PhysicsCore26 min read

Thermal Properties of Matter

Temperature, expansion, specific heat capacity and latent heat

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01

Temperature is not the same as heat

Definition

Thermal equilibrium — The state reached when two objects in contact are at the same temperature, so there is no longer any net flow of energy between them.

These two words are used interchangeably in everyday speech and mean quite different things in physics. Temperature is a measure of the average kinetic energy of the particles. Thermal energy, or internal energy, is the total energy of all of them added together.

A cup of tea at 80 °C is hotter than a swimming pool at 25 °C. But the pool holds far more thermal energy, because it contains vastly more particles. Temperature tells you the average; thermal energy depends on the average and on how much substance there is.

Energy always flows from higher temperature to lower temperature, never the other way of its own accord. That flow is what we call heating, and it continues until the two are at the same temperature — thermal equilibrium.

02

Specific heat capacity

Definition

Specific heat capacity — The energy needed to raise the temperature of 1 kg of a substance by 1 °C, measured in J kg⁻¹ °C⁻¹.

Different substances need very different amounts of energy for the same temperature rise. Water needs 4200 J to warm one kilogram by one degree; the same kilogram of copper needs only 385 J. Water is unusually hard to heat.

That high value has real consequences. It is why water is used in central heating systems and car radiators — it carries a great deal of energy for a modest temperature change. It is also why coastal areas have milder weather than inland ones: the sea warms and cools far more slowly than the land.

The word "specific" simply means "per kilogram". Multiply by the mass and the temperature change and you have the energy.

E = m c Δθc = E / (m Δθ)water: c = 4200 J kg⁻¹ °C⁻¹ — the highest of any common substance
E
energy transferredJ
m
masskg
c
specific heat capacityJ kg⁻¹ °C⁻¹
Δθ
temperature change°C
Worked example 14 marks

A 2.0 kW kettle contains 0.50 kg of water at 20 °C. Calculate how long it takes to reach 100 °C, assuming no energy is lost.

  1. Temperature change Δθ = 100 − 20 = 80 °C.A change, not a final value.
  2. E = mcΔθ = 0.50 × 4200 × 80.
  3. E = 168 000 J.
  4. t = E/P = 168 000 / 2000 = 84 s.Power is energy per second, so dividing gives the time.

84 s, about a minute and a half

03

Changing state and latent heat

Definition

Specific latent heat — The energy needed to change the state of 1 kg of a substance without any change in temperature.

Heat a solid steadily and its temperature climbs — until it starts to melt. Then, remarkably, the temperature stops rising even though energy is still going in. It stays put until every last bit has melted, and only then does it start climbing again.

The energy has not vanished. It is being used to break the forces holding the particles in their fixed positions, not to make them move faster. Since temperature measures average kinetic energy, and the kinetic energy is not changing, the thermometer does not move.

The same happens at the boiling point, and the energy required there is much larger — separating particles completely takes far more work than merely letting them slide past one another. For water, melting takes 334 000 J per kilogram and boiling takes 2 260 000 J per kilogram.

This is why a steam burn is so much worse than a burn from boiling water at the same temperature. The steam must first condense on your skin, releasing all of that latent heat before it even begins to cool.

E = m LL_f (water) = 3.34 × 10⁵ J kg⁻¹L_v (water) = 2.26 × 10⁶ J kg⁻¹no Δθ term — the temperature does not change during a change of state
E
energyJ
m
masskg
L
specific latent heatJ kg⁻¹

The two flat sections are melting and boiling. Energy is still being supplied throughout, but the temperature holds steady because that energy is breaking bonds rather than increasing the particles' speed. Notice how much longer the boiling plateau is.

04

Thermal expansion

Almost everything expands when heated. The particles vibrate more vigorously, so on average they sit slightly further apart, and the object grows very slightly in every direction.

The effect is small but the forces involved are enormous, so engineers must design for it. Bridges are built with expansion joints, and railway lines with small gaps, so that summer heat does not buckle them. Overhead power lines are hung with a deliberate sag in winter, because they contract and tighten in cold weather.

Gases expand most, liquids less, and solids least — a direct consequence of how strongly the particles are held. A bimetallic strip exploits the difference between two metals: they expand by different amounts, so the strip curls when heated, and that movement can switch a circuit. This is how a simple thermostat works.

Water is the famous exception. Between 0 °C and 4 °C it contracts as it warms, and ice is less dense than the water it forms from. That is why ice floats and why ponds freeze from the top down, leaving fish alive underneath.

Key points

  1. Temperature is average kinetic energy; thermal energy is the total.
  2. E = mcΔθ for a temperature change; E = mL for a change of state.
  3. During melting or boiling the temperature stays constant.
  4. Water has an unusually high specific heat capacity, at 4200 J kg⁻¹ °C⁻¹.
  5. Solids expand least, gases most, and water behaves oddly below 4 °C.

Practice questions

7 questions · 25 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Differentiate between heat and temperature.
Model answer

Temperature is the average kinetic energy of the particles, measured in °C or K. Heat is the total thermal energy of all the particles, measured in joules, and depends on how many there are.

Examiner tip. This is the single most repeated short question in the chapter. Give both definitions and both units.

SQ2[2 marks]
Why does a bimetallic strip bend on heating?
Model answer

The two metals expand by different amounts for the same temperature rise, so the strip curves toward the metal that expands less.

Examiner tip. "Different rates of expansion" earns the first mark; naming the direction of the bend earns the second.

SQ3[2 marks]
Why does the temperature remain constant while ice is melting?
Model answer

The energy supplied is used to break the bonds holding the particles in the lattice, not to increase their kinetic energy. Since temperature measures kinetic energy, it does not change.

Examiner tip. The words "breaks bonds" and "not kinetic energy" are the two marking points.

Solved numericals

1 · 5 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[5 marks]
Calculate the energy needed to convert 0.20 kg of ice at 0 °C completely into water at 20 °C. Take the specific latent heat of fusion of ice as 3.34 × 10⁵ J kg⁻¹ and the specific heat capacity of water as 4200 J kg⁻¹ °C⁻¹.

Given. m = 0.20 kg, L_f = 3.34 × 10⁵ J kg⁻¹, c = 4200 J kg⁻¹ °C⁻¹, Δθ = 20 °C

Full working
  1. Recognises two stages: melting, then warminga single-stage answer cannot score more than two[1]
  2. Melting: E₁ = mL = 0.20 × 3.34 × 10⁵ = 6.68 × 10⁴ J[1]
  3. Warming: E₂ = mcΔθ = 0.20 × 4200 × 20[1]
  4. E₂ = 1.68 × 10⁴ J[1]
  5. Total = 6.68 × 10⁴ + 1.68 × 10⁴ = 8.36 × 10⁴ Jadding the two stages[1]

8.36 × 10⁴ J

Examiner tip. Any question that crosses a change of state needs two or more stages. Sketch the heating curve in the margin and count the sections — each one is a separate calculation.

Exam questions

3 · 14 marks

Multi-part questions with a full mark scheme.

Q1[5 marks]
An electric heater of power 2.0 kW is used to heat 1.5 kg of water. The specific heat capacity of water is 4200 J kg⁻¹ °C⁻¹.
  1. Calculate the energy needed to raise the temperature of the water from 18 °C to 88 °C.
  2. Calculate the minimum time this would take.
  3. In practice the heater takes longer than your answer to (b). Give one reason.
Mark scheme
  1. Uses E = mcΔθ with Δθ = 70 °C88 − 18, not 88[1]
  2. E = 1.5 × 4200 × 70 = 4.41 × 10⁵ J[1]
  3. Uses t = E/P with P = 2000 Wconverting kW to W[1]
  4. t = 441000 / 2000 = 220 saccept 220.5 s[1]
  5. Energy is lost to the surroundings / the container is also heatedeither reason accepted[1]

(a) 4.4 × 10⁵ J (b) 220 s (c) heat lost to surroundings

Examiner tip. Part (c) is one mark for one sentence and needs no calculation. Questions that say "in practice" or "explain why this is a minimum" are always asking about energy losses.

Q2[5 marks]
A student heats a block of ice at −10 °C steadily until it becomes steam. Sketch and describe the shape of the temperature–time graph.
Mark scheme
  1. Temperature rises from −10 °C to 0 °C[1]
  2. Horizontal section at 0 °C while the ice melts[1]
  3. Temperature rises from 0 °C to 100 °C[1]
  4. Longer horizontal section at 100 °C while the water boilsthe boiling plateau must be longer than the melting one[1]
  5. During the flat sections the energy supplied breaks bonds between particles rather than raising kinetic energythe explanation mark[1]

Examiner tip. The boiling plateau must be drawn noticeably longer than the melting one — the latent heat of vaporisation of water is about seven times its latent heat of fusion, and examiners look for that difference.

Q3[4 marks]
Explain, in terms of particles, why a bimetallic strip bends when heated, and state one use for it.
Mark scheme
  1. Heating makes particles vibrate more and push further apart, so each metal expands[1]
  2. The two metals expand by different amounts for the same temperature risethis is the essential point[1]
  3. The strip bends toward the metal that expands less[1]
  4. Used in a thermostat / fire alarm / oven switchany valid use[1]

Examiner tip. "In terms of particles" is an instruction, not decoration. An answer that never mentions particle vibration or spacing cannot score the first mark however correct the rest of it is.