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The Carnot Cycle (HL) (HL IB Physics)

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Katie M

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Katie M

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The Carnot Cycle

  • A thermodynamic system that runs at its greatest possible efficiency follows a cycle called the Carnot cycle

2-4-8-carnot-cycle-diagram

The four stage Carnot cycle of a gas in a piston

2-4-8-carnot-cycle-pv-diagram

p-V diagram for the Carnot cycle. The enclosed area equals the work done after one cycle

  • The Carnot cycle is an idealised and reversible process
  • It consists of four stages:

1.  Isothermal expansion

    • The gas absorbs heat Q subscript H from a hot reservoir at temperature T subscript H
    • Work is done by the gas as it expands i.e. volume increases, temperature is constant open parentheses increment T space equals space 0 close parentheses
    • Work done by the gas = heat gained Q subscript H

2.  Adiabatic expansion

    • The gas continues to expand
    • The gas does work on the surroundings as its volume increases and pressure decreases
    • The gas cools down from T subscript H to T subscript C, but no thermal energy is transferred open parentheses increment Q space equals space 0 close parentheses

3.  Isothermal compression

    • The gas is compressed and transfers heat Q subscript C to a cold reservoir at temperature T subscript C
    • Work is done on the gas as it is compressed i.e. volume decreases, temperature is constant open parentheses increment T space equals space 0 close parentheses
    • Work done on the gas = heat lost Q subscript C

4.  Adiabatic compression

    • The gas continues to be compressed
    • Work is done on the gas as its volume decreases and pressure increases
    • The gas heats up from T subscript C to T subscript H, but no thermal energy is transferred open parentheses increment Q space equals space 0 close parentheses
  • At the end of the fourth stage, the gas has returned to its original state and the cycle can be repeated as many times as needed

2-4-8-carnot-cycle-ts-diagram

The variation of temperature and entropy throughout the Carnot cycle

  • As the efficiency of a thermodynamic system increases, the difference between the temperatures of the hot and cold reservoirs increases
  • The maximum theoretical efficiency of a heat engine using the Carnot cycle is:

eta subscript C space equals space 1 space minus space T subscript C over T subscript H

  • Where:
    • eta subscript C = maximum theoretical efficiency (Carnot cycle only)
    • T subscript C = temperature in the cold reservoir (K)
    • T subscript H = temperature in the hot reservoir (K)

Worked example

In an idealised heat engine, the hot and cold reservoirs are held at temperatures of T subscript H and T subscript C respectively.

Using the equation for the change in entropy

increment S space equals space fraction numerator increment Q over denominator T end fraction

Show that the maximum theoretical efficiency of a heat engine is given by

eta subscript C space equals space 1 space minus space T subscript C over T subscript H

Answer:

Step 1: Determine the change in entropy during isothermal expansion

  • In isothermal expansion (AB):  the gas absorbs heat Q subscript H from a hot reservoir at temperature T subscript H
  • Therefore, the increase in entropy is:

increment S subscript A B end subscript space equals space Q subscript H over T subscript H

Step 2: Determine the change in entropy during isothermal compression

  • In isothermal compression (CD):  the gas transfers heat Q subscript C to a cold reservoir at temperature T subscript C
  • Therefore, the decrease in entropy is:

increment S subscript C D end subscript space equals space minus Q subscript C over T subscript C

Step 3: Consider the net change in entropy over the cycle

2-4-8-carnot-cycle-entropy-worked-example-ma

  • During adiabatic expansion (BC) and compression (DA), entropy does not change as there is no thermal energy gained or lost

increment S subscript B C end subscript space equals space increment S subscript D A end subscript space equals space 0

  • We know that the overall entropy of the system does not change in a cyclic process, therefore

increment S space equals space increment S subscript A B end subscript space plus space increment S subscript B C end subscript space plus space increment S subscript C D end subscript space plus space increment S subscript D A end subscript

increment S space equals space Q subscript H over T subscript H space plus space 0 space plus space open parentheses negative Q subscript C over T subscript C close parentheses space plus space 0 space equals space 0

Q subscript H over T subscript H space equals space Q subscript C over T subscript C space space space space space rightwards double arrow space space space space space Q subscript C over Q subscript H space equals space T subscript C over T subscript H

Step 4: Substitute the expression into the equation for the efficiency of a heat engine

  • The efficiency of a heat engine is given by:

eta space equals space fraction numerator u s e f u l space w o r k over denominator i n p u t space e n e r g y end fraction space equals space W over Q subscript H

  • Where useful work is W space equals space Q subscript H space minus space Q subscript C

eta space equals space fraction numerator Q subscript H space minus space Q subscript C over denominator Q subscript H end fraction space equals space 1 space minus space Q subscript C over Q subscript H

  • Combining with the expression derived above gives:

eta subscript C space equals space 1 space minus space T subscript C over T subscript H

Worked example

An engineer designs a heat engine that has an inlet temperature of 500 K and an outlet temperature of 300 K. The engineer claims that 100 kJ of thermal energy flows out of the hot reservoir and 25 kJ of thermal energy flows into the cold reservoir.

Determine, with reference to the second law of thermodynamics, whether this engine is thermodynamically possible.

Answer:

Step 1: Determine the efficiency of the proposed engine

  • The efficiency of this engine would be

eta space equals space W over Q subscript H space equals space fraction numerator Q subscript H space minus space Q subscript C over denominator Q subscript H end fraction space equals space 1 space minus space Q subscript C over Q subscript H

  • Where:
    • Heat transferred in, Q subscript H = 100 kJ
    • Heat transferred out, Q subscript C = 25 kJ

Efficiency = 1 space minus space 25 over 100 = 0.75 = 75%

Step 2: Determine the maximum theoretical (Carnot) efficiency of the proposed engine

  • A Carnot engine operating between the same temperatures would have an efficiency of

eta subscript C space equals space 1 space minus space T subscript C over T subscript H

  • Where:
    • Inlet temperature, T subscript H = 500 K
    • Outlet temperature, T subscript C = 300 K

Carnot efficiency:  eta subscript C space equals space 1 space minus space 300 over 500 = 0.4 = 40%

Step 3: Discuss the proposed engine in relation to the second law

  • The Clausius form of the second law states: it is impossible for heat to flow from a cooler body to a hotter body without performing work
  • This law sets an upper limit on the maximum possible efficiency of the transfer of thermal energy to mechanical energy in a heat engine
  • The maximum possible efficiency of the proposed engine is 40%, but the engineer is proposing an efficiency of 75% i.e. an efficiency greater than the Carnot efficiency
  • This violates the Clausius form of the second law, hence the proposed engine is impossible

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    Katie M

    Author: Katie M

    Expertise: Physics

    Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.