The Carnot Cycle and the Ceiling on Efficiency
The reversible Carnot cycle has two isothermal and two adiabatic legs. It sets the maximum efficiency for engines exchanging heat only with two reservoirs at constant temperatures. The limit depends on their temperatures in kelvin.
What a learner can do afterwards
- Describes the four Carnot legs, their heat flows and the work represented by a clockwise pV loop
- Computes the Carnot efficiency from the two reservoir temperatures
- Argues why no engine between the same reservoirs can do better
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An isothermal leg keeps temperature fixed; an adiabatic leg exchanges no heat. For an ideal gas, the Carnot engine follows four reversible legs: hot isothermal expansion absorbs heat; adiabatic expansion cools the gas; cold isothermal compression releases heat; adiabatic compression warms it back to its starting state. A reversible process can be undone with no net change left in the system and surroundings.
On a pV diagram, pressure is vertical and volume is horizontal. Expansion moves right; compression moves left. For an ideal-gas Carnot engine, the hot isotherm lies above the cold one. Adiabats connect them and are steeper than the isotherms at their meeting points. The engine follows a clockwise loop. The enclosed area gives net work done by the gas per cycle, not efficiency.
Let Th and Tc be the hot and cold reservoir temperatures, with 0 < Tc < Th. For heat exchange only with these two fixed-temperature reservoirs, maximum efficiency is 1 - Tc/Th. Use kelvin: add 273.15 to each Celsius value. If Qh is absorbed heat, maximum work is (1 - Tc/Th) times Qh. A reversible engine reaches the limit. Raising Th at fixed Tc, or lowering Tc at fixed Th, raises the limit.
Between 900 K and 300 K, maximum efficiency is 1 - 300/900 = 2/3. With 900 J absorbed per cycle, maximum work is 600 J and minimum rejected heat is 900 - 600 = 300 J. The bars show operation at this limit. An engine producing less work from the same heat input rejects more heat. Energy amounts are in joules; efficiency at the limit is 600/900.
Suppose a rival beats Carnot between the same reservoirs. Run it forward and drive a reversed Carnot machine with all its work. Match their work amounts. For the same work, the rival absorbs less heat from the hot reservoir and rejects less heat to the cold reservoir. Together, the machines move heat from cold to hot with no net work. This violates the second law. Reversibility allows this comparison. All reversible engines share the limit; irreversible engines fall below it.
Two isotherms plus two adiabats set the ceiling of one minus cold over hot, measured in kelvin.
This lesson and its questions were written for this dataset. OpenStax thermodynamics material was consulted for scientific facts; its prose and figures are not reproduced.
The replacement readings were written for this dataset after factual study of OpenStax thermodynamics material. The retained video reference points to Khan Academy. Video terms have not been verified.
More to read
Following a reversible engine cycle
Picture an ideal gas under a movable piston. A hot reservoir and a cold reservoir keep fixed temperatures. The gas starts by expanding at the hot temperature while absorbing heat. It then expands under insulation, doing work while its temperature falls to the cold temperature.
The return has two compression legs. At the cold temperature, compression releases heat to the cold reservoir. An insulated compression then raises the gas temperature and restores the initial state. Both insulated legs exchange zero heat.
Each leg must be reversible for this to be a Carnot cycle. Reversibility concerns the system and its surroundings together: reversing a process must undo its effects without leaving another net change.
On a pressure-volume diagram, expansion moves toward larger volume and compression toward smaller volume. The engine follows a clockwise loop. Its enclosed area is the difference between expansion work and compression work. This measures net work per cycle, not the fraction called efficiency.
Calculating the work limit
For a cyclic engine, absorbed heat Qh and rejected heat Qc are positive amounts. Net work is W = Qh - Qc, and efficiency is W/Qh.
A reversible engine exchanging heat only with two fixed-temperature reservoirs has Qc/Qh = Tc/Th. Its efficiency is therefore 1 - Tc/Th. No engine using only those reservoirs can exceed this value.
For example, reservoirs at 1000 K and 400 K give a ceiling of 1 - 400/1000 = 0.60. With 700 J absorbed per cycle, maximum work is 0.60 × 700 = 420 J. At that limit, 280 J is rejected.
Use absolute temperatures. A Celsius value becomes a kelvin value by adding 273.15. Raising the hot temperature while holding the cold temperature fixed raises the ceiling. Lowering the cold temperature at fixed hot temperature also raises it. If both temperatures change, calculate their ratio; comparing their difference is not enough.
Why a better rival would break the second law
Suppose an engine were more efficient than Carnot between the same two reservoirs. Run that engine forward and use its entire work output to run a reversible Carnot machine backward. Choose their operating amounts so the work supplied equals the work required.
For equal work, the proposed engine would absorb less heat from the hot reservoir. It would also reject less heat to the cold reservoir, because rejected heat equals absorbed heat minus work.
The backward Carnot machine would deliver more heat to the hot reservoir than the proposed engine removes. It would remove more heat from the cold reservoir than the proposed engine returns. With the work amounts canceled, the combined devices would transfer heat from cold to hot without any other net effect.
The refrigerator form of the second law forbids that result. The proposed engine cannot beat Carnot. Comparing reversible engines in both directions also shows that they must have equal efficiencies between the same reservoirs.
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