Thermodynamics · Lesson 5 of 7
The Second Law and Reversibility
“Explain why energy conservation is not enough, interpret engines and refrigerators, and identify the conditions for reversibility.”
• State the Kelvin–Planck and Clausius forms of the second law with their qualifications. • Use cyclic energy balances to define engine efficiency and refrigerator performance. • Distinguish returning a system to its starting state from reversing an entire process. • Explain irreversibility from gradients, free expansion and dissipative effects. • Explain why quasi-static operation is necessary but not sufficient for reversibility.
Energy conservation does not decide every possibility
Imagine a book lying on a table suddenly rising because the table cools and gives up just enough internal energy to supply the book’s gain of gravitational potential energy. The total energy could balance, yet such a spontaneous conversion is not observed. Energy conservation alone does not specify the direction or achievable performance of every process. The second law supplies an additional restriction on macroscopic thermal processes.
A familiar example is spontaneous heat transfer from a hot vessel base toward cooler parts of the vessel. Energy can be conserved in either imagined direction, but the reverse separation into a hotter base and colder upper part does not occur by itself. A refrigerator can move energy from colder to warmer surroundings, but only with another effect, such as a work input. That qualification is central to the law.
An engine must be considered over a complete cycle
A heat engine uses a working substance that repeatedly returns to its starting state. During a cycle it absorbs a positive heat magnitude QH from a hot reservoir, rejects a positive magnitude QC to a colder reservoir and delivers net work Wout. Since ΔUcycle = 0, the absorbed heat equals the rejected heat plus the delivered work. We use QH and QC as positive magnitudes here; under the signed heat convention the cold-side transfer to the working substance is −QC.
Problem
A cyclic engine absorbs 1200 J from its hot reservoir and releases 800 J to its cold reservoir in each cycle. Find work and efficiency.
- 1.Wout = QH − QC = 1200 − 800 = 400 J per cycle.
- 2.η = 400/1200 = 1/3 ≈ 33.3%.
- 3.The rejected 800 J is not missing energy. It reaches the cold reservoir. The efficiency describes the fraction of the input heat delivered as net work.
- 4.This first-law calculation alone does not prove that the engine is physically possible for any chosen pair of reservoir temperatures. The second-law temperature limit must also be checked.
Two statements of the second law
The Kelvin–Planck statement forbids a process whose sole overall result is to extract heat from one reservoir and convert all of it into work. For a cyclic heat engine with no other net effects, this means η cannot be 1: some heat must be rejected. “Sole overall result” and the cyclic condition matter. A single ideal-gas isothermal expansion can have Q = W, but the gas ends at a different state; it is not an engine producing work repeatedly with no other change.
The Clausius statement forbids a process whose sole overall result is to transfer heat from a colder body to a hotter body. It does not forbid a powered refrigerator: the surroundings supply work, so there is an additional effect. Both statements express the same second law, though they appear to focus on different devices.
Their equivalence can be understood by coupling devices. If a perfect one-reservoir engine existed, it could power an ordinary refrigerator; after balancing the hot-side exchange, the combined sole result could be cold-to-hot heat transfer without work. Conversely, a device that moved heat cold-to-hot without work could return an ordinary engine’s rejected heat to its hot reservoir, leaving complete conversion of the net heat extracted into work. Either violation would permit the other. This argument depends on the total effects of the coupled apparatus, not on any one leg in isolation.
Refrigeration measures a different useful output
A refrigerator takes positive heat magnitude QC from the cold region and receives work Win. It releases QH = QC + Win to the warmer surroundings. The useful output is cooling, so its coefficient of performance α is QC/Win. This is not engine efficiency: α can be greater than 1 because the device moves thermal energy rather than turning work entirely into new energy. A refrigerator’s warm rear coils release both the extracted cold-side energy and the input work.
Problem
A refrigerator removes 600 J from its cold space while consuming 200 J of work in each cycle. Find heat rejected and α.
- 1.QH = QC + Win = 600 + 200 = 800 J per cycle.
- 2.α = QC/Win = 600/200 = 3.
- 3.The coefficient 3 does not violate energy conservation: 200 J of work enables transport of 600 J already present in the colder region.
- 4.A refrigerator operating entirely within a closed room and left with its door open cannot cool the whole room by this process. It releases the removed heat back into the room plus the input work.
Reversing the system is not enough
A thermodynamic process is reversible only if it can be run backward so that both the system and all its surroundings return to their original states, with no other net change anywhere. Merely restoring the gas to its original pressure, volume and temperature is weaker. That restoration may require work or may leave a reservoir with extra energy.
After a gas freely expands into vacuum, a piston can compress it back into its starting volume. But the compression requires work and may require heat removal to restore its initial temperature. The surroundings have changed. That is why free expansion is irreversible even when an ideal gas has the same initial and final temperatures and the forward expansion exchanges neither heat nor work.
Common irreversible processes include diffusion of leaking gas through a room, a petrol–air combustion event, heat conduction across a finite temperature difference, and mechanical energy lost through friction or viscosity. A stirred liquid warms as the stirrer’s work increases internal energy. If the liquid is maintained at fixed temperature by a reservoir, that energy ultimately enters the reservoir. Recovering exactly the original work solely from that reservoir while restoring everything else would violate the second law.
Two broad sources of irreversibility recur. First, finite imbalances can drive the system through non-equilibrium states, as in free expansion or an explosive reaction. Second, dissipative effects such as friction and viscosity convert organised mechanical energy into disordered internal energy. Neither effect is removed simply by naming a process “slow”.
Conditions for an ideal reversible path
A reversible piston process is quasi-static and free from dissipative effects. Mechanical driving differences approach zero; when heat is exchanged, temperature differences also approach zero. With a frictionless piston, changing an external load by an infinitesimal amount can reverse the direction. In a reversible isothermal expansion, heat flows in through a vanishingly small temperature difference and the reverse compression releases the same heat to the same reservoir in the ideal limit.
A piston sliding very slowly against friction is not reversible: friction still converts work into internal energy, and reversing the motion does not undo that dissipation. Conversely, reversible adiabatic motion involves no heat exchange at all, so equality with a surrounding thermal reservoir is not a requirement during its insulated leg. The relevant requirement is near-equilibrium internal states and mechanical balance without dissipation. Real processes can approximate the reversible limit, but cannot operate infinitely slowly while producing finite power.
Problem
Compare (a) slow frictionless isothermal expansion with an almost equal-temperature reservoir, (b) slow expansion with piston friction, and (c) rapid insulated expansion into vacuum.
- 1.Process (a) approaches the reversible ideal when both temperature and mechanical driving differences tend to zero.
- 2.Process (b) can be slow but remains irreversible because friction dissipates energy.
- 3.Process (c) is adiabatic but irreversible. The transient gas is not in uniform equilibrium and no reversed infinitesimal driving change retraces the expansion.
- 4.Quasi-static, adiabatic and reversible therefore describe different properties; none should be treated as a synonym for another.
Quick Quiz
Quiz
Which does the Kelvin–Planck statement forbid?
The prohibited result is complete cyclic conversion from one reservoir with no other net effect.
A refrigerator moves 600 J from cold space while receiving 200 J work. It releases
Its cyclic energy balance is QH = QC + Win = 800 J.
Which condition by itself is insufficient for reversibility?
Slow motion may still involve friction or finite-temperature-difference heat transfer.
A heat engine absorbs 1000 J and rejects 700 J per cycle. Its efficiency is
Net work is 300 J, so η = 300/1000 = 0.30.
Why does compressing freely expanded gas back to its initial state not prove reversibility?
Reversibility requires restoration of the system and all surroundings, not merely the system’s endpoints.
Practice Problems
- State the Kelvin–Planck and Clausius forms of the second law in your own words. Explain the role of the phrase “sole overall result” in each.
- A cyclic engine takes 2.5 kJ from a hot reservoir and rejects 1.7 kJ to a cold reservoir per cycle. Find work output and efficiency. Does the first law alone establish that this is possible between any chosen temperatures?
- A refrigerator removes 1.2 kJ from a cold space per cycle and has α = 4. Find work input and heat released to the room. Explain why α greater than 1 is allowed.
- Explain why restoring a gas to its initial P, V and T after free expansion may still leave the overall process irreversible. Identify the likely changes in the surroundings.
- Compare a frictionless quasi-static expansion, a slow expansion with friction and a rapid insulated free expansion. For each, discuss equilibrium description, dissipation and reversibility.
- Use a liquid stirred by a rotating blade as an example to explain work-to-internal-energy conversion. Why would an exact reverse with no other change violate the second law?
- Explain why an open refrigerator operating inside a closed room adds net energy to the room despite removing energy from the region near its cooling coils.
Key Takeaways
• The first law conserves energy; the second law restricts the possible overall effects. • A cyclic engine has Wout = QH − QC and η = Wout/QH. • A refrigerator requires work to move heat cold-to-hot; α = QC/Win can exceed 1. • Reversibility requires restoration of the system and all surroundings. • Finite imbalances and dissipation make processes irreversible; slow motion alone is insufficient.