
Content Curator
Carnot's Theorem is a basic principle of thermodynamics that tells us how efficient a heat engine can be at its best.
- Carnot’s theorem is also known as Carnot’s rule or Carnot’s principle.
- The theorem states that any heat engine does not have 100% efficiency. For maximum efficiency of the heat engine, the processes involved should be reversible.
- This theorem is the basis of modern thermodynamics. It is based on the idea that some processes can be reversed.
- It's used in engineering, physics, and chemistry, among other fields, and it's still a key idea in the study of energy and heat transfer.
| Table of Content |
Key Terms: Carnot’s theorem, Heat engine, Entropy change, Second law of thermodynamics, Carnot cycle, Efficiency of heat engine, Reversible and irreversible process.
Carnot’s Theorem
[Click Here for Sample Questions]
In 1924, French scientist Nicolas Leonard Sadi Carnot developed a theory that specifies limits on the maximum efficiency that any heat engine can obtain, which is famously known as Carnot’s theorem. He is also known as the "Father of Thermodynamics" because of his work on the Carnot theorem and cycle.
Carnot’s theorem states that
- All heat engines operating between the two heat reservoirs cannot have efficiencies greater than a reversible heat engine operating between the same reservoirs.
- The efficiency of the Carnot engine is independent of the working substance.
All reversible heat engine operating between two temperatures is called Carnot engine and the sequences of steps constituting one cycle are called the Carnot cycle.
The Maximum efficiency of a heat engine, also known as the efficiency of a Carnot engine is given as,
ηmax = ηcarnot = \(\frac{T_H-T_C}{T_H} = 1-\frac{T_C}{T_H}\)
Where
- ƞ is the ratio of the work done by the engine to the heat drawn out of the hot reservoir.
- TC is the absolute temperature of the cold reservoir.
- TH is the absolute temperature of the hot reservoir.
Reversible Process
A reversible process is a process in which at any stage of the process it can be traversed back in the opposite direction in such a way that the system passes through exactly the same conditions at every step in the reverse process as in the direct process.
The basic requirement for a process to be reversible is:
- The process should take place very slowly.
- The pressure difference between the working substance and surrounding at any stage of the process should be very small.
- There should be no friction.
- There should be no loss of energy.
Irreversible Process
Irreversible process may be defined as the process in which the system cannot be retraced to its initial state.
Thermodynamics processes that occur in nature are all irreversible processes. Some examples of irreversible process are
- Heat generation during the flow of current through a resistor.
- Work done against friction.
- Diffusion of two gases.
- Exchange of heat between bodies at different temperatures.
Carnot’s Theorem Proof
[Click Here for Sample Questions]
Let's say that there is a heat engine X that works between two reservoirs at temperatures T1 and T2 respectively. The heat engine X makes work W by moving heat Q1 from the hot reservoir (source) at T1 to the cold reservoir (sink) at temperature T2.

Heat Engine
Also, consider a Carnot engine that works between the same two reservoirs at temperatures T1 and T2 respectively. The Carnot engine goes through two processes that are isothermal and two processes that are adiabatic.
The efficiency of the Carnot engine is given by:
ƞcarnot = 1 – \(\frac{T_2}{T_1}\) ………………(i)
Let ΔS be the net change in universe entropy.
At steady temperature T1, the source emits heat Q1, and the change in its entropy is given by
ΔS1 = – \(\frac{Q_1}{T_1}\)
At steady temperature T2, the sink receives heat Q2, and the change in its entropy is given by
ΔS2 = \(\frac{Q_2}{T_2}\)
Net change in universe’s entropy is given by
ΔS = ΔS1 + ΔS2
⇒ ΔS = – \(\frac{Q_1}{T_1}\) + \(\frac{Q_2}{T_2}\)
From Second law of thermodynamics, we have
ΔS ≥ 0
- \(\frac{Q_1}{T_1}\) + \(\frac{Q_2}{T_2}\) ≥ 0
\(\frac{Q_2}{T_2}\) ≥ \(\frac{Q_1}{T_1}\)
\(\frac{Q_2}{Q_1}\) ≥ \(\frac{T_2}{T_1}\)
Subtracting both sides by 1, we get
1 – \(\frac{Q_2}{Q_1}\) ≥ 1 – \(\frac{T_2}{T_1}\)
Left side equation of the above expression shows the efficiency of heat engine X, and from equation (i), right side equation represents the efficiency of Carnot engine. i.e.
ƞX ≥ ƞcarnot
So, we can say that a heat engine running between two heat reservoirs can't be more efficient than a Carnot engine running between the same two reservoirs.
Limitations of Carnot Theorem
[Click Here for Sample Questions]
The Carnot Theorem is crucial to understanding heat engine thermodynamics, but its limits must be considered when applying it to real-world systems.
- Only idealized thermodynamic cycles use the Carnot Theorem. That doesn't work for imperfect systems that lose energy differently.
- The theory assumes thermal equilibrium, which may not be true in practice.
- The Carnot Theorem implies the engine's working fluid is a perfect gas, which many fluids are not.
- The theorem states that heat transfer between the system and reservoirs can always be bidirectional, which is not necessarily true.
- The Carnot Theorem only applies to systems between two reservoirs with fixed temperatures. It's unsuitable for multi-temperature systems.
- The theorem ignores friction, heat leaks, and other energy losses in real-world systems.
Also Read:
| Related Articles | ||
|---|---|---|
| Types of Motors | Types of Switches | Types of DC Motor |
| Types of Transistors | Types of Friction | Balanced Force |
Applications of Carnot Theorem
[Click Here for Sample Questions]
The Carnot Theorem has many applications in engineering, physics, and materials research, despite its limits.
- The Carnot Theorem is needed to develop and improve heat engines like steam engines and gas turbines by figuring out how efficient they can be at their theoretical best.
- The Carnot Theorem is a way to figure out how much energy is needed or given off during phase changes like melting and boiling.
- This Theorem can be used to study thermoelectric materials, which directly convert temperature differences into electricity, and to improve thermoelectric devices.
Things to Remember
- Carnot’s theorem states that no heat engine has greater efficiency than a Carnot engine.
- Carnot's theory gives engineers a theoretical maximum heat engine efficiency to use when constructing and enhancing engines.
- The efficiency of the Carnot engine is given by: ηcarnot = 1 – \(\frac{T_2}{T_1}\)
- Entropy is defined as the system’s thermal energy per unit temperature that is unavailable for doing useful work.
- Carnot's theorem calculates thermoelectric devices' maximum efficiency.
- Carnot's theorem can be used to study solid-to-liquid transitions and reservoir temperatures' effects on heat engine efficiency.
- Carnot's theorem can improve solar and wind power efficiency.
- Carnot's theorem can study Earth's climate system's efficiency and temperature changes.
- The efficiency of a heat engine is given by η = 1 – \(\frac{Q_2}{Q_1}\)
Sample Questions
Ques. What is the efficiency of a Carnot engine operating between two heat reservoirs at temperatures of 340 oC and 70 oC? (2 Marks)
Ans. The efficiency of a Carnot engine operating between two heat reservoirs at temperatures of T1 = 340 oC and T2 = 70 oC is given by the formula
ηcarnot = 1 - T2/T1
⇒ ηcarnot = 1 - 343.15/613.15 = 0.440, or 44.0%.
Ques. A heat engine with a 55% efficiency performs 70 kJ of work every Carnot cycle. What amount of heat is rejected in each cycle? (3 Marks)
Ans. Use the formula for the efficiency of a heat engine to find the amount of heat rejected in each cycle:
η = W/Q1
where
- η is the efficiency,
- W is the work performed, and
- Q1 is the heat absorbed.
The efficiency is 55%, then:
0.55 = 70kJ/Q1
Solving for Q1,
Q1 = 70kJ/0.55 = 127.27 kJ
Therefore, the heat engine absorbs 127.27 kJ of heat every Carnot cycle, and since it performs 70 kJ of work, the amount of heat rejected in each cycle is
Q2 = Q1 - W = 127.27 kJ - 70 kJ = 57.27 kJ
So, the heat engine rejects 57.27 kJ of heat every Carnot cycle.
Ques. What is the amount of heat absorbed by a Carnot engine operating between heat reservoirs at temperatures of T1 = 400 K and T2 = 560 K that produces 2 kJ of mechanical work during each cycle? (5 Marks)
Ans. The Carnot engine operates between heat reservoirs at temperatures of T1 = 400K and T2 = 560K and produces 2 kJ of mechanical work during each cycle.
Using the formula for the efficiency of a Carnot engine, η = 1 – T2/T1, calculate that the maximum theoretical efficiency of the engine is
η = 1 – 560K/400K = 0.4, or 40%.
The work produced during each cycle is 2 kJ, so the amount of heat absorbed by the engine from the hot reservoir during each cycle is:
Q1 = W/η = 2 kJ/0.4 = 5 kJ
Since the engine operates in a cycle, it must also reject some heat to the cold reservoir. Therefore, the amount of heat rejected by the cold reservoir during each cycle is:
Q2 = Q1 – W = 5 kJ – 2 kJ = 3 kJ
Therefore, the amount of heat transmitted to the engine by the hot reservoir is 5 kJ, and the amount of heat transmitted to the cold reservoir is 3 kJ.
Ques. What working fluid is used in a Carnot cycle? (2 Marks)
Ans. In a Carnot cycle, the working fluid can be any substance that can be compressed and expanded in an isentropic way and then compressed and expanded in an isothermal way. In real life, ideal gases are often used as working fluids in Carnot engines because they closely match how the Carnot cycle works in theory. But as a working fluid, you can use anything that meets the requirements of a Carnot cycle.
Ques. What is the COP of a reversible heat pump? (5 Marks)
Ans. The coefficient of performance (COP) of a reversible heat pump is given by:
COP = Qh / W
where
- Qh is the heat absorbed from the hot reservoir and
- W is the work done on the system.
In a reversible heat pump, the cycle can be run in reverse to act as a refrigerator, so the COP can also be expressed as
COP = Qc / W
where Qc is the heat rejected to the cold reservoir.
For a reversible heat pump, the COP is given by the Carnot efficiency, which is:
COP = Th / (Th - Tc)
where
- Th is the absolute temperature of the hot reservoir and
- Tc is the absolute temperature of the cold reservoir.
Therefore, the COP of a reversible heat pump is limited by the temperature difference between the hot and cold reservoirs, and it can approach but never exceed the Carnot efficiency.
Ques. What is the efficiency of the reversible heat engine? (3 Marks)
Ans. The efficiency of a reversible heat engine operating between two heat reservoirs at temperatures T1 (the high-temperature reservoir) and T2 (the low-temperature reservoir) is given by the Carnot efficiency, which is given by:
ηcarnot = 1 - T2 / T1
where ηcarnot is the efficiency of the engine expressed as a decimal or percentage.
Ques. Give examples of a reversed heat engine. (3 Marks)
Ans. A reversible heat engine theoretically converts all heat energy from a hot reservoir into work and returns all heat energy from a cold reservoir. Some systems are nearly reversible, but heat engines cannot be.
- An idealized heat engine, a Carnot engine works between two temperature reservoirs to maximize efficiency.
- Compressing and expanding gas under pressure in gas turbine Brayton cycle engines converts heat energy into mechanical work.
- The Seebeck effect converts temperature differences into electrical energy in a thermoelectric generator, which is reversible under certain conditions.
- Quantum mechanics power reversible nanoscale quantum heat engines.
Ques. What is a diesel cycle? (3 Marks)
Ans. Diesel engines transform fuel chemical energy into mechanical work using the thermodynamic Diesel cycle. Rudolf Diesel's late-1800s internal combustion engine cycle powers several diesel engines.
Diesel has four stages:
- Air is pulled into the engine cylinder and compressed as the piston rises.
- Compression: Squeezing air increases its pressure and temperature, improving combustion.
- Combustion: Fuel ignites and burns in compressed air, expanding the air and pushing the piston down.
- Exhaust: The piston expels combustion gases during the exhaust stroke. Completes the cycle.
Ques. What is the efficiency and work ratio of a simple gas turbine cycle? (5 Marks)
Ans. Compression ratio, turbine inlet temperature, and working fluid-specific heat ratio affect basic gas turbine cycle efficiency and work ratio. Simple gas turbine cycle efficiency and work ratio equations:
Efficiency of a simple gas turbine cycle:
η = (T3 - T2) / (T3 / (r0.4 – 1) - T2)
where
- η is the efficiency of the cycle,
- T2 is the temperature of the ambient air,
- T3 is the turbine inlet temperature, and
- r is the compression ratio.
Work ratio of a simple gas turbine cycle:
WR = (h3 – h2) / (h4 - h3)
where WR is the cycle's work ratio, h2 is the compressed air's enthalpy at the combustor's input, h3 is the exhaust gases at the turbine's inlet, and h4 is at the turbine's outlet.
Ques. Name the gas that has a minimum molecular mass. (2 Marks)
Ans. Hydrogen (H2) has the minimum molecular mass of any gas, with a molecular weight of approximately 2.
Ques. What is the equation for the first law of thermodynamics? (3 Marks)
Ans. The first law of thermodynamics asserts that energy cannot be generated or destroyed but can be altered. The first law of thermodynamics equation:
ΔU = Q - W
where
- U is the system's internal energy change,
- Q is its heat addition, and
- W is its work.
This equation asserts that a system's internal energy change is equal to its heat contributed minus its work. Science and engineering apply the first law of thermodynamics to conserve energy.
Ques. What is the entropy change for an ideal Carnot cycle? (3 Marks)
Ans. Isothermal and adiabatic processes have negligible entropy change in an ideal Carnot cycle.
- The system's entropy change is compensated by the hot reservoir's entropy change throughout the isothermal expansion process.
- No heat is exchanged during adiabatic expansion, hence the system's entropy remains constant.
- The cold reservoir absorbs all the heat emitted by the system during isothermal compression, resulting in no net change in entropy.
- No heat is exchanged during adiabatic compression, hence the system's entropy remains constant.
- An ideal Carnot cycle has zero net entropy change.
For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates
Do Check Out:






Comments