
Optional extension · Interactive lesson
The second law of thermodynamics
Energy conservation is blind to direction: it licenses the reverse of every process you have ever seen. This lesson pins down exactly what the second law forbids, in engine language and in refrigerator language, shows the two prohibitions are one, and turns them into a ceiling you can test a claim against.
UPII-E3.07 · Lesson 7 of 9 in this extension. Position is not a completion score.Optional extension. Use this unnumbered support sequence where your university’s syllabus or prerequisite needs require it. It is not an additional numbered unit. Optional work is not a required completion gate.
Build the model
The first law says what balances; the second says which way it runs.
The key idea
The second law of thermodynamics is the assertion that no cyclic device can have as its sole effect either the complete conversion of heat from one reservoir into work, or the transfer of heat from a colder body to a hotter one.
Example. A fridge lifts 120 W out of its 277 K cabinet into a 295 K kitchen, but only because a 40 W compressor pays for it and 160 W leaves at the back; pull the plug and the same cold-to-hot transfer is forbidden, though energy would still balance.
The complete physical model and assumptions
The first law is an equality, and equalities run both ways: every process it permits, it permits in reverse. Yet heat never drifts from the cold end of a bar to the hot end, and a warm bench never cools by a thirtieth of a kelvin to throw a book back into your hand. The second law is the separate assertion that direction is real, and it is stated as two prohibitions. Kelvin and Planck: no cyclic device can have, as its sole effect, the complete conversion of heat drawn from a single reservoir into work. Clausius: no cyclic device can have, as its sole effect, the transfer of heat from a colder body to a hotter one. They appear to be claims about different machines, but couple a hypothetical violator of one to an ordinary device of the other kind and you build a violator of the other, so denying either denies both. Its content is direction, not accounting, which is why it cannot be derived from energy bookkeeping and has to be asserted from experience. And it stops being a mere prohibition once you notice that the best allowed device is the reversible one: η ≤ 1 − TC/TH then converts the word impossible into the phrase at most.
1Energy conservation permits far too much
The first law, ΔU = Q − W, is an equality, and it says nothing about direction. Let a 2.0 kg book fall 1.5 m onto a bench. It gives up mgh = 2.0 × 9.81 × 1.5 = 29 J, which ends up as thermal energy in the book and the bench. Now run the film backwards: the bench and book cool by 29 J — with a combined heat capacity near 800 J K⁻¹ that is a drop of 29/800 = 0.036 K, far too small to notice — and the book rises 1.5 m into your hand. Energy is conserved to the last joule in both films. One happens daily, the other has never been seen. That gap is not a detail the first law forgot to mention; it is a second, independent law, and what it constrains is direction rather than totals. Nothing in mechanics or in energy accounting implies it, so it has to be asserted from experience and then tested.
2The engine statement, word by word
Kelvin and Planck: no cyclic device can have, as its sole effect, the absorption of heat from a single reservoir and its complete conversion into work. Three words carry the weight. Cyclic means the working substance returns to its starting state, so ΔU = 0 over the cycle and W = QH − |QC| exactly. Single reservoir means nothing colder is available to dump into. Sole effect means no permanent change is left anywhere else. Put together, they say |QC| > 0 strictly, so η = W/QH < 1 for every engine ever built or ever to be built — not because of friction or poor insulation, but because of this law. The stakes are large: the ocean holds about 1.4 × 10²¹ kg of water at roughly 4.0 kJ kg⁻¹ K⁻¹, so cooling it by one millikelvin would release 5.6 × 10²¹ J, some nine years of world energy use. A ship that ran on that alone is called a perpetual-motion machine of the second kind, and the engine statement is the claim that none exists.
3The refrigerator statement, and what sole effect rules in
Clausius: no cyclic device can have, as its sole effect, the transfer of heat from a colder body to a hotter one. Read carelessly, this says heat never flows from cold to hot — which your kitchen refutes every night. A fridge lifting 120 W out of a 277 K cabinet into a 295 K room while drawing 40 W of electrical work does exactly that, and delivers 160 W to the room. It survives the statement because the transfer is not its sole effect: 40 W of work has been consumed and the surroundings carry a permanent record of it. Strip the payment away and the transfer is forbidden. So the statement is not heat never goes from cold to hot; it is heat never goes from cold to hot for free. The same clause is what saves an air conditioner, a heat pump and every cryostat, and the size of the required payment is what the quantitative form of the law fixes.
4One law in two dialects
Suppose someone breaks the engine statement, with a cycle that takes 400 J per cycle from a 500 K reservoir and delivers 400 J of work while rejecting nothing. Feed that work to an ordinary refrigerator running between 300 K and 500 K with a coefficient of performance of 1.0, comfortably inside the reversible limit of 300/(500 − 300) = 1.5. It lifts 400 J from the cold reservoir and delivers 400 + 400 = 800 J to the hot one. Draw one box around both machines. The hot reservoir gives up 400 J and receives 800 J, a net gain of 400 J; the cold reservoir loses 400 J; no work crosses the boundary at all. The box moves 400 J from cold to hot as its sole effect, which is precisely what Clausius forbids. The reverse coupling works too, so the two statements stand or fall together. They are not two laws that happen to agree; they are one law read through an engine or through a refrigerator.
5Reversible is the boundary, irreversible is everything real
A process is reversible if at every instant the system and its surroundings are only infinitesimally out of balance, so the whole path can be retraced leaving no residue anywhere. Nothing real qualifies. Heat crossing a finite temperature difference, friction and viscous drag, a gas expanding into vacuum, two gases mixing, current warming a resistor, a metal bent past its elastic limit — each of these has a direction you can spot on film, and each is irreversible. Reversibility demands infinite time, so it is a limit, not a procedure. Its value is that it is the boundary: every real process falls on one side of it, so the reversible device sets the record that real ones approach and never beat. That is the origin of the inequality ∮ dQ/T ≤ 0, whose equality is reserved for the reversible cycle, and it is why the next topic can turn the whole law into a single statement about entropy.
6From a prohibition to a number
In practice nobody uses the second law to say impossible; they use it to say at most. Because the reversible cycle is the boundary case, no cycle between two reservoirs can beat η = 1 − TC/TH, and no refrigerator can beat K = TC/(TH − TC). A supercritical steam plant taking steam at 873 K and condensing at 303 K has a ceiling of 1 − 303/873 = 0.653; a real net efficiency of 46% is therefore 70% of the best the law allows, which is a fair verdict on the engineering and a hard stop on the marketing. Both ceilings demand absolute temperatures: writing 1 − 30/600 for a plant quoted in degrees Celsius gives 0.95 and is nonsense. The habit worth forming is to audit any claim in two passes — first the first law, does the energy balance, then the second law, does the work or the cooling sit under the ceiling.
Key equations
Read the units, assumptions and limits before substituting.
Engine statement (Kelvin–Planck)
W ≤ 0 for any cycle exchanging heat with one reservoir
Kills η = 1 outright: every engine must reject heat to something colder than its source.
W is the net work out per cycle in J; the reservoir holds one temperature T in K throughoutRefrigerator statement (Clausius)
Qᵤₚ ≤ 0 for any cycle running with W = 0
Cooling always costs work; nothing pumps heat up a temperature difference for free.
Qᵤₚ is the heat delivered per cycle to the hotter body at TH from the colder at TC, in J; W is the work supplied, in JEfficiency ceiling on any engine
η ≤ 1 − TC/TH
Converts the engine statement from a prohibition into a number a claim can be tested against.
TC and TH are the two reservoir temperatures in kelvin, never degrees Celsius; equality holds only for a reversible cycleCooling ceiling on any refrigerator
K ≤ TC/(TH − TC)
The Clausius statement made quantitative: the colder you pump from, the dearer each joule of cooling gets.
K = |QC|/W is the coefficient of performance, heat lifted per cycle (J) over work supplied (J); T in kelvinClausius inequality
∮ dQ/T ≤ 0
The single inequality both statements collapse to, with equality reserved for a reversible cycle.
dQ is heat entering the working substance (J) at the boundary temperature T (K), summed once round a closed cycleInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Start at the default, where the claimed 560 J overshoots the 500 J ceiling: drag the work slider down until the block past the dashed line disappears, then raise TH or lower TC and watch the ceiling slide right without ever reaching the arrow's full 1000 J.
Live results
- CLAIMED EFFICIENCY
- 56.0 %
- SECOND-LAW CEILING
- 50.0 %
- LARGEST ALLOWED W
- 500 J
- CLAIM − CEILING
- 60 J
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Hot reservoir TH: 600 K; Cold reservoir TC: 300 K; Claimed work per cycle: 560 J
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The engine statement is not the claim that heat can never be turned entirely into work. Expand 1.0 mol of an ideal gas slowly and isothermally at 400 K from 2.0 L to 6.0 L: ΔU = 0, so every joule absorbed leaves as work, Q = W = nRT ln 3 = 8.314 × 400 × 1.0986 = 3.65 kJ, a 100% conversion that breaks nothing. The catch is that the gas ends up three times bigger, so the trick cannot be repeated. The word cyclic carries the whole prohibition: a machine has to return to its starting state to keep working, and the moment it must, some heat has to be rejected. The same fine print protects the fridge, which moves heat from cold to hot every day, but never as its sole effect.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
An inventor demonstrates a sealed unit standing in a 293 K workshop. Over one hour it draws 7.2 MJ of heat from the workshop air, delivers 2.0 kWh of electrical work to the grid, and returns to exactly its starting state every cycle, with nothing else in the room changing. Test the claim against the first law, then against the second.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Put the work in joules: 2.0 kWh = 2.0 × 3.6 MJ = 7.2 MJ, the same number as the heat drawn in.
- First law over a whole number of cycles: the unit returns to its starting state, so ΔU = 0 and Q = W. With Q = 7.2 MJ and W = 7.2 MJ the books balance exactly, and the first law raises no objection.
- Identify the reservoirs. Heat enters from the workshop air at 293 K and nothing is rejected anywhere, so |QC| = 0 and the device touches one reservoir only.
- That is the forbidden combination. A cyclic device exchanging heat with a single reservoir cannot deliver net work, so the claimed η = 7.2/7.2 = 1 is out of reach.
- The fault is not a leak or a stiff bearing. No improvement in engineering reaches η = 1, because what blocks it is a law and not a loss.
Answer. The claim conserves energy exactly and still cannot work: it is a perpetual-motion machine of the second kind, forbidden by the Kelvin–Planck statement because it draws on one 293 K reservoir and rejects nothing.
MediumExample 2
A laboratory cooler is said to hold a 265 K chamber steady against a 4.0 W heat leak while drawing 0.40 W of electrical power, standing in a 298 K room. Does it break the first law? Does it break the Clausius statement as that statement is worded? Does the second law allow it, and if not, what is the least power that would?
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Energy first. Each second 4.0 J enters from the chamber and 0.40 J enters as electrical work, and 4.4 J is delivered to the room. The books balance, so the first law is satisfied.
- Now the wording. The Clausius statement forbids cold-to-hot transfer as the sole effect of a cycle. Work is supplied here and the surroundings are permanently changed, so the transfer is not the sole effect and the bare statement does not condemn it.
- Test it against the ceiling the law puts on any refrigerator: K ≤ TC/(TH − TC) = 265/(298 − 265) = 265/33 = 8.03.
- The claimed coefficient of performance is K = |QC|/W = 4.0/0.40 = 10.0, which is above that ceiling, so no machine of any design can do it.
- The least power consistent with the second law is W = |QC|/Kₘₐₓ = 4.0/8.03 = 0.50 W, and that is the reversible limit, so a real cooler needs several times as much.
Answer. The first law is satisfied and the bare Clausius wording is not breached, but K = 10.0 exceeds the ceiling of 8.03, so the cooler cannot exist. The reversible limit would need 0.50 W.
HardExample 3
A workshop advertises a heat multiplier: running in a cycle, it moves 480 J from a 280 K cold store to a 350 K hot store with no work input and no other change. Accept the claim for a moment, couple the device to an ordinary heat engine working between the same two stores, and find the net work per cycle the combination delivers and the net heat it draws.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Pick a legitimate partner. Between 350 K and 280 K the second law allows any engine with η ≤ 1 − 280/350 = 0.200, and the reversible one sits exactly at that limit.
- Size it so its exhaust matches the multiplier's intake. Taking QH = 600 J per cycle at η = 0.200 gives W = 0.200 × 600 = 120 J and rejects |QC| = 600 − 120 = 480 J to the 280 K store, exactly the 480 J the multiplier claims to lift.
- Couple them and audit the cold store: it receives 480 J from the engine and gives 480 J to the multiplier, so it ends each cycle unchanged and may as well not be there.
- Audit the hot store: it gives 600 J to the engine and receives 480 J back from the multiplier, a net loss of 120 J per cycle.
- So the sealed combination draws 120 J per cycle from the 350 K store alone and delivers 120 J of work with nothing else changed — a cyclic engine of efficiency 1 running on one reservoir, which the engine statement forbids.
- The conclusion survives a worse partner. A real engine at η = 0.12 would need QH = 480/0.88 = 545 J to reject the same 480 J, leaving a net 65 J drawn from the hot store and 65 J of work: still a perfect single-reservoir engine.
Answer. The combination delivers 120 J of work per cycle while drawing 120 J from the 350 K store and leaving the 280 K store untouched. A free cold-to-hot pump and a perfect single-reservoir engine stand or fall together.
Quick check
Select an answer and explain it before checking.
Review & continue
The second law of thermodynamics is the assertion that no cyclic device can have as its sole effect either the complete conversion of heat from one reservoir into work, or the transfer of heat from a colder body to a hotter one.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics II completion tasks.