the work output and heat rejected.
Solution
η = 1 − 300/600 = 0.5 = 50%
Wnet = η × QH = 0.5 × 1000 = 500 kJ
QL = QH − Wnet = 1000 − 500 = 500 kJ
The engine converts half the input heat into work and rejects the other half to the cold reservoir.
Example 3: Carnot Refrigerator COP
Problem: A Carnot refrigerator maintains a cold space at −23°C while rejecting heat to the environment at 27°C. Calculate the COP and the work input needed to remove 5 kJ of heat from the cold space.
Solution
TL = −23 + 273 = 250 K, TH = 27 + 273 = 300 K
COPR = TL / (TH − TL) = 250 / (300 − 250) = 250/50 = 5
W = QL / COPR = 5 / 5 = 1 kJ
Only 1 kJ of work is needed to remove 5 kJ of heat — a COP of 5 means the refrigerator moves 5 times more energy than it consumes.
Example 4: GATE-Style Problem — Unknown Temperature
Problem: A Carnot engine has an efficiency of 40%. If the temperature of the hot reservoir is increased by 100 K while the cold reservoir temperature remains unchanged, the new efficiency becomes 50%. Find both reservoir temperatures.
Solution
From the first condition: 0.4 = 1 − TL/TH → TL/TH = 0.6 → TL = 0.6 TH … (i)
From the second condition: 0.5 = 1 − TL/(TH + 100) → TL/(TH + 100) = 0.5 → TL = 0.5(TH + 100) … (ii)
From (i) and (ii): 0.6 TH = 0.5 TH + 50
0.1 TH = 50 → TH = 500 K
TL = 0.6 × 500 = TL = 300 K
Verification: η&sub1; = 1 − 300/500 = 40% ✓ and η&sub2; = 1 − 300/600 = 50% ✓
8. Why Real Engines Fall Short
No real engine achieves Carnot efficiency. Here is why:
| Source of Irreversibility | Effect on Efficiency |
|---|---|
| Friction in pistons, bearings, and turbine blades | Converts useful work into waste heat, reducing net work output |
| Finite-rate heat transfer (heat flows across a temperature difference) | Requires Tgas < TH during heat addition and Tgas > TL during rejection, reducing effective ΔT |
| Non-quasi-static processes (rapid compression/expansion) | Creates pressure and temperature gradients within the gas, generating entropy |
| Heat loss through engine walls and insulation gaps | Reduces the heat available for conversion to work |
| Combustion irreversibility (in IC engines) | Chemical reactions are inherently irreversible, destroying availability |
Typical real-world efficiencies compared to Carnot:
| Engine Type | Typical Actual Efficiency | Carnot Efficiency (same T range) |
|---|---|---|
| Petrol (gasoline) engine | 25–30% | ~55–60% |
| Diesel engine | 35–45% | ~60–65% |
| Coal-fired power plant | 33–40% | ~60–65% |
| Combined cycle gas turbine | 55–62% | ~70–75% |
| Nuclear power plant | 30–35% | ~45–50% |
9. Common Mistakes Students Make
- Using Celsius instead of Kelvin: This is the single most common error in Carnot problems. η = 1 − 27/527 = 94.9% is WRONG. Correct: η = 1 − 300/800 = 62.5%. Always convert to Kelvin first.
- Claiming 100% efficiency is possible: Carnot efficiency equals 100% only when TL = 0 K (absolute zero), which is physically unattainable according to the Third Law of Thermodynamics.
- Confusing COP with efficiency: COP applies to refrigerators and heat pumps and can exceed 1. Thermal efficiency applies to heat engines and is always less than 1. These are different performance measures.
- Assuming real engines can reach Carnot efficiency: Carnot efficiency is a theoretical upper bound. Real engines always fall short due to irreversibilities. Stating that a real engine achieves exactly Carnot efficiency is physically incorrect.
- Forgetting that Carnot efficiency is independent of working fluid: The formula η = 1 − TL/TH applies regardless of whether the engine uses air, steam, helium, or any other substance. This is a direct consequence of Carnot’s theorem.
10. Frequently Asked Questions
What is the Carnot cycle?
The Carnot cycle is a theoretical thermodynamic cycle consisting of four reversible processes (isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression) that achieves the maximum possible efficiency for a heat engine operating between two thermal reservoirs. It was proposed by Sadi Carnot in 1824 and remains the fundamental benchmark for all heat engine performance.
What is the formula for Carnot efficiency?
ηCarnot = 1 − TL/TH, where TL and TH are the absolute temperatures (in Kelvin) of the cold and hot reservoirs respectively. This formula gives the upper limit of efficiency — no heat engine can exceed this value for the given temperature limits.
Why can no real engine achieve Carnot efficiency?
The Carnot cycle requires all processes to be perfectly reversible — zero friction, infinitely slow operation, and heat transfer across zero temperature difference. These are physical impossibilities. Every real engine has friction in moving parts, operates at finite speeds, and transfers heat across finite temperature differences. Each of these generates entropy and pushes the efficiency below the Carnot limit.