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

This study guide follows the IB Physics guide, first assessment 2025, and covers the first half of topic B.1 (Thermal energy transfers) in the theme on the particulate nature of matter. It applies to Standard Level and Higher Level alike; B.1 contains nothing marked as additional higher level. The second half of B.1 (conduction, convection and thermal radiation) is taught separately.

Pair it with the revision notes and the practice set. The course page is at IB DP Physics and the printable list is the IB DP Physics checklist. To find weak spots first, try the free 10-minute diagnostics.

What this unit covers

B.1 understanding	What you must be able to do	Level
Molecular theory	Explain solids, liquids and gases in terms of particle arrangement and motion	SL and HL
Density	Use ρ = m/V	SL and HL
Temperature scales	Use Kelvin and Celsius, convert, and know that a change is the same on both	SL and HL
Kinetic energy and temperature	Use Ek = 3/2 kB T	SL and HL
Internal energy	Describe it as potential plus random kinetic energy of the molecules	SL and HL
Direction of transfer	Use temperature difference to say which way energy flows	SL and HL
Phase change	Explain a change of state at constant temperature	SL and HL
Heating calculations	Use Q = mcΔT and Q = mL for fusion and vaporization	SL and HL

The guide also expects the words melting, freezing, boiling, condensing and evaporation to be familiar.

The molecular model

Matter is made of particles (atoms or molecules) that are always moving and that attract each other at short range. The three states differ in how those two facts balance.

	Solid	Liquid	Gas
Spacing	Close together	Close together	Far apart compared with their size
Arrangement	Fixed positions, often a regular pattern	No fixed pattern, can slide past each other	Random, no pattern
Motion	Vibrate about fixed positions	Vibrate and move around, always in contact	Rapid, random, straight-line motion between collisions
Forces	Strong, hold particles in place	Strong enough to keep contact, not fix position	Weak except during collisions
Shape and volume	Fixed shape, fixed volume	Takes shape of container, fixed volume	Fills the container

This explains why solids and liquids are hard to compress (particles already touch) and gases are easy to compress (large gaps). A liquid flows because particles can move past each other.

Density

Density links mass and volume: ρ = m/V, the mass in each unit of volume. The SI unit is kg m⁻³. Solids and liquids have densities of order 10³ kg m⁻³, and gases are about a thousand times lower because the particles are far apart.

Worked example 1. A rectangular block measures 0.12 m by 0.050 m by 0.040 m and has mass 1.62 kg. Find its density.

V = 0.12 × 0.050 × 0.040 = 2.4 × 10⁻⁴ m³
ρ = m/V = 1.62 / 2.4 × 10⁻⁴ = 6.75 × 10³ kg m⁻³

Worked example 2. An oil has density 920 kg m⁻³. Find the mass of 350 cm³.

V = 350 cm³ = 350 × 10⁻⁶ m³ = 3.5 × 10⁻⁴ m³
m = ρV = 920 × 3.5 × 10⁻⁴ = 0.322 kg

Convert volume to m³ before substituting. One cm³ is 10⁻⁶ m³, not 10⁻² m³.

Temperature scales

The Celsius scale is set by everyday reference points. The Kelvin scale starts at absolute zero, where the particles have the lowest possible energy, and is the SI scale. To convert, use

T / K = θ / °C + 273 (273.15 is more exact)

Worked example 3. (a) Convert 37 °C to kelvin. (b) Convert 77 K to degrees Celsius.

(a) T = 37 + 273 = 310 K
(b) θ = 77 − 273 = −196 °C

A kelvin and a degree Celsius are the same size, so a temperature change is identical on both scales. Heating water from 20 °C to 45 °C is ΔT = 25 °C = 25 K. This is why you can put a Celsius change straight into Q = mcΔT, but you must never put a Celsius temperature into Ek = 3/2 kB T.

Temperature and molecular kinetic energy

The kelvin temperature indicates the mean random kinetic energy of a particle, through:

Ek = 3/2 kB T

where kB is the Boltzmann constant, 1.38 × 10⁻²³ J K⁻¹ (given in the questions here). Ek is the average kinetic energy per particle and T must be in kelvin. Doubling the kelvin temperature doubles the average kinetic energy. It does not double when the Celsius temperature doubles.

Worked example 4. Helium atoms of mass 6.64 × 10⁻²⁷ kg are in a gas at 240 K. Work out the mean kinetic energy per atom and the speed it corresponds to.

Ek = 3/2 × 1.38 × 10⁻²³ × 240 = 4.97 × 10⁻²¹ J
Ek = ½mv²  →  v = √(2Ek/m) = √(2 × 4.968 × 10⁻²¹ / 6.64 × 10⁻²⁷)
v = 1.22 × 10³ m s⁻¹

The speed found this way is a typical speed that corresponds to the average kinetic energy, because individual atoms move at a spread of speeds. At the same temperature a heavier atom has the same average kinetic energy but a lower speed.

Internal energy

Internal energy has two parts, added together:

1. the total intermolecular potential energy, which comes from the forces between the molecules
2. the summed random kinetic energy of the molecules, which comes from their random motion

Both are totals over every molecule in the system, so a larger sample at the same temperature has more internal energy. This is different from the kinetic energy of the whole object moving through space, which is not part of it.

Two links to temperature follow.

• Raising the temperature raises the average random kinetic energy, so internal energy rises.
• Potential energy rises when the particles are pulled further apart against the attractive forces. This is why internal energy can change with no temperature change (next sections).

Direction of thermal transfer

Energy is transferred as heat between bodies at different temperatures. The temperature difference decides the direction: the net transfer is from the hotter body to the cooler one, and it continues until the temperatures are equal.

The direction does not depend on which body holds more internal energy. Energy flows from a small cup of tea at 90 °C into a large lake at 12 °C, because the tea is hotter, even though the lake has far more internal energy overall.

Phase changes

During melting, freezing, boiling and condensing, the temperature stays constant while energy is supplied or removed.

• Melting (solid to liquid) and freezing (liquid to solid) happen at a fixed temperature for a pure substance.
• Boiling (liquid to gas throughout the liquid) and condensing (gas to liquid) also happen at a fixed temperature at a given pressure.
• Evaporation is liquid turning to gas at the surface only, and it can happen well below the boiling point.

At a phase change the energy supplied does not increase the average random kinetic energy. It increases the potential energy by moving particles apart against the forces between them. The particle behaviour changes while the temperature stays constant. When the substance freezes or condenses, the same energy is given out as the potential energy falls.

Evaporation cools the liquid left behind. The fastest molecules at the surface are the ones that escape, so the average kinetic energy of the remaining molecules falls.

Heating calculations

Specific heat capacity c is the energy needed per unit mass to change the temperature by one kelvin: Q = mcΔT, with c in J kg⁻¹ K⁻¹.

Specific latent heat L is the energy per unit mass needed to change state with no temperature change: Q = mL. Use L of fusion (L_f) for melting or freezing and L of vaporization (L_v) for boiling or condensing.

Worked example 5. Find the energy needed to heat a 1.2 kg copper block from 20 °C to 95 °C. Use c = 385 J kg⁻¹ K⁻¹.

ΔT = 95 − 20 = 75 K
Q = mcΔT = 1.2 × 385 × 75 = 3.47 × 10⁴ J

Worked example 6. A heater of power 1.8 kW boils away 0.12 kg of water already at 100 °C. Use L_v = 2.26 × 10⁶ J kg⁻¹ and assume all the energy goes into the water.

Q = mL = 0.12 × 2.26 × 10⁶ = 2.71 × 10⁵ J
t = Q/P = 2.712 × 10⁵ / 1800 = 151 s

Worked example 7. Find the total energy to turn 0.180 kg of ice at −18 °C into water at 40 °C. Use c(ice) = 2100, c(water) = 4180 J kg⁻¹ K⁻¹ and L_f = 3.34 × 10⁵ J kg⁻¹.

Split the process into stages, because the formula changes at 0 °C.

Stage 1, warm ice −18 → 0 °C:   0.180 × 2100 × 18 = 6804 J
Stage 2, melt at 0 °C:           0.180 × 3.34 × 10⁵ = 60 120 J
Stage 3, warm water 0 → 40 °C:  0.180 × 4180 × 40 = 30 096 J
Total = 6804 + 60 120 + 30 096 = 97 020 J = 9.70 × 10⁴ J

Melting takes by far the largest share, which shows how much potential energy must be added to separate the particles.

Common errors

• Putting a Celsius temperature into Ek = 3/2 kB T instead of kelvin.
• Using cm³ or g without converting to m³ or kg in ρ = m/V and Q = mcΔT.
• Applying Q = mcΔT straight through a phase change instead of splitting the calculation into stages.
• Saying that the temperature rises during melting. It stays constant.
• Saying that temperature is the same thing as internal energy. Temperature is linked to average kinetic energy only.
• Giving the direction of transfer from the body with more internal energy rather than from the hotter body.

Official syllabus

Source: the Physics guide for the IB Diploma Programme (first assessment 2025), International Baccalaureate Organization. This page covers topic B.1 up to and including quantitative analysis using Q = mcΔT and Q = mL.