IB DP Physics: Entropy, the second law, heat engines and the Carnot limit (DP Physics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover the second half of sub-topic B.4, Thermodynamics, in the IB Physics guide (first assessment 2025): entropy, the second law, heat engines and the Carnot limit. All of B.4 is additional higher level, so every question is HL only. Constants are supplied where needed: k_B = 1.38 × 10⁻²³ J K⁻¹, and 0 °C = 273.15 K. Learn the ideas first in the study guide and the revision notes.

Related pages: the course hub and the printable checklist.

Questions

1. Six identical coins are tossed and land independently, each heads or tails with equal probability. For the macrostate “3 heads”: find the number of microstates, the probability of the macrostate, and the entropy S. [3]

2. State the Clausius form and the Kelvin form of the second law of thermodynamics. [2]

3. A metal strap carries 900 J of thermal energy from a large reservoir at 360 K to a large reservoir at 280 K. Calculate the entropy change of each reservoir and of the combined system. [3]

4. A 0.15 kg sample of lead melts at its melting point of 600.6 K. The specific latent heat of fusion of lead is 2.3 × 10⁴ J kg⁻¹ (supplied data). Calculate the entropy change of the lead, and explain the sign of the change in terms of microstates. [4]

5. A heat engine produces 410 J of useful work per cycle with an efficiency of 0.34. Calculate the energy taken in per cycle and the energy rejected per cycle. [3]

6. A Carnot engine operates between reservoirs at 650 K and 290 K and takes in 4.8 kJ per cycle from the hot reservoir. Calculate its efficiency and the work it does per cycle. [3]

7. Two proposals are made for an engine working between 650 K and 290 K and taking in 5000 J per cycle. Proposal A produces 2600 J of work. Proposal B produces 3000 J of work. Decide whether each is consistent with the second law, and for B support your decision by calculating the entropy change of the two reservoirs per cycle. [4]

8. In a model of a gas, each of N = 3.0 × 10²³ particles is equally likely to be found in either half of a sealed box. A partition between the halves is removed from a gas that was entirely in one half. Calculate the entropy increase, and explain why the gas is never seen to return to one half. [4]

9. A freezer converts 0.20 kg of water at 273.15 K to ice at the same temperature. The room is at 297 K. The specific latent heat of fusion of water is 3.34 × 10⁵ J kg⁻¹ (supplied data).

(a) Calculate the entropy change of the water. [2] (b) Calculate the least energy that must be delivered to the room. [1] (c) Hence find the least work the freezer must do, and explain why the second law allows the water’s entropy to fall. [2]

10. A power station has steam at 823 K and cooling water at 293 K. It delivers 500 MW of electrical power with an efficiency of 0.38. Treat the temperatures as 823.15 K and 293.15 K.

(a) Calculate the Carnot limit for these temperatures. [2] (b) Calculate the rate at which energy is supplied to the steam and the rate at which energy is rejected. [2] (c) Calculate the rate of entropy production in the reservoirs. [2] (d) Show that a Carnot engine with the same energy input would deliver about 347 MW more than this station. [2]

11. An engine takes 1200 J per cycle from a hot reservoir at 600 K and rejects 800 J per cycle to a cold reservoir at 300 K. It completes 20 cycles per second.

(a) Calculate the work per cycle and the efficiency. [2] (b) Calculate the Carnot limit and comment on the comparison. [2] (c) Calculate the entropy change of the reservoirs per cycle. [2] (d) Calculate the power output. [1] (e) Explain why the engine cannot be changed so that no energy is rejected. [1]

12. A Carnot engine works between 600 K and 300 K. Compare two upgrades: raising the hot reservoir by 50 K, or lowering the cold reservoir by 50 K. State which gives the larger Carnot limit, and why a real engine will not reach either value. [5]

Answers

1. Ω = 6!/(3! × 3!) = 20 [1]. Total microstates = 2⁶ = 64, so probability = 20/64 = 0.3125 [1]. S = k_B ln 20 = 1.38 × 10⁻²³ × 3.00 = 4.14 × 10⁻²³ J K⁻¹ [1] Examiner insight: Show the combination formula before the number; a bare “20” gives the reader nothing to credit if it is wrong, and S needs the natural logarithm.

2. Clausius: thermal energy does not flow of its own accord from a colder body to a hotter one (it needs work to be done) [1]. Kelvin: no cyclic process can take energy from a single reservoir and convert all of it into work [1] Examiner insight: Each form is a separate statement. Kelvin answers are stronger with both “cyclic” and “all of it into work”, so a vague “no engine is 100% efficient” is weaker.

3. Hot reservoir: ΔS = −900/360 = −2.50 J K⁻¹ [1]. Cold reservoir: ΔS = +900/280 = +3.21 J K⁻¹ [1]. Total = −2.50 + 3.21 = +0.714 J K⁻¹ [1] Examiner insight: The sign of each term matters. State which reservoir loses energy before dividing, since a correct size with the wrong sign can lose credit for the total.

4. Q = mL = 0.15 × 2.3 × 10⁴ = 3450 J [1]. ΔS = 3450/600.6 = 5.74 J K⁻¹ [1]. The change is positive [1] because the liquid has far more possible arrangements of its particles than the solid, so Ω increases and S = k_B ln Ω increases [1] Examiner insight: Link the sign to Ω explicitly. “More disorder” on its own is acceptable only if you also say what is being counted.

5. Q_h = W/η = 410/0.34 = 1.21 × 10³ J [1]. Q_c = Q_h − W = 1206 − 410 [1] = 796 J [1] Examiner insight: Keep unrounded Q_h for the subtraction. Rounding to 1210 first gives 800 J and can cost the final answer.

6. η = 1 − T_c/T_h = 1 − 290/650 [1] = 0.554 [1]. W = ηQ_h = 0.554 × 4.8 = 2.66 kJ [1] Examiner insight: Temperatures are already in kelvin here, but say so. Writing the formula before substituting makes any later slip easier to credit.

7. Carnot limit = 1 − 290/650 = 0.554 [1]. Proposal A: η = 2600/5000 = 0.52, below the limit, so it is consistent with the second law [1]. Proposal B: η = 0.60, above the limit, so it is not allowed [1]. Support: Q_c = 5000 − 3000 = 2000 J, and ΔS = 2000/290 − 5000/650 = 6.90 − 7.69 = −0.796 J K⁻¹, a decrease in total entropy, which the second law forbids [1] Examiner insight: For A, say “consistent with” rather than “possible in practice”; being under the Carnot limit does not mean the engine can be built.

8. Ω_f/Ω_i = 2^N [1]. ΔS = N k_B ln 2 = 3.0 × 10²³ × 1.38 × 10⁻²³ × 0.693 [1] = 2.87 J K⁻¹ [1]. Returning to one half is one microstate among 2^N equally likely ones, which is overwhelmingly improbable, and it would decrease the entropy of an isolated system [1] Examiner insight: Mention both probability and entropy. Saying only “it would break the second law” does not explain why.

9. (a) Q = 0.20 × 3.34 × 10⁵ = 6.68 × 10⁴ J removed [1]. ΔS = −6.68 × 10⁴/273.15 = −245 J K⁻¹ [1] (b) The room must gain at least 245 J K⁻¹, so energy delivered is at least 245 × 297 = 7.26 × 10⁴ J [1] (c) Minimum work = 7.26 × 10⁴ − 6.68 × 10⁴ = 5.8 × 10³ J [1]. The water is not isolated, so its entropy can fall locally if the surroundings gain at least as much [1] Examiner insight: Part (c) needs the idea of compensation by the surroundings. Name the system and the surroundings explicitly.

10. (a) η_Carnot = 1 − 293.15/823.15 [1] = 0.644 [1] (b) Input = 500/0.38 = 1.32 × 10³ MW [1]. Rejected = 1316 − 500 = 816 MW [1] (c) Rate = 816/293.15 − 1316/823.15 = 2.783 − 1.598 [1] = 1.18 MW K⁻¹ [1] (d) Carnot output = 0.644 × 1316 = 847 MW [1], and 847 − 500 = 347 MW (equal to 293.15 × 1.18) [1] Examiner insight: A “show that” needs the intermediate value displayed. Use unrounded numbers from (b) so the final figure lands on 347.

11. (a) W = 1200 − 800 = 400 J [1]. η = 400/1200 = 0.333 [1] (b) η_Carnot = 1 − 300/600 = 0.50 [1]. The engine is below the limit, as every real engine must be [1] (c) ΔS = 800/300 − 1200/600 = 2.67 − 2.00 = +0.667 J K⁻¹ [1] per cycle, and the sign is positive because the process is irreversible [1] (d) P = 400 × 20 = 8.0 kW [1] (e) The Kelvin form forbids converting all the energy from one reservoir into work in a cycle, so some must be rejected [1] Examiner insight: In (c) state that the reservoirs are being analysed, not the engine, whose own entropy is unchanged after a full cycle.

12. Baseline: η = 1 − 300/600 = 0.50 [1]. Raise T_h to 650 K: η = 1 − 300/650 = 0.538 [1]. Lower T_c to 250 K: η = 1 − 250/600 = 0.583 [1]. Lowering the cold reservoir gives the larger limit [1]. A real engine has friction, finite temperature differences and other irreversibilities, which produce entropy and keep efficiency below the Carnot value [1] Examiner insight: Give both numbers and the comparison. A real engine is not limited only by temperatures, so name an irreversibility rather than saying “losses”.

Where marks are usually lost

• Using °C in 1 − T_c/T_h or in ΔQ/T.
• Mixing up which reservoir loses and which gains energy, giving the wrong sign for each ΔS.
• Using log base 10 for S = k_B ln Ω.
• Not stating the macrostate count as a combination (N! over k! and (N − k)!).
• Rounding intermediate values and drifting from the expected answer in multi-step engine questions.
• Giving the entropy statement without “isolated”, or ignoring the surroundings in the freezer type of question.
• Treating a Carnot limit as an achievable efficiency.
• Saying “it breaks the second law” without a number or a named form.
• Omitting units on ΔS (J K⁻¹) or on rates of entropy production (MW K⁻¹).

Next steps

• Review the ideas in the revision notes and the study guide.
• Go to the course hub or tick items off the printable checklist.
• Take any of the free 10-minute diagnostics to find your next topic.
• Book a free trial class if you would like help from a teacher.

Official syllabus

IB Physics guide, first assessment 2025, published by the International Baccalaureate Organization. This resource follows the second half of sub-topic B.4, Thermodynamics, which is entirely additional higher level.