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Thermodynamics · Chapter 6 of 10 · Intermediate

The Second Law of Thermodynamics

The first law says energy is conserved; the second says it cannot all be turned into work. That single limit sets the best possible efficiency of every engine and the best possible performance of every refrigerator.

01

Readiness check

This chapter is about limits and ratios. Tick only what you can do closed-notes.

  • Form a ratio of two energies as an efficiency.
  • Convert °C to kelvin and use absolute temperature.
  • State the first law for a cycle (net work equals net heat).
  • Distinguish a heat engine from a refrigerator.
  • Compute a percentage and a reciprocal.
0 or 1 weak itemsContinue with this chapter.
2 weak itemsRevisit the energy balance in Chapter 2.
3 or more weak itemsReview the first law in Physics: First Law.
02

The core idea

Heat will not flow uphill on its own, and no engine can turn all its heat into work. A device working between a hot and a cold reservoir is bounded by the Carnot limit, set only by the two temperatures.

ηth = W/QH = 1 − QL/QHηCarnot = 1 − TL/TH

The first law treats a kilojoule of work and a kilojoule of heat as equal, but experience says they are not: work converts fully to heat, while heat never converts fully to work. The Kelvin-Planck statement forbids an engine that produces work from a single reservoir; the Clausius statement forbids heat moving from cold to hot with no input. Both lead to the Carnot limit, the highest efficiency any engine can reach between two temperatures, achieved only by a reversible cycle.

The skill works when: you identify the two reservoirs and compare the device against its Carnot limit.
The skill breaks down when: temperatures are left in °C, or an efficiency above the Carnot value is accepted.
The concept. A heat engine takes QH from the hot reservoir, delivers work W, and must reject QL to the cold reservoir. The rejected heat is unavoidable; only its size, set by the Carnot limit, can be reduced.
03

The skills, taught in order

Five skills state the second law, define efficiency and COP, and fix the Carnot limit that bounds them.

6.1 The two statements

The Kelvin-Planck statement says no engine can take heat from one reservoir and turn all of it into work; some must be rejected to a colder reservoir. The Clausius statement says heat cannot flow from cold to hot without a work input. The two are equivalent: violating one would let you violate the other.

6.2 Heat engines and thermal efficiency

A heat engine operates in a cycle, taking QH from a hot reservoir, producing net work W, and rejecting QL to a cold reservoir. Since a cycle returns to its start, W = QH − QL, and the thermal efficiency is ηth = W/QH = 1 − QL/QH.

6.3 Refrigerators and heat pumps

Run the engine backwards and a work input moves heat from cold to hot. The same device is a refrigerator if you value the heat removed from the cold space, and a heat pump if you value the heat delivered to the warm space. Their measures are coefficients of performance, which can exceed one.

DevicePerformance measureCarnot (reversible) limit
Heat engineηth = 1 − QL/QH1 − TL/TH
RefrigeratorCOPR = QL/WTL/(TH − TL)
Heat pumpCOPHP = QH/WTH/(TH − TL)

Temperatures must be absolute (kelvin). Note COPHP = COPR + 1 for the same reservoirs.

6.4 Reversibility

A reversible process can be undone with no trace left in the surroundings. Real processes are irreversible because of friction, unrestrained expansion, and heat transfer across a finite temperature difference. Reversibility is an idealisation, but it sets the benchmark every real device is measured against.

6.5 The Carnot limit

The Carnot principles state that no engine between two reservoirs can beat a reversible one, and that all reversible engines between the same two reservoirs share the same efficiency, η = 1 − TL/TH. This depends only on the temperatures, not the working fluid, which is why raising TH or lowering TL is the only way to raise the ceiling.

Engineering connection: the Carnot limit is the yardstick for the real power and refrigeration cycles of Chapters 9 and 10, and the second-law efficiency of Chapter 8.

04

Worked example 1: a heat engine against its Carnot limit

A heat engine receives 1000 kW from a furnace at 800 K and rejects heat to the surroundings at 300 K, producing 400 kW of power. Find its thermal efficiency, the rejected heat, the Carnot efficiency for these reservoirs, and the second-law efficiency.

Figure 1. Of 1000 kW supplied, 400 kW becomes work and 600 kW is rejected. The Carnot limit for 800 K and 300 K shows how much of the shortfall is fundamental.
  1. ProblemFind ηth, QL, ηCarnot, and the second-law efficiency for the engine in Figure 1.
  2. Given / findQ̇H = 1000 kW, Ẇ = 400 kW, TH = 800 K, TL = 300 K. Find ηth, Q̇L, ηCarnot, ηII.
  3. AssumptionsSteady cyclic operation between two fixed-temperature reservoirs.
  4. ModelThermal efficiency is work over heat in; the rejected heat closes the energy balance; Carnot sets the ceiling; the ratio of the two efficiencies is the second-law efficiency.
  5. Equationsηth = Ẇ/Q̇HQ̇L = Q̇H − ẆηCarnot = 1 − TL/THηII = ηth/ηCarnot
  6. Solveηth = 400/1000 = 0.40. Q̇L = 1000 − 400 = 600 kW. ηCarnot = 1 − 300/800 = 0.625. ηII = 0.40/0.625 = 0.64.
  7. CheckThe actual efficiency is below the Carnot ceiling, as the second law demands. A second-law efficiency of 0.64 says the engine captures 64 percent of what a reversible engine could.
  8. ConclusionThe Carnot value separates the inevitable loss (the 37.5 percent a perfect engine still rejects) from the avoidable loss due to irreversibility. Real plants live in this gap.
Result. ηth = 0.40, Q̇L = 600 kW, ηCarnot = 0.625, ηII = 0.64.
05

Worked example 2: a refrigerator and its heat-pump twin

A refrigerator keeps a cold space at −5 °C while the kitchen sits at 25 °C, and must remove heat at 5 kW. Find the best possible (Carnot) coefficient of performance, the minimum power needed, and the heat-pump COP for the same two temperatures.

Figure 2. A refrigerator uses work to pump heat from the cold space up to the kitchen. The smaller the temperature gap, the less work each kilowatt of cooling needs.
  1. ProblemFind the Carnot COP, the minimum power, and the heat-pump COP for the refrigerator in Figure 2.
  2. Given / findTL = −5 °C = 268.15 K, TH = 25 °C = 298.15 K, Q̇L = 5 kW. Find COPR, Ẇmin, COPHP.
  3. AssumptionsReversible (Carnot) operation between two fixed-temperature reservoirs gives the best case.
  4. ModelThe Carnot refrigerator COP depends only on the temperatures; the minimum power is the cooling load divided by that COP.
  5. EquationsCOPR = TL/(TH − TL)Ẇmin = Q̇L/COPRCOPHP = TH/(TH − TL)
  6. SolveCOPR = 268.15/(298.15 − 268.15) = 268.15/30 = 8.94. Ẇmin = 5/8.94 = 0.56 kW. COPHP = 298.15/30 = 9.94.
  7. CheckCOPHP = COPR + 1, exactly as it must, because the heat delivered up top is the heat removed plus the work in. A small 30 K gap gives a high COP.
  8. ConclusionCooling needs surprisingly little work when the temperature lift is small, which is why a real fridge with COP near 3 still leaves room to improve.
Result. COPR = 8.94, Ẇmin = 0.56 kW, COPHP = 9.94.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Temperatures in °CCarnot efficiency or COP comes out wrong"Did I convert to kelvin?"Carnot formulas require absolute temperature.
Expecting 100 percent efficiencyNo rejected heat in the balance"Where does QL go?"Some heat must always be rejected to a colder reservoir.
COP above the Carnot limitA device that beats reversibility"Is my COP under TL/(TH − TL)?"No real device can exceed the Carnot COP.
Confusing the two COPsFridge and heat pump performance swapped"Is the useful effect cooling or heating?"Use QL/W for cooling, QH/W for heating.
07

Practice ladder

Level 1 · Direct skill

An engine takes 500 kJ from a hot reservoir and rejects 320 kJ. Find its work output and thermal efficiency.

Show answer

W = QH − QL = 500 − 320 = 180 kJ. ηth = W/QH = 180/500 = 0.36, or 36 percent.

Level 2 · Mixed concept

For the engine of Worked Example 1, what is the maximum power it could produce from the same 1000 kW if it ran reversibly?

Show answer

Wmax = ηCarnot Q̇H = 0.625 × 1000 = 625 kW. The real engine gives 400 kW, so 225 kW of work potential is lost to irreversibility.

Level 3 · Independent problem

A heat pump delivers 12 kW of heating to a house at 22 °C from outdoor air at 2 °C. Find the Carnot COP and the minimum power it needs.

Show answer

COPHP = TH/(TH − TL) = 295.15/(295.15 − 275.15) = 295.15/20 = 14.76. Wmin = 12/14.76 = 0.81 kW. A heat pump beats a resistance heater because COP exceeds one.

Level 4 · Transfer to real engineering

Look up the real efficiency of a car engine or a power plant and compare it with the Carnot limit between its hot and cold temperatures. Where does the gap come from?

What good work looks like

Sensible hot and cold temperatures in kelvin, a Carnot value above the quoted real efficiency, and named irreversibilities (combustion, friction, finite-temperature heat transfer) that explain the gap.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that my efficiency is below the Carnot limit for these temperatures."
"Give me four devices; I will say whether each is an engine, fridge, or heat pump."
"What is the Carnot efficiency?" Computing 1 − TL/TH yourself is the skill.
"Is this engine possible?" Comparing against the Carnot ceiling is the point.

Portfolio task

Pick a real engine, refrigerator, or heat pump. Estimate its hot and cold temperatures, compute the Carnot limit, find its real performance, and report the second-law efficiency.

Must include: two absolute temperatures, a Carnot limit, a real value below it, and a second-law-efficiency ratio.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. State the Kelvin-Planck statement.

No cyclic engine can take heat from a single reservoir and convert all of it to work; some heat must be rejected.

2. Write the thermal efficiency of a heat engine.

ηth = W/QH = 1 − QL/QH.

3. Give the Carnot efficiency.

ηCarnot = 1 − TL/TH, with temperatures in kelvin.

4. Relate the two coefficients of performance.

COPHP = COPR + 1 for the same reservoirs.

5. What makes a process reversible?

It can be undone leaving no change in the surroundings; real processes are irreversible due to friction and finite-temperature heat transfer.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive the engine and refrigerator results from a blank page.
+3 daysCompare two real devices with their Carnot limits.
+7 daysCarry the second law into entropy, Chapter 7.
+30 daysReuse the Carnot limit to judge the cycles of Chapters 9 and 10.
10

Textbook mapping

ItemMapping
Primary sourceBorgnakke and Sonntag, Fundamentals of Thermodynamics, Chapter 7 (The Second Law of Thermodynamics)
Cross-referenceÇengel and Boles, Ch. 6 · Moran, Ch. 5
Core topics6.1 Kelvin-Planck and Clausius · 6.2 Heat engines · 6.3 Refrigerators and heat pumps · 6.4 Reversibility · 6.5 Carnot limit
Engineering connectionThe Carnot limit is the benchmark for every real cycle and for second-law efficiency.
Read nextChapter 7: Entropy.