AP Physics 2 · Topic 9.6

Topic 9.6: Entropy and the Second Law of Thermodynamics

Unit 9: Thermodynamics15-18% of the multiple-choice section

The second law says the total entropy of an isolated system can never decrease, and stays constant only when every process it undergoes is reversible. AP Physics 2 treats entropy qualitatively: the tendency of energy to spread, and the unavailability of some of a system's energy to do work.

AP Physics: Unit 9 (topics 9.6 Entropy and the Second Law of Thermodynamics). AP Physics 2 Unit 9, Topic 9.6. One learning objective, 9.6.A, describe the change in entropy for a given system over time, supported by eight essential knowledge statements: 9.6.A.1 (the second law of thermodynamics states that the total entropy of an isolated system can never decrease and is constant only when all processes the system undergoes are reversible), 9.6.A.2 (entropy can be qualitatively described as the tendency of energy to spread or the unavailability of some of the system's energy to do work), 9.6.A.2.i (localized energy will tend to disperse and spread out), 9.6.A.2.ii (entropy is a state function and therefore only depends on the current state or configuration of a system, not how the system reached that state), 9.6.A.2.iii (maximum entropy occurs when a system is in thermodynamic equilibrium), 9.6.A.3 (the change in a system's entropy is determined by the system's interactions with its surroundings), 9.6.A.3.i (isolated systems spontaneously move toward thermodynamic equilibrium), and 9.6.A.3.ii (the entropy of an isolated system never decreases, but the entropy of a closed system can decrease because energy can be transferred into or out of the system). The topic's boundary statement reads in full: only qualitative treatment of the second law of thermodynamics is within the scope of AP Physics 2. The CED's suggested skills here are 1.A, 2.C, 3.B, and 3.C, with no computation skill listed. The word disorder does not appear anywhere in the AP Physics 2 Course and Exam Description, and no entropy equation is printed on the AP Physics 2 equation sheet. Unit 9 carries 15 to 18 percent of the multiple-choice section and a suggested 10 to 16 class periods.

What Topic 9.6 requires

Topic 9.6 closes Unit 9, Thermodynamics, weighted at 15 to 18 percent of the multiple-choice section with a suggested 10 to 16 class periods. It carries one learning objective, eight essential knowledge statements, and one boundary statement. That is the whole of the required course content, and worth reading in full before reading anything else about entropy.

9.6.A, describe the change in entropy for a given system over time.

  • 9.6.A.1 states that the second law of thermodynamics states that the total entropy of an isolated system can never decrease and is constant only when all processes the system undergoes are reversible.
  • 9.6.A.2 states that entropy can be qualitatively described as the tendency of energy to spread or the unavailability of some of the system's energy to do work.
  • 9.6.A.2.i states that localized energy will tend to disperse and spread out.
  • 9.6.A.2.ii states that entropy is a state function and therefore only depends on the current state or configuration of a system, not how the system reached that state.
  • 9.6.A.2.iii states that maximum entropy occurs when a system is in thermodynamic equilibrium.
  • 9.6.A.3 states that the change in a system's entropy is determined by the system's interactions with its surroundings.
  • 9.6.A.3.i states that isolated systems spontaneously move toward thermodynamic equilibrium.
  • 9.6.A.3.ii states that the entropy of an isolated system never decreases, but the entropy of a closed system can decrease because energy can be transferred into or out of the system.

Boundary statement, in full: "Only qualitative treatment of the second law of thermodynamics is within the scope of AP Physics 2."

The CED lists four suggested skills: 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

No computation skill is listed. Every other Unit 9 topic pairs with 2.A, 2.B, or 2.D; 9.6 pairs with a representation skill, a comparison skill, and two claim skills. The boundary statement is showing up in the skills table. The CED does caveat that Progress Check questions are built on this pairing but "AP Exam questions can pair the content with any of the skills.\"

The second law, in the CED's own words (9.6.A.1)

The CED states it in one sentence, and both halves are examinable:

"The second law of thermodynamics states that the total entropy of an isolated system can never decrease and is constant only when all processes the system undergoes are reversible."

Three words carry that sentence.

  • Total. The entropy of the whole isolated system, not of any part of it. The parts are free to go either way, and 9.6.A.3.ii exists to say so.
  • Isolated. The qualifier that popular statements of the second law drop, and dropping it makes the law false. More on that below, because it is where most exam questions live.
  • Reversible. The only condition under which the total holds still. So "entropy increases" is the ordinary case and "entropy is constant" is the special one.

Read the sentence for what it does not say. It does not say entropy always increases, it does not say the entropy of every system increases, and it gives you no number for anything. Nor is "can never decrease" the same as "must increase": constant is a permitted outcome, and the CED names exactly one condition for it.

One honest limit. The word "reversible" appears exactly once in the entire AP Physics 2 Course and Exam Description, in this sentence, and the CED never defines it. What you are responsible for is the conditional structure of the statement, not a technical account of reversibility. The defensible exam move is the one 9.6.A.1 licenses: for an isolated system, claim the total entropy did not decrease, and say it increased where a process was clearly not reversible.

What entropy is in this course, and why disorder is not it (9.6.A.2)

Essential knowledge 9.6.A.2 gives one description of entropy, and it is explicitly labelled qualitative:

"Entropy can be qualitatively described as the tendency of energy to spread or the unavailability of some of the system's energy to do work."

Two descriptions, joined by "or", and both are about energy. Neither is about disorder, tidiness, messiness, or information.

That matters more than vocabulary. The word "disorder" appears nowhere in the AP Physics 2 Course and Exam Description: not in Unit 9's required course content, not in the unit opener, not on either page of Topic 9.6. It is the framing most of the web uses for entropy, and it is not this course's.

The practical problem with "disorder" is that it gives you nothing to reason with: you cannot inspect a gas and score its tidiness. The CED's two descriptions both point at something you can track:

  • Did the energy spread out? Energy that started concentrated in one place and ended distributed over more of the system spread out.
  • Is less of the energy now available to do work? The first law says none of the energy went anywhere, but that is not the same as saying all of it is still usable.

Those questions have answers you can defend from the setup in front of you, which is exactly what skills 3.B and 3.C ask for.

The second description is the one students skip and the one that connects to the rest of the unit. Topic 9.4 says the first law is a restatement of conservation of energy, so after any process the energy is all still accounted for. An increase in entropy says less of it can now be used to do work on anything. Both are true at once, and questions that look paradoxical almost always turn on holding both.

Energy spreads, and equilibrium is where it stops

Three of the essential knowledge statements chain together into the only story this topic tells.

  1. 9.6.A.2.i: localized energy will tend to disperse and spread out.
  2. 9.6.A.3.i: isolated systems spontaneously move toward thermodynamic equilibrium.
  3. 9.6.A.2.iii: maximum entropy occurs when a system is in thermodynamic equilibrium.

In order, they say that spontaneous change and increasing entropy run the same way, and that equilibrium is where the increase stops. That is why 9.6.A.2.iii says "maximum": equilibrium is not one state among many, it is the end of the road for an isolated system.

Unit 9 tells this story twice. Topic 9.3 tells it as an observation: essential knowledge 9.3.A.3 says energy is transferred through thermal processes spontaneously from a higher-temperature system to a lower-temperature system, and 9.3.A.4 says thermal equilibrium results when no net energy is transferred between two systems in thermal contact. Topic 9.6 supplies a principle with the same content. The CED does not derive one from the other, but they point the same way and a 3.B question can be answered from either.

Topic 9.3 is also where the course puts its only probabilistic language, and it is worth seeing where it is not. Essential knowledge 9.3.A.3.i says that in collisions between atoms from different systems, energy is most likely to be transferred from higher-energy atoms to lower-energy atoms, and 9.3.A.3.ii says that after many collisions the most probable state is one in which both systems have the same temperature. Those two statements are about atomic collisions in Topic 9.3, not about entropy in Topic 9.6, and neither comes with a formula. AP Physics 2 does not import a statistical-mechanics definition of entropy, and an answer that reaches for one is answering a different course's question.

Entropy is a state function (9.6.A.2.ii)

Essential knowledge 9.6.A.2.ii gives the definition and its consequence in a single sentence: entropy is a state function and therefore only depends on the current state or configuration of a system, not how the system reached that state.

Three things that buys you.

  • Same final state, same entropy. Two processes ending at the same state end with the same entropy, so once you establish that the final states match, the entropy comparison is settled without knowing either path.
  • A closed cycle returns the entropy to where it started, whatever the system did in between. That is the same logic that makes ΔU\Delta U zero around a cycle on a PV diagram.
  • It separates entropy from quantities that are not state functions. Work is the clear case. Essential knowledge 9.4.B.2.ii says the absolute value of the work done on a gas when it expands or compresses is equal to the area underneath the curve of a plot of pressure vs. volume, and two curves between the same two points enclose different areas, so WW depends on the path. Because ΔU=Q+W\Delta U = Q + W with ΔU\Delta U fixed by the endpoints, QQ inherits that dependence too.

The second worked example runs exactly that comparison with numbers: two paths between the same two states, differing by 2000 J in both the work and the heating, with identical ΔU\Delta U and therefore identical change in entropy. The PV diagrams guide covers the first-law bookkeeping behind those numbers.

A caution on how far to push this. Being a state function means the entropy change has a definite value set by the endpoints. It does not mean AP Physics 2 asks you for that value.

Isolated versus closed: the distinction the topic is built on

Essential knowledge 9.6.A.3 says the change in a system's entropy is determined by the system's interactions with its surroundings. Essential knowledge 9.6.A.3.ii then makes that concrete, and it is the sentence to memorize whole rather than in halves:

"The entropy of an isolated system never decreases, but the entropy of a closed system can decrease because energy can be transferred into or out of the system."

Half of that sentence is the thing everybody says; the other half is what makes the first half correct.

Unit 9 defines the two kinds of system in Topic 9.4, not in 9.6, and 9.6 leans on those definitions. Essential knowledge 9.4.B.1.i says that for an isolated system, the total energy is constant. Essential knowledge 9.4.B.1.ii says that for a closed system, the change in internal energy is the sum of energy transferred to or from the system by heating, or work done on the system. A closed system exchanges energy with its surroundings; an isolated one does not. That is precisely why 9.6.A.3 makes a system's entropy change a matter of its interactions with those surroundings.

isolated systemclosed system
energy crossing the boundarynone (9.4.B.1.i)allowed, by heating or work (9.4.B.1.ii)
what happens to its entropynever decreases (9.6.A.1, 9.6.A.3.ii)can decrease (9.6.A.3.ii)
where it ends upthermodynamic equilibrium, at maximum entropy (9.6.A.2.iii, 9.6.A.3.i)wherever its surroundings drive it

Every apparent violation of the second law a student finds is a closed system wearing the wrong label. A refrigerator cools its contents, an air conditioner cools a room, a block cools while the block beside it warms. In each case the thing whose entropy fell was exchanging energy with something else. Draw the boundary wide enough that nothing crosses it, and the total inside goes up.

That is also why the CED's first suggested skill here is 1.A, create diagrams, tables, charts, or schematics to represent physical situations. The diagram that answers most 9.6 questions is the one with the system boundary drawn on it and the energy transfers marked crossing it or not.

The boundary statement: qualitative only, and what that rules out

"Only qualitative treatment of the second law of thermodynamics is within the scope of AP Physics 2."

Concretely, here is what that means.

There is no entropy equation on the sheet. The AP Physics 2 equation sheet has 129 entries across seven groups, and none of them contains an entropy. The Thermal Physics column holds exactly eight: P=FAP = \dfrac{F_{\perp}}{A}, Kavg=32kBT=12mvrms2K_{\text{avg}} = \frac{3}{2}k_B T = \frac{1}{2}mv_{\text{rms}}^2, QΔt=kAΔTL\dfrac{Q}{\Delta t} = \dfrac{kA\Delta T}{L}, PV=nRT=NkBTPV = nRT = Nk_B T, U=32nRT=32NkBTU = \frac{3}{2}nRT = \frac{3}{2}Nk_B T, W=PΔVW = -P\Delta V, ΔU=Q+W\Delta U = Q + W, and Q=mcΔTQ = mc\Delta T. Counting them is the check; not one is an entropy relation.

Formulas you may have met elsewhere are not part of this course. An expression for entropy change as an energy transfer divided by a temperature, and a logarithmic expression counting configurations, are both outside the required course content and are printed nowhere on the sheet. If a question could only be answered by computing a numerical entropy change, it is not an AP Physics 2 question as the CED defines the course.

What is expected instead is a direction and a justification: say whether the entropy of a named system increased, decreased, or stayed the same, and support that from the essential knowledge statements, with skill 3.B applying the law and 3.C supplying the evidence. Comparative answers are fair game, which is what 2.C is for, but they compare direction and scenario, not computed values.

This is one place where reading the current CED is worth real marks. Material built on an older treatment will hand you a formula and a calculation. The Effective Fall 2024 course description fences both off in one sentence.

How Topic 9.6 is tested

The four suggested skills, and the question shapes each produces:

  1. Say what happened to the entropy of a stated system, and justify it (3.B and 3.C). The system is named for a reason. Check first whether it is isolated or closed, and answer 9.6.A.3.ii accordingly.
  2. Compare two scenarios (2.C). Which of two processes produces a larger increase, or does one of them leave the total unchanged. Compare directions and reasoning, not numbers.
  3. Resolve an apparent violation (3.B). A described device or process seems to break the second law. Name the system, check the qualifier, widen the boundary.
  4. Draw or annotate a representation (1.A). System, surroundings, and the direction of every energy transfer across the boundary.
  5. Explain why a process runs one way and not the other when both directions conserve energy (3.B). This is the question the first law cannot answer, and it is the reason the topic exists.

The unit opener flags that the fourth free-response question is the Qualitative/Quantitative Translation, which "first requires students to make a claim and provide evidence and reasoning to support their claim without reference to equations". That is exactly the register Topic 9.6 is written in, and the CED notes that students used mainly to numerical problem solving often struggle with the format. Practise writing the argument in sentences.

A structure that works for the claim questions, and maps onto 3.B and 3.C: claim, the entropy of the named system increased, decreased, or stayed the same; evidence, the energy transfers that actually crossed that system's boundary, and whether it is isolated or closed; reasoning, the essential knowledge statement that licenses the claim, accurately paraphrased. Use 9.6.A.1 for an isolated total, 9.6.A.3.ii for a closed part, 9.6.A.2.iii for an equilibrium endpoint.

Where Topic 9.6 sits in Unit 9

The unit opener says Unit 9 "also acquaints students with the second law of thermodynamics, including entropy". "Acquaints" is the CED's word, and the boundary statement backs it up: this topic is the unit's qualitative capstone, not a new calculation.

The order of dependence runs cleanly:

  • Topic 9.1 ties temperature to atomic motion, and Topic 9.2 gives the ideal gas law.
  • Topic 9.3 gives spontaneous transfer from hot to cold, and thermal equilibrium.
  • Topic 9.4 gives internal energy, the first law, and the isolated and closed vocabulary that 9.6 borrows.
  • Topic 9.5 gives you the numbers: an equilibrium temperature out of Q=mcΔTQ = mc\Delta T, a conduction rate out of kAΔTL\dfrac{kA\Delta T}{L}.
  • Topic 9.6 says which of the processes the first law permits actually happen.

That last line is the point of the topic. The first law is a filter that lets far too much through: every reversal of a spontaneous process conserves energy perfectly, and the first law objects to none of them. The second law rules them out, and it is the only thing in the course that does.

The unit's remaining essential questions are worth reading with that in mind. "How cold can something get?" is answered in Topic 9.2, whose essential knowledge 9.2.A.4 says a temperature at which an ideal gas has zero pressure can be extrapolated from a graph of pressure as a function of temperature. The two questions about heat engines and about the limitations and efficiency of technological systems have no essential knowledge attached to them: Unit 9's required course content contains no heat-engine statement and no efficiency equation. The CED calls essential questions thought-provoking questions that motivate students and inspire inquiry, so treat those two as context rather than as content to revise.

Two blocks in an insulated box: which entropies go up and which go down

Block A (0.400 kg, specific heat 900J/(kgK)900 \, \mathrm{J/(kg \cdot K)}) starts at 350 K. Block B (0.600 kg, specific heat 400J/(kgK)400 \, \mathrm{J/(kg \cdot K)}) starts at 290 K. They are placed in contact inside a rigid, perfectly insulating container and left until nothing changes. (a) Find the final temperature and the energy transferred. (b) Say what happens to the entropy of block A, of block B, and of the two-block system, and justify each. (c) Explain why the reverse process is not ruled out by the first law but is ruled out by the second.

  1. Sign convention: QQ is positive for energy transferred into an object, matching the sheet's definition. The container is rigid and insulating, so no energy crosses its wall and no work is done: QA+QB=0Q_A + Q_B = 0.

  2. (a) Energy per kelvin: mAcA=(0.400)(900)=360J/Km_A c_A = (0.400)(900) = 360 \, \mathrm{J/K} and mBcB=(0.600)(400)=240J/Km_B c_B = (0.600)(400) = 240 \, \mathrm{J/K}. Then 360(T350)+240(T290)=0360(T - 350) + 240(T - 290) = 0, so 600T=195,600600\,T = 195{,}600 and T=326.0KT = 326.0 \, \mathrm{K}.

  3. Transfers: QA=360(326.0350)=8640JQ_A = 360(326.0 - 350) = -8640 \, \mathrm{J} and QB=240(326.0290)=+8640JQ_B = 240(326.0 - 290) = +8640 \, \mathrm{J}, summing to zero as the insulation requires.

  4. (b) Block A on its own is a closed system: 8640 J left it, and essential knowledge 9.6.A.3.ii permits its entropy to decrease for exactly that reason. Block B is also closed, with 8640 J entering, and its entropy increased.

  5. The two blocks together, inside a rigid insulating container, are an isolated system, so by 9.6.A.1 the total entropy can never decrease. Transfer across a 60 K difference is not reversible, so this is not the constant case and the total increased. By 9.6.A.2.iii, 326.0 K is where this isolated system's entropy is at its maximum.

  6. (c) The reverse process moves 8640 J from B back to A, returning A to 350 K and B to 290 K. Total energy is unchanged, so 9.4.B.1.i still holds and the first law has no objection.

  7. It is 9.6.A.1 that forbids it. Running the process backwards returns the isolated pair to its earlier state, which by 9.6.A.2.iii has less entropy than equilibrium, so the total entropy of an isolated system would have to decrease. Essential knowledge 9.3.A.3 says the same thing as an observation: spontaneous transfer runs from higher temperature to lower.

(a) T=326.0KT = 326.0 \, \mathrm{K}, with 8640 J transferred from A to B. (b) Block A's entropy decreased, which 9.6.A.3.ii permits because A is closed and energy left it; block B's increased; the isolated pair's total increased, and 326.0 K is its maximum-entropy state. (c) The reverse conserves energy perfectly, so the first law permits it. 9.6.A.1 forbids it, because the total entropy of an isolated system can never decrease.

Two paths, same endpoints: what changes and what does not

One mole of a monatomic ideal gas goes from P1=2.00×105PaP_1 = 2.00 \times 10^5 \, \mathrm{Pa}, V1=1.00×102m3V_1 = 1.00 \times 10^{-2} \, \mathrm{m^3} to P2=1.00×105PaP_2 = 1.00 \times 10^5 \, \mathrm{Pa}, V2=3.00×102m3V_2 = 3.00 \times 10^{-2} \, \mathrm{m^3}. Path A expands at constant pressure, then drops the pressure at constant volume. Path B drops the pressure at constant volume first, then expands at constant pressure. Find WW, QQ, and ΔU\Delta U for each path, and say what each does to the gas's entropy.

  1. Sign convention: WW is the work done on the gas, per the sheet's W=PΔVW = -P\Delta V and its symbol list, so an expansion has ΔV>0\Delta V > 0 and W<0W < 0.

  2. ΔU\Delta U is fixed by the endpoints: the sheet prints U=32nRTU = \frac{3}{2}nRT and PV=nRTPV = nRT, so U=32PVU = \frac{3}{2}PV. With P1V1=(2.00×105)(1.00×102)=2000JP_1V_1 = (2.00 \times 10^5)(1.00 \times 10^{-2}) = 2000 \, \mathrm{J} and P2V2=(1.00×105)(3.00×102)=3000JP_2V_2 = (1.00 \times 10^5)(3.00 \times 10^{-2}) = 3000 \, \mathrm{J}, ΔU=32(1000)=+1500J\Delta U = \frac{3}{2}(1000) = +1500 \, \mathrm{J} for both paths.

  3. Path A, isobaric leg at 2.00×105Pa2.00 \times 10^5 \, \mathrm{Pa}: W=PΔV=(2.00×105)(2.00×102)=4000JW = -P\Delta V = -(2.00 \times 10^5)(2.00 \times 10^{-2}) = -4000 \, \mathrm{J}; the isovolumetric leg has ΔV=0\Delta V = 0, so W=0W = 0. Total WA=4000JW_A = -4000 \, \mathrm{J}, and QA=ΔUWA=+5500JQ_A = \Delta U - W_A = +5500 \, \mathrm{J}.

  4. Path B, isovolumetric leg: W=0W = 0. Isobaric leg at 1.00×105Pa1.00 \times 10^5 \, \mathrm{Pa}: W=(1.00×105)(2.00×102)=2000JW = -(1.00 \times 10^5)(2.00 \times 10^{-2}) = -2000 \, \mathrm{J}. Total WB=2000JW_B = -2000 \, \mathrm{J}, and QB=+3500JQ_B = +3500 \, \mathrm{J}.

  5. Check the difference against the diagram: the paths differ by 2000 J in both WW and QQ, and the rectangle they enclose on a PV plot has area (1.00×105)(2.00×102)=2000J(1.00 \times 10^5)(2.00 \times 10^{-2}) = 2000 \, \mathrm{J}. Work is a path quantity, which 9.4.B.2.ii puts under the curve, and QQ inherits that dependence because ΔU\Delta U has none.

  6. Entropy. Essential knowledge 9.6.A.2.ii makes entropy a state function, depending only on the current state and not on how the system got there. Both paths end at the same state, so both produce the same change in the gas's entropy, even though one transferred 2000 J more by heating.

  7. What the course does not ask is a number for that change: the boundary statement limits the second law to qualitative treatment, and the sheet prints nothing to compute an entropy with. The examinable point is that the internal energy and the entropy are path independent while QQ and WW are not.

ΔU=+1500J\Delta U = +1500 \, \mathrm{J} on both paths. Path A: W=4000JW = -4000 \, \mathrm{J}, Q=+5500JQ = +5500 \, \mathrm{J}. Path B: W=2000JW = -2000 \, \mathrm{J}, Q=+3500JQ = +3500 \, \mathrm{J}. The change in entropy is the same for both, because 9.6.A.2.ii makes entropy a state function. Work and heating are not state quantities and differ by 2000 J, the rectangle between the two paths.

A refrigerator does not break the second law: writing the argument

A student argues: a refrigerator makes the food inside it colder, so the food's energy is less spread out and its entropy has decreased; the second law says entropy can never decrease; therefore refrigerators are impossible. Identify the flaw and write the corrected argument in the claim, evidence, and reasoning form that skills 3.B and 3.C ask for.

  1. Name the candidate systems first and draw a boundary for each: the food alone, then the food together with the refrigerator and the room it stands in.

  2. Test the first against the wording of 9.6.A.1, which restricts the second law to an isolated system. The food is not isolated: energy is transferred out of it, which is what cooling it means.

  3. So the student's first claim survives and the law they invoked does not apply to it. Essential knowledge 9.6.A.3.ii says so directly: the entropy of an isolated system never decreases, but the entropy of a closed system can decrease because energy can be transferred into or out of the system. The food's entropy really does decrease, and nothing is violated. That is 9.6.A.3 in action.

  4. Now widen the boundary to the food, the refrigerator, and the room. Treat that as isolated and 9.6.A.1 applies: its total entropy cannot decrease.

  5. Account for the energy with the first law, 9.4.B.1. In steady operation the refrigerator's internal energy is not changing, so the energy it removes from the food plus the work the electricity supply delivers ends up in the room. Energy that arrived concentrated in one cable is spread through the air of the kitchen, which is 9.6.A.2.i.

  6. Write it as claim, evidence, reasoning. Claim: the food's entropy decreases and the total entropy of the food, refrigerator, and room does not. Evidence: energy is transferred out of the food and into the room, and the wider system exchanges essentially nothing with anything else. Reasoning: 9.6.A.3.ii permits a closed system's entropy to fall, and 9.6.A.1 governs only the isolated total. Do not attempt a number: the boundary statement puts any quantitative entropy change outside the course, and this argument is the complete expected answer.

The flaw is a missing qualifier. Essential knowledge 9.6.A.1 restricts the second law to an isolated system, and food in a refrigerator is not isolated. Essential knowledge 9.6.A.3.ii permits exactly this: a closed system's entropy can decrease because energy can be transferred into or out of it. Redraw the boundary around food, refrigerator, and room and that system is isolated, so its total entropy does not decrease. AP Physics 2 asks for the argument in words, not for a calculation.

Frequently asked questions

What is the second law of thermodynamics in AP Physics 2?

Essential knowledge 9.6.A.1 states it in one sentence: the total entropy of an isolated system can never decrease and is constant only when all processes the system undergoes are reversible. Both halves matter. The law is about an isolated system, so the entropy of a part of that system, or of any system exchanging energy with its surroundings, is free to fall. And a constant total is permitted, but only in the reversible case, so an increase is the ordinary outcome.

Is entropy the same as disorder?

Not in AP Physics 2. The word disorder does not appear anywhere in the AP Physics 2 Course and Exam Description. Essential knowledge 9.6.A.2 describes entropy qualitatively as the tendency of energy to spread, or the unavailability of some of the system's energy to do work. Both descriptions are about energy, and both give you something you can actually reason from: where the energy went, and how much of it is still available to do work. Disorder is a popular framing from outside the course and it will not support an answer.

Can entropy ever decrease?

Yes, for the right kind of system. Essential knowledge 9.6.A.3.ii says the entropy of an isolated system never decreases, but the entropy of a closed system can decrease because energy can be transferred into or out of the system. A can of drink cooling in a refrigerator has decreasing entropy and breaks no law, because it is not isolated. What can never decrease is the total entropy of a system with nothing crossing its boundary.

What is the difference between an isolated and a closed system in thermodynamics?

AP Physics 2 draws the line in Topic 9.4. Essential knowledge 9.4.B.1.i says that for an isolated system, the total energy is constant. Essential knowledge 9.4.B.1.ii says that for a closed system, the change in internal energy is the sum of energy transferred to or from the system by heating, or work done on the system. So a closed system exchanges energy with its surroundings and an isolated one does not. That distinction is what decides whether the second law applies, so identify which one a question is describing before answering it.

Do you need to calculate entropy in AP Physics 2?

No. The Topic 9.6 boundary statement reads in full: only qualitative treatment of the second law of thermodynamics is within the scope of AP Physics 2. There is no entropy equation anywhere on the AP Physics 2 equation sheet, whose Thermal Physics column holds eight equations and none of them an entropy relation. What is expected is a direction and a justification: say whether a named system's entropy increased, decreased, or stayed the same, and support it from the essential knowledge statements.

What does it mean that entropy is a state function?

Essential knowledge 9.6.A.2.ii says entropy is a state function and therefore only depends on the current state or configuration of a system, not how the system reached that state. Two processes that finish in the same state finish with the same entropy, and a system returned to its starting state is back to its starting entropy. Work and energy transferred by heating are not state functions: the work done on a gas is the area under its path on a pressure-volume plot, and different paths between the same two states enclose different areas.