IB DP Physics: Molecular model, temperature, internal energy and heating calculations (DP Physics) – Revision Notes

These revision notes condense the first half of topic B.1 (Thermal energy transfers) from the IB Physics guide, first assessment 2025, for Standard Level and Higher Level. Nothing in B.1 is marked additional higher level. The study guide gives the full explanations, and the practice set tests them.

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Molecular model in one table

State	Spacing	Motion	Forces
Solid	Touching, fixed positions	Vibration about a point	Strong
Liquid	Touching, no fixed positions	Vibration plus sliding past neighbours	Strong but not locking
Gas	Far apart	Fast random motion between collisions	Negligible between collisions

• Solids and liquids resist compression because the particles already touch.
• Gases fill their container and compress easily because of the large gaps.

Formulas and rules

Quantity	Relationship	Notes
Density	ρ = m/V	kg m⁻³, volume in m³
Temperature conversion	T / K = θ / °C + 273	273.15 for more precision
Temperature change	ΔT in K = ΔT in °C	Same size of unit
Average kinetic energy	Ek = 3/2 kB T	T in kelvin only
Heating	Q = mcΔT	c in J kg⁻¹ K⁻¹, no phase change
Phase change	Q = mL	L_f fusion, L_v vaporization
Constant power	Q = Pt	Energy equals power times time

Key ideas to recall

• Internal energy is the sum, over all molecules, of the intermolecular potential energy and the random kinetic energy.
• Temperature is a measure of the average random kinetic energy per particle (in kelvin). It is not a measure of total internal energy.
• Direction of transfer is from the higher temperature to the lower temperature, whatever the sizes of the two bodies.
• Phase change: temperature constant, kinetic energy per particle unchanged, potential energy changes.
• Evaporation is a surface effect that needs no particular temperature. Boiling takes place in the bulk of the liquid at its boiling point.

Must-know distinctions

Idea A	Idea B
Temperature (average kinetic energy)	Internal energy (total of kinetic and potential)
Specific heat capacity (change temperature)	Specific latent heat (change state)
Boiling (bulk, fixed temperature)	Evaporation (surface, any temperature)
Melting (solid to liquid)	Freezing (liquid to solid)
Condensing (gas to liquid)	Evaporation (liquid to gas)

Method for a heating calculation

1. List the stages. A new stage starts at every melting or boiling point.
2. For each rise in temperature use Q = mcΔT with the c for that state.
3. For each change of state use Q = mL with the right L.
4. Add the stages. If a heater is involved, use t = Q/P.
5. Check units (kg, J, K) and give the answer to three significant figures unless told otherwise.

Method for a mixing problem

1. State the assumption: no energy is lost to the surroundings.
2. The hotter body loses energy and the colder body gains the same amount: m₁c₁(T₁ − T) = m₂c₂(T − T₂).
3. Solve for T. A sensible answer sits between the two starting temperatures.
4. Check that no phase change has been crossed.

Worked reminders

Density. A 2.0 litre container (2.0 × 10⁻³ m³) holds 5.4 kg of a liquid: ρ = 5.4 / 2.0 × 10⁻³ = 2.7 × 10³ kg m⁻³.

Average kinetic energy. At 620 K, Ek = 3/2 × 1.38 × 10⁻²³ × 620 = 1.28 × 10⁻²⁰ J. At 1500 K it is 3.11 × 10⁻²⁰ J.

Heating water. Warming 0.60 kg of water by 50 K needs 0.60 × 4180 × 50 = 1.25 × 10⁵ J. A 500 W heater takes 251 s.

Latent heat. Melting 0.90 kg of ice at 0 °C needs 0.90 × 3.34 × 10⁵ = 3.01 × 10⁵ J.

Heating curves

A heating curve plots temperature against time for a sample heated at constant power.

Part of the graph	What is happening	Energy goes into
Sloping line, solid	Temperature rising	Random kinetic energy of vibrating particles
Flat line at melting point	Melting	Potential energy as particles separate
Sloping line, liquid	Temperature rising	Random kinetic energy, mostly
Flat line at boiling point	Boiling	Potential energy as particles move far apart
Sloping line, gas	Temperature rising	Random kinetic energy

• Length of a flat section times power gives Q, and Q/m gives L.
• Gradient of a sloping section is P/(mc), so a steeper line means a smaller specific heat capacity for the same mass and power.
• The boiling stage is much longer than the melting stage because L_v is much larger than L_f.
• On cooling, the same curve is read backwards: condensing and freezing release the energy.

Two short worked reminders

Two-stage heating. Warm 0.40 kg of water from 90 °C to 100 °C, then boil away 0.050 kg of it.

Stage 1: 0.40 × 4180 × 10 = 1.67 × 10⁴ J
Stage 2: 0.050 × 2.26 × 10⁶ = 1.13 × 10⁵ J
Total = 1.30 × 10⁵ J

Using the kinetic energy relation backwards. A gas has atoms with an average kinetic energy of 6.9 × 10⁻²¹ J. Then T = 2Ek / 3kB = 2 × 6.9 × 10⁻²¹ / (3 × 1.38 × 10⁻²³) = 333 K, which is about 60 °C.

Explaining in words

Answers score best when they name the correct energy store. Use these sentence frames.

• Warming a solid: the average random kinetic energy of the particles increases, so the temperature rises.
• Melting: the energy supplied increases the potential energy of the particles as the bonds between them are loosened, while the average kinetic energy and temperature stay constant.
• Compressing a gas is easy, compressing a liquid is hard: the gas particles are far apart, the liquid particles are already in contact.

Quick self-test

1. Convert 20 °C to kelvin and 500 K to degrees Celsius.
2. A body is heated from 300 K to 340 K. What is the change in temperature in degrees Celsius?
3. A 0.75 kg block of material occupies 3.0 × 10⁻⁴ m³. Find its density.
4. Find the average kinetic energy of a particle at 750 K (kB = 1.38 × 10⁻²³ J K⁻¹).
5. By what factor does the average kinetic energy change when the temperature rises from 150 K to 450 K?
6. A student says the average kinetic energy doubles when a gas goes from 10 °C to 20 °C. Is that correct?
7. Name the two parts of the internal energy of a substance.
8. A 0.25 kg lead block (c = 129 J kg⁻¹ K⁻¹) cools from 300 °C to 20 °C. How much energy does it lose?
9. How much energy is needed to boil away 1.5 kg of water at 100 °C? (L_v = 2.26 × 10⁶ J kg⁻¹)
10. State what happens to the potential energy and the average kinetic energy of the molecules as ice melts at 0 °C.
11. Block P at 60 °C and a much larger block Q at 40 °C touch. In which direction is the net energy transfer?

Answers

1. 293 K; 227 °C.
2. 40 °C (a change of 40 K is 40 °C).
3. ρ = 0.75 / 3.0 × 10⁻⁴ = 2.5 × 10³ kg m⁻³.
4. Ek = 1.5 × 1.38 × 10⁻²³ × 750 = 1.55 × 10⁻²⁰ J.
5. It triples (450/150 = 3).
6. No. In kelvin the temperature goes from 283 K to 293 K, a rise of under 4 percent.
7. Intermolecular potential energy and random kinetic energy of the molecules.
8. Q = 0.25 × 129 × 280 = 9.03 × 10³ J.
9. Q = 1.5 × 2.26 × 10⁶ = 3.39 × 10⁶ J.
10. Potential energy rises; average kinetic energy stays the same because the temperature is constant.
11. From P to Q, because P has the higher temperature, even though Q is larger.

Where marks are usually lost

• Using °C in Ek = 3/2 kB T, or doubling a Celsius temperature and calling it a doubling of energy.
• Substituting grams or cm³ without converting to kg or m³.
• Using the specific heat capacity of water for ice or the other way round.
• Leaving out a stage, usually the melting or boiling stage, in a multi-stage heating calculation.
• Writing that the temperature rises during a phase change.
• Explaining a phase change as an increase in kinetic energy of the particles, rather than potential energy.
• Confusing temperature with internal energy, or heat with temperature, in written answers.
• Leaving the unit off the last line, or quoting too many or too few significant figures.

Official syllabus

Source: the Physics guide for the IB Diploma Programme (first assessment 2025), International Baccalaureate Organization. Topic B.1, Thermal energy transfers, first half. Keep to the guide’s own wording of each statement when you write answers.