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NotesPhysics HLTopic 2.4Entropy and system evolution
Back to Physics HL Topics
2.4.25 min read

Entropy and system evolution (Physics HL)

IB Physics • Unit 2

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Contents

  • What entropy measures
  • Calculating entropy change
  • The second law of thermodynamics
  • Time's arrow — why heat flows hot to cold
  • In the exam
The big idea: Stir a spoon of sugar into your coffee and it spreads out and dissolves — it never gathers itself back into a neat cube. Entropy (symbol S) is the measure behind that one-way spreading: the number of microscopic ways (microstates) the particles can be arranged while still looking the same on the outside.

More ways to arrange the particles ⇒ higher entropy. A tidy, ordered state has low entropy; a spread-out, jumbled state has high entropy.
Microstates in plain words: Imagine gas atoms in a box. There are vastly more ways for them to be spread evenly throughout the box than to be bunched in one corner. So the spread-out state has far more microstates — it is the high-entropy state, and it is the one you actually find.

Entropy is measured in joules per kelvin (J K⁻¹). It is a property of the whole system, like temperature or internal energy.

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When an amount of heat ΔQ flows into or out of a body at constant temperature T, the body's entropy changes by a definite amount. The temperature must always be in kelvin.

Given in the data booklet. Heat in (ΔQ > 0) raises entropy; heat out (ΔQ < 0) lowers it. T must be in kelvin.
entropy change (J K⁻¹)
heat transferred (J): positive in, negative out
absolute temperature (K)
IB-style questionCalculate[2 marks]

600 J of heat flows into a large reservoir held at a constant temperature of 300 K. Find the change in entropy of the reservoir.

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The second law: The second law of thermodynamics states:

The entropy of an isolated system never decreases.

It stays the same for an ideal (reversible) process, and it increases for every real (irreversible) process. The entropy of the universe as a whole therefore always goes up.
Look at the whole, isolated system: One part can lose entropy — but only if another part gains more. The law applies to the total entropy of an isolated system (one that exchanges no heat with its surroundings).
IB-style questionDetermine[3 marks]

1200 J of heat flows from a hot body at 400 K to a cold body at 300 K. Treating the two bodies together as an isolated system, find the net change in entropy.

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Why real processes only run one way: The second law gives time a direction ('time's arrow'). A process happens spontaneously only if it increases the total entropy of the universe.

Heat flowing hot → cold raises total entropy (the worked example gave +1.0 J K⁻¹), so it happens by itself. The reverse — heat flowing cold → hot on its own — would lower total entropy, so it never happens.

Happens by itself (ΔS_total > 0)

  • Heat flows from a hot body to a cold body
  • A gas expands to fill an empty space
  • A drop of ink spreads through water
  • These are irreversible — they never un-happen on their own

Never happens by itself (would give ΔS_total < 0)

  • Heat flowing from a cold body to a hot body unaided
  • A spread-out gas collecting back into one corner
  • Ink un-mixing out of the water
  • Each would decrease total entropy, so it is forbidden

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How this is tested — entropy and the second law are HL only (B.4) and appear at two levels:

Paper 1A

  • A one-step 'what is ΔS?' using ΔS = ΔQ/T.
  • 'What does the second law state?' multiple-choice.

Paper 2

  • Calculate a net entropy change for two bodies.
  • Then comment on whether the process is possible (look for ΔStotal > 0).
The classic trap: Leaving temperature in °C. ΔS = ΔQ/T needs T in kelvin (add 273). Forgetting to convert throws every entropy value out — and if your total ΔS comes out negative for a real process, a sign or a unit has slipped.
Three easy marks: (1) Always put T in kelvin (add 273 to °C). (2) Heat in is +ΔQ, heat out is −ΔQ. (3) For a real process the total ΔS must come out positive — if yours is negative, check a sign.
IB-style questionDetermine[4 marks]

A copper block at 500 K is placed in contact with a water bath at 250 K. In a short interval 2000 J of heat flows from the block to the water, and both temperatures stay essentially constant. Determine the net entropy change of the block–water system and state whether the process can occur.

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When the water in the previous part freezes, its entropy decreases. A student claims this breaks the second law of thermodynamics.

why the freezing of the water is in fact consistent with the second law.
[2 marks]

Related Physics HL Topics

Continue learning with these related topics from the same unit:

2.1.1Internal energy and the particle model
2.1.2Specific heat capacity
2.1.3Latent heat and calorimetry
2.1.4Conduction, convection and radiation
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