OCR A Level Physics A: Temperature, kinetic model, internal energy, specific heat capacity and latent heat (H556)

This study guide follows the OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017. It covers sections 5.1.1 (Temperature), 5.1.2 (Solid, liquid and gas) and 5.1.3 (Thermal properties of materials), all part of the full A level. The ideal gas statements in 5.1.4 are taught separately. Work through this guide, then use the revision notes and the practice set. The course hub is OCR A Level Physics, the printable list is the checklist, and the free 10-minute diagnostics show which topics need work. For the units and uncertainty skills used in the calculations, see Foundations of physics.

What this unit covers

Section	What you must be able to do
5.1.1(a)-(d)	Explain thermal equilibrium; describe the absolute (thermodynamic) scale; use °C and K; use T (K) ≈ θ (°C) + 273
5.1.2(a)-(c)	Describe solids, liquids and gases by spacing, ordering and motion; use the simple kinetic model; explain Brownian motion and its smoke demonstration
5.1.2(d)-(g)	Define internal energy; explain absolute zero; link temperature rise and change of phase to internal energy
5.1.3(a)-(b)	Use E = mcΔθ; estimate c by the method of mixtures; describe electrical methods for a metal block and a liquid
5.1.3(c)-(d)	Use E = ml for fusion and vaporisation; describe electrical methods for a solid and a liquid

5.1.1 Temperature

Thermal equilibrium (a). Two objects in thermal contact exchange energy until there is no net energy transfer between them. They are then in thermal equilibrium and at the same temperature. Energy transfers spontaneously from the object at higher temperature to the one at lower temperature. This is why a thermometer must be left in a liquid for a while: it reads correctly only once it has reached equilibrium with the liquid.

The absolute scale (b). Scales built on one property of one substance, such as the expansion of mercury or the resistance of a thermistor, can disagree between their fixed points. The thermodynamic scale is different. It does not depend on the property of any particular substance. Its lowest value, 0 K, is absolute zero.

Celsius and kelvin (c, d). Kelvin is the SI unit of temperature. The two scales have the same size of degree, so a temperature change has the same value in both. The conversion is T (K) ≈ θ (°C) + 273.

Worked example
Liquid nitrogen boils at 77 K.  θ = 77 − 273 = −196 °C
A room at 20 °C:  T = 20 + 273 = 293 K
A rise of 25 K is also a rise of 25 °C.

In equations such as E = mcΔθ you may use either scale, because only the change matters. When temperature appears on its own, as in later work on gases, it must be in kelvin.

5.1.2 Solids, liquids and gases

The simple kinetic model (a, b)

State	Spacing	Ordering	Motion
Solid	Touching, close together	Regular pattern (fixed positions)	Vibrate about fixed positions
Liquid	Close together, about as close as a solid	No long-range pattern	Move past each other randomly
Gas	Far apart, many times the molecular size	None	Rapid, random, in straight lines between collisions

The model treats atoms or molecules as small particles in constant motion. The faster they move, the higher the temperature. In a solid the particles keep their places because of strong forces between neighbours. In a gas the forces only matter during collisions.

Worked example (data supplied: density of liquid water is 1670 times
that of steam at 100 °C)
Same mass, so the steam occupies 1670 times the volume.
Volume per molecule ∝ (spacing)³, so spacing ratio = 1670^(1/3) = 11.9
The molecules in steam are about 12 times further apart than in the liquid.

Brownian motion (c)

Smoke particles in a glass cell are lit from the side and viewed through a microscope. You see small bright specks that move in short, irregular, zig-zag paths. The specks are smoke particles, not air molecules.

The explanation uses the kinetic model. Air molecules are moving randomly and rapidly. They collide with a smoke particle from all sides. At any instant the collisions on one side do not exactly balance those on the other, so there is a small resultant force and the particle jerks in a random direction. The motion never stops, which is evidence that the molecules are in continuous random motion.

Internal energy (d)

The internal energy of a system is the total of the randomly distributed kinetic and potential energies of its molecules. The kinetic part comes from random motion. The potential part comes from the forces between molecules. The kinetic energy of the whole object moving, or its gravitational potential energy, is not part of its internal energy.

Absolute zero (e)

Absolute zero, 0 K, is the lowest limit of temperature. It is the temperature at which a substance has minimum internal energy.

Temperature rise and change of phase (f, g)

When a body is heated and its temperature rises, the average kinetic energy of its molecules increases, so its internal energy increases.

During a change of phase the temperature stays constant. Energy is still supplied (melting) or removed (freezing), and the internal energy changes. The energy goes into the potential energy of the molecules as they are pulled apart, not into their average kinetic energy. This is why the temperature does not change.

Stage on heating a solid at constant power	Temperature	Kinetic energy	Potential energy
Solid warming	Rises	Rises	Small rise
Melting	Constant	Constant	Rises
Liquid warming	Rises	Rises	Small rise
Boiling	Constant	Constant	Rises a lot

Vaporisation takes much more energy per kilogram than fusion because the molecules must be separated completely, not just loosened.

5.1.3 Thermal properties of materials

Specific heat capacity (a)

The specific heat capacity c of a substance is the energy that must be supplied to 1 kg of the substance to warm it by 1 K, with no change of phase. Its unit is J kg⁻¹ K⁻¹.

E = m c Δθ

Worked example
An aluminium block of mass 0.85 kg (c = 900 J kg⁻¹ K⁻¹, data supplied)
is heated from 18 °C to 64 °C.
Δθ = 64 − 18 = 46 K
E = 0.85 × 900 × 46 = 3.5 × 10⁴ J   (35 190 J before rounding)

Method of mixtures. A hot sample is placed in cooler water in an insulated container. Energy lost by the sample equals energy gained by the water, if no energy escapes.

Worked example
A metal sample of mass 0.150 kg is heated to 98 °C and dropped into
0.200 kg of water at 18 °C. The final temperature is 23.2 °C.
(c of water = 4200 J kg⁻¹ K⁻¹)
Energy gained by water = 0.200 × 4200 × (23.2 − 18) = 4368 J
Energy lost by sample  = 0.150 × c × (98 − 23.2) = 11.22 c
c = 4368 / 11.22 = 389 J kg⁻¹ K⁻¹

Losses to the air and the container make the water’s temperature rise too small, so the value of c is an estimate that is slightly low.

Electrical determination of c (b)

Metal block. Use a block with two holes: one for an electric heater, one for a thermometer. Put a drop of oil in each hole for good thermal contact. Measure the mass with a balance. Wrap the block in insulation. Connect the heater to a power supply with an ammeter in series and a voltmeter across the heater. Record the starting temperature, switch on, and time the heating with a stopwatch while the current and potential difference stay constant. Switch off and keep watching the thermometer, because the temperature continues to rise for a short time as energy spreads through the block. Use the highest reading.

The electrical energy is E = VIt, so

c = VIt / (m Δθ)

Worked example
m = 0.800 kg, V = 12.0 V, I = 3.50 A, t = 420 s, temperature 18.5 °C to 42.0 °C
E = 12.0 × 3.50 × 420 = 17 640 J
Δθ = 42.0 − 18.5 = 23.5 K
c = 17 640 / (0.800 × 23.5) = 9.4 × 10² J kg⁻¹ K⁻¹

Energy lost to the surroundings means the block warms less than it should, so this c is too large. Better insulation, or starting a few degrees below room temperature and finishing the same amount above it, reduces the effect.

Liquid. Use an immersion heater fully covered by the liquid in an insulated container. Stir gently so the temperature is uniform. Measure the mass of liquid and take temperature readings at regular intervals while the power is constant. A graph of θ against t is a straight line for steady heating. Its gradient is P / (mc), so c = P / (m × gradient). This uses all the readings, so one poor point has little effect.

Worked example
P = 12.0 × 4.0 = 48 W, m = 0.300 kg, gradient = 0.080 K s⁻¹
c = 48 / (0.300 × 0.080) = 2.0 × 10³ J kg⁻¹ K⁻¹

Specific latent heat (c)

The specific latent heat of fusion is the energy required to turn 1 kg of solid into liquid at its melting point, with no change of temperature. The specific latent heat of vaporisation is the energy required to turn 1 kg of liquid into gas at its boiling point. The unit is J kg⁻¹.

E = m l

Worked example (data supplied: c of ice 2100, l of fusion 3.34 × 10⁵,
c of water 4200)
Energy to take 0.200 kg of ice at −15 °C to water at 20 °C:
warm the ice   0.200 × 2100 × 15 = 6300 J
melt the ice   0.200 × 3.34 × 10⁵ = 66 800 J
warm the water 0.200 × 4200 × 20 = 16 800 J
Total = 89 900 J = 9.0 × 10⁴ J
A 150 W heater with no losses needs 89 900 / 150 = 600 s.

Do the stages separately. Never multiply m by a single c across a change of phase.

Electrical determination of l (d)

Fusion (solid). Put crushed ice in a funnel with a heater in it, and collect the meltwater in a beaker on a balance. Run a second, identical set-up with the heater off. The second set-up shows how much ice melts from the warm room alone. Run both for the same time.

Worked example
Heater: 12.0 V, 3.00 A for 300 s, so E = 36.0 × 300 = 10 800 J
Heater on: 0.0412 kg collected. Heater off: 0.0079 kg collected.
Mass melted by the heater = 0.0412 − 0.0079 = 0.0333 kg
l = 10 800 / 0.0333 = 3.2 × 10⁵ J kg⁻¹

Vaporisation (liquid). Boil a liquid with an immersion heater in a well-lagged container on a balance. Measure the mass lost in a known time once the liquid is boiling steadily. To remove the effect of losses, repeat at a second power for the same time. The extra energy only boils the extra mass:

l = (P₂ − P₁) t / (m₂ − m₁)

Worked example
P₁ = 40 W, P₂ = 80 W, t = 600 s, m₁ = 0.0098 kg, m₂ = 0.0204 kg
l = 40 × 600 / (0.0204 − 0.0098) = 24 000 / 0.0106 = 2.3 × 10⁶ J kg⁻¹

Common errors

• Writing T = θ − 273 instead of T = θ + 273.
• Saying that the temperature stays constant “because no energy is supplied” during a phase change. Energy is supplied; it increases potential energy.
• Describing Brownian motion as the motion of air molecules. It is the motion of visible particles caused by them.
• Calling internal energy “heat” or “thermal energy of the whole object”.
• Using m c Δθ for the stage where a phase change occurs.
• Forgetting the control run in the fusion experiment, or the second power in the vaporisation experiment.

Official syllabus

OCR Advanced GCE in Physics A (H556), specification Version 3.0 (March 2026), first assessment 2017, published by OCR: sections 5.1.1, 5.1.2 and 5.1.3.