What is the change of state in Physics, and how does it relate to solids, liquids, and gases?

In Physics, a change of state refers to the transformation of a substance from one state of matter to another, such as from solid to liquid or liquid to gas. This process involves the absorption or release of energy, typically in the form of heat. For example, melting is the change from solid to liquid, while evaporation is the change from liquid to gas. Understanding these changes is crucial in the Edexcel IGCSE curriculum as it explains the behavior of particles in different states of matter.

What are the main types of changes of state, and what are their characteristics?

The main types of changes of state include melting, freezing, evaporation, condensation, sublimation, and deposition. Melting is the change from solid to liquid, requiring heat absorption. Freezing is the reverse, where a liquid becomes a solid, releasing heat. Evaporation is the change from liquid to gas, occurring at the surface of a liquid. Condensation is the reverse, where gas becomes liquid. Sublimation is the direct change from solid to gas, bypassing the liquid state, while deposition is the reverse process. Each change involves energy transfer and particle movement, key concepts in the Edexcel IGCSE Physics syllabus.

How does temperature affect the change of state in substances?

Temperature plays a crucial role in the change of state. It affects the kinetic energy of particles in a substance. For instance, increasing temperature provides energy for particles to overcome intermolecular forces, leading to melting or evaporation. Conversely, decreasing temperature reduces particle energy, causing freezing or condensation. Understanding the relationship between temperature and state changes is essential for Edexcel IGCSE students to grasp how energy influences matter.

What is the role of latent heat in the change of state?

Latent heat is the energy absorbed or released during a change of state without a change in temperature. There are two types: latent heat of fusion (solid to liquid) and latent heat of vaporization (liquid to gas). This energy is used to break or form intermolecular bonds rather than increasing temperature. In the Edexcel IGCSE Physics curriculum, understanding latent heat helps explain why substances require energy to change state even when their temperature remains constant.

Why does sublimation occur, and can you give an example?

Sublimation occurs when a solid changes directly into a gas without passing through the liquid state. This happens when the particles in a solid gain enough energy to overcome intermolecular forces and enter the gas phase. An example of sublimation is dry ice (solid carbon dioxide) turning into carbon dioxide gas at room temperature. This process is significant in the Edexcel IGCSE Physics curriculum as it illustrates the unique conditions under which certain substances change state.

Why is "Saying the temperature rises while a solid is melting" a common mistake in Change of State?

Why it happens: Intuitive — heat is being added, so temperature 'should' rise. How to avoid it: During a change of state the temperature stays CONSTANT. The supplied heat goes into BREAKING BONDS (increasing potential energy), not increasing kinetic energy. Once all the solid has melted, T resumes rising. Source: 4PH1/2P Examiner Reports 2022-2024

Why is "Confusing 'closely packed' (liquid) with 'fixed positions' (solid)" a common mistake in Change of State?

Why it happens: Both look similar in a diagram. How to avoid it: Solid = REGULAR, fixed positions, only vibrate. Liquid = close-packed but RANDOM arrangement, slide past each other.

Why is "Using $\Delta Q = mc\Delta T$ across a state change" a common mistake in Change of State?

Why it happens: Forgetting the latent-heat plateau. How to avoid it: mc T applies only when temperature is CHANGING. During melting/boiling, use latent heat (beyond 4PH1 scope but still worth knowing the principle). For multi-step problems: heat the solid up; melt; heat the liquid up; etc.

Why is "Quoting specific heat capacity in J/kg or J/°C instead of J/(kg °C)" a common mistake in Change of State?

Why it happens: Compound unit confusion. How to avoid it: c has units of J per kilogram per degree. Always write J/(kg °C) — both 'per kg' AND 'per °C'. Source: 4PH1/2P Examiner Reports 2022-2024

Why is "Surprised when an experimental $c$ comes out lower than the textbook value" a common mistake in Change of State?

Why it happens: Not anticipating heat loss to surroundings. How to avoid it: Heat losses → some of Q goes to the surroundings rather than the block → experimental T is smaller for a given Q. Insulating reduces this; the spec 5.14P practical accepts a final c slightly off from the standard value.

Solids, Liquids and GasesEdexcel IGCSE Physics (4PH1)

Change of State

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Change of State — Pearson Edexcel International GCSE Physics 4PH1 Study Notes (2026 onwards, Paper 2 only — specs 5.8P-5.14P)

Paper-2 only. How heating changes energy/temperature/state, particle arrangement and motion in s/l/g, the constant-temperature plateau during melting/boiling (required practical 5.11P), specific heat capacity $\Delta Q = mc\Delta T$ and the required practical for measuring $c$ (5.14P).

What you’ll learn

Mapped to the Cambridge IGCSE 4PH1 syllabus (2026 onwards).

5.8P — Understand that heating a system changes the energy stored within the system by increasing the energy of the particles that make up the system; this either raises the temperature OR produces a change of state. (Paper 2)
5.9P — Describe the changes when matter changes state — solid to liquid (melting), liquid to gas (boiling/evaporating), and the reverse. (Paper 2)
5.10P — Describe the arrangement and motion of particles in solids, liquids and gases. (Paper 2)
5.11P — Required practical: obtain a temperature-time graph to show the constant temperature during a change of state. (Paper 2)
5.12P — Define specific heat capacity. (Paper 2)
5.13P — Know and use the relationship: change in thermal energy = mass × specific heat capacity × change in temperature, $\Delta Q = mc\Delta T$ . (Paper 2)
5.14P — Required practical: investigate the specific heat capacity of materials including water and some solids. (Paper 2)

Heating: temperature or state? (spec 5.8P, 5.9P)

Two ways for energy to enter a system.

Spec 5.8P principle. When you heat a system, energy is transferred TO the particles. That energy can do one of two things:

Raise the temperature — increases the average kinetic energy of the particles (and they vibrate or move faster).
Cause a change of state — increases the potential energy of the particles by separating them against the intermolecular forces; temperature stays CONSTANT during the change.

Spec 5.9P — the three changes.

Solid → liquid (melting). Particles gain enough KE to break out of the rigid lattice. They can now slide past each other. Temperature stays constant at the MELTING POINT.
Liquid → gas (boiling). Particles gain enough KE to escape the liquid surface and move freely. Temperature stays constant at the BOILING POINT.
Liquid → gas (evaporation). Slower process — fastest particles escape from the surface AT ANY TEMPERATURE below boiling. Causes cooling of the remaining liquid (the average KE of the remainder is lower because the fast ones left).

Reverse changes.

Gas → liquid (condensation): particles slow enough for attractive forces to pull them together. Releases energy (= the latent heat absorbed during boiling).
Liquid → solid (freezing): particles slow enough to lock into a regular lattice. Releases energy.

Mark-scheme phrasing. Edexcel rewards descriptions that mention BOTH the energy change AND the structural rearrangement.

Heat → either T rises OR state changes (not both at once).
Melting / boiling: T constant; energy breaks bonds (PE rise).
Evaporation: surface escape at any T; cools the remaining liquid.

Particles in solids, liquids and gases (spec 5.10P)

Three states, three arrangements.

Solid.

ARRANGEMENT: regular, ordered lattice. Particles close together.
MOTION: vibrate about fixed positions; they do not move past each other.
PROPERTIES: fixed shape, fixed volume, slightly compressible.

Liquid.

ARRANGEMENT: close-packed but RANDOM (no fixed lattice).
MOTION: particles move past each other → liquids flow.
PROPERTIES: takes the shape of its container, fixed volume, only very slightly compressible.

Gas.

ARRANGEMENT: particles WIDELY SPACED, random.
MOTION: rapid, random motion; particles collide elastically with each other and the container walls.
PROPERTIES: fills the container, can be easily compressed (lots of empty space).

Numerical comparison. Spacing in a gas is roughly 10× the spacing in the solid/liquid → volume of a gas is ~1000× the volume of the same mass of liquid (e.g. 1 g of water becomes ~1700 cm³ of steam at 100 °C).

Diagram cues for exam answers.

Solid: regular 2D grid of touching circles.
Liquid: closely packed but disordered cluster of circles.
Gas: very few circles spread out widely with arrows for motion.

Edexcel will accept any clear annotated sketch — make sure spacings differ noticeably between liquid and gas.

Solid: fixed lattice + vibration only.
Liquid: random + close-packed; can flow.
Gas: widely spaced + rapid random motion.

See the full worked example for change of state →

Required practical: T-time graph for change of state (spec 5.11P)

Plateau at the melting/boiling point.

Aim. Show that the temperature stays constant during a change of state.

Typical apparatus. A test tube containing a sample of solid that melts at a manageable temperature (commonly stearic acid, mp ≈ 69 °C, or water, mp = 0 °C / bp = 100 °C). Thermometer. Stopwatch. Water bath as a heat source (Bunsen-heated beaker of water).

Method (melting stearic acid).

Place a test tube of solid stearic acid in a beaker of cold water on a tripod.
Insert a thermometer into the stearic acid.
Heat the water bath gently with a Bunsen burner.
Every 30 s, record the temperature shown by the thermometer.
Continue until all the stearic acid has melted and the temperature has risen well into the liquid region.

Method (cooling water from above 100 °C — alternative).

Boil a beaker of water; record temperature every 30 s as it cools.
The temperature will plateau around 0 °C if ice forms in equilibrium (slower process, requires care).

Expected graph. Plot temperature (y-axis) vs time (x-axis). Three regions:

SLOPE UP (solid warming) → temperature rising.
HORIZONTAL PLATEAU at the melting point.
SLOPE UP again (liquid warming).

Explanation of the plateau. During the change of state, supplied heat goes into BREAKING the bonds between particles (increases their potential energy by separating them) — not into increasing their kinetic energy. Since temperature ≡ average kinetic energy, T stays constant.

Sources of error. Heat lost to surroundings from the test tube (curve may droop); thermometer calibration; non-uniform heating (stirring helps).

T-time graph: rise → plateau → rise.
Plateau because energy breaks bonds (PE↑), not KE.
Sample: stearic acid (mp ≈ 69 °C) common in 4PH1 labs.

See the full worked example for change of state →

Specific heat capacity (spec 5.12P, 5.13P)

$\Delta Q = mc\Delta T$ .

Definition (spec 5.12P). The specific heat capacity $c$ of a substance is the energy required to raise the temperature of 1 kg of that substance by 1 °C (or 1 K).

Unit: J/(kg °C) — joules per kilogram per degree Celsius.

Formula (spec 5.13P).

$\Delta Q = m c \Delta T$

$\Delta Q$ = energy supplied (J)
$m$ = mass (kg)
$c$ = specific heat capacity (J/(kg °C))
$\Delta T$ = temperature change (°C — same number whether you use K or °C since it's a DIFFERENCE)

Standard values to recognise.

Water: 4200 J/(kg °C) (unusually high — high heat capacity reason oceans moderate climate).
Aluminium: ~900 J/(kg °C).
Iron / steel: ~450 J/(kg °C).
Copper: ~390 J/(kg °C).
Air: ~1000 J/(kg °C).

Worked example. How much energy to heat 2.0 kg of water from 15 °C to 60 °C? $\Delta Q = m c \Delta T = 2.0 \times 4200 \times 45 = 3.78 \times 10^{5}$ J.

Implications of water's high $c$ .

A kettle uses a lot of electricity to boil water (vs the same volume of, say, alcohol).
Coastal regions have milder climates than inland — the ocean acts as a thermal reservoir.
Car radiators use water as the cooling fluid.
The human body is mostly water → relatively stable internal temperature.

$\Delta Q = mc\Delta T$ .
$c$ in J/(kg °C); $\Delta Q$ in J.
Water $c$ = 4200 J/(kg °C) — unusually high.
Applies only when T is changing (NOT during state change).

See the full worked example for change of state →

Required practical: measuring $c$ (spec 5.14P)

Immersion heater + insulated block / water.

Aim. Determine the specific heat capacity of a material (solid block or water).

Solid block method.

Drill TWO holes in a metal block (e.g. aluminium): one for an electric immersion heater, one for a thermometer. Add a drop of oil to the thermometer hole to improve thermal contact.
Mass the block: $m$ in kg.
INSULATE the block — wrap in cotton wool / felt to reduce heat loss.
Record the initial temperature $T_1$ .
Connect the heater via an ammeter (current $I$ ) and voltmeter (voltage $V$ ). Power $P = VI$ . Alternatively, use a joulemeter that measures $\Delta Q$ directly.
Switch on and start a stopwatch.
Heat for a measured time $t$ (e.g. 600 s).
Switch off; wait for the temperature to equilibrate; record final temperature $T_2$ .

Calculation.

Energy supplied: $\Delta Q = P \times t = VIt$ .
Specific heat capacity: $c = \Delta Q / (m \Delta T) = VIt / [m(T_2 - T_1)]$ .

Water method. Use a known mass of water in an insulated polystyrene cup (calorimeter). Same apparatus. Stir gently during heating to ensure uniform temperature. Calculate $c$ the same way.

Common sources of error.

HEAT LOSS to surroundings — biggest source, causes $c$ to come out TOO HIGH (need more energy than the formula predicts) when corrected naively. Insulation mitigates.
Thermometer calibration / parallax.
Energy absorbed by the heater and thermometer themselves (small).

Reducing error.

Insulate well.
Start with the block BELOW room temperature so cooling balances heating around room T (heat-gain = heat-loss at end-points cancel).
Repeat 3× and take the mean.

$c = VIt / (m\Delta T)$ .
Insulate to reduce heat loss.
Common to find $c$ slightly too high in practice.
Repeat for accuracy.

See the full worked example for change of state →

How it’s examined

Change of state appears only on Paper 2 (4PH1/2P), typically 6-10 marks total. Highest-frequency items: T-time graph + explanation of plateau (5-6 marks); particle arrangement and motion in s/l/g (4-5 marks); specific heat capacity calculations (3-4 marks); required practical methods (5-6 marks).

Sources: Pearson Edexcel International GCSE Physics 4PH1 specification — Issue 4, September 2024 (valid for 2026 onwards); Pearson Edexcel 4PH1 Examiner Reports 2022-2024; 4PH1/2P May/Jun 2024 question paper and mark scheme; 4PH1/2P January 2024 question paper and mark scheme. Last reviewed 2026-05-12.

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Step-by-step worked examples — Change of State

Step-by-step solutions to past-paper-style questions on change of state, written exactly the way a tutor would explain them at the board.

1Particle arrangement and motion (5 marks, Paper 2)
Core• Adapted from 4PH1/2P May/Jun 2024• particles, states, spec-5.10P, Paper 2
Question
Compare the arrangement and motion of particles in a solid, a liquid and a gas. (5 marks) [Paper 2 only]
Step-by-step solution
1. Step 1
  Solid. Particles are in a REGULAR, CLOSELY-PACKED arrangement (often a fixed lattice). Particles vibrate ABOUT FIXED POSITIONS but do not move past each other.
2. Step 2
  Liquid. Particles are closely packed but in a RANDOM (disordered) arrangement. They have enough KE to slide past each other; the liquid can flow and take the shape of its container.
3. Step 3
  Gas. Particles are FAR APART, in random arrangement. They move RAPIDLY in random directions, occasionally colliding with each other and the walls.
4. Step 4
  Spacing comparison. Solid spacing ≈ liquid spacing (both close-packed); gas spacing is ~10× larger in each direction — so a gas has ~1000× the volume of the same mass of liquid.
5. Step 5
  Energy of motion. Going solid → liquid → gas, particles gain kinetic energy. In a solid: vibrational only. In a liquid: vibrational + translational (sliding). In a gas: translational + rotational + faster.
Answer
Solid: regular fixed arrangement, particles vibrate about fixed points. Liquid: close-packed random, particles slide. Gas: far apart, random rapid motion.
Examiner tip
Edexcel mark scheme awards: 1 mark per state for ARRANGEMENT; 1 mark per state for MOTION. Diagrams should show solid (regular rows), liquid (closely packed but disordered), gas (widely spaced random). Don't confuse 'close-packed' (liquid) with 'fixed positions' (solid).
2Temperature-time graph showing change of state (6 marks, Paper 2)
Extended• Adapted from 4PH1/2P January 2024• required practical, change of state, spec-5.11P, Paper 2
Question
Stearic acid initially at room temperature is heated continuously until it melts. The temperature is recorded every 30 s. (a) Sketch the shape of a temperature-time graph for the heating process. (b) Explain why the temperature stays constant during the change of state. (6 marks) [Paper 2 only]
Step-by-step solution
1. Step 1
  (a) Graph shape (3 marks). Three regions: (i) temperature rises (solid warming) — slope upward; (ii) HORIZONTAL plateau at the melting point (e.g. 69 °C for stearic acid); (iii) temperature rises again (liquid warming) — slope upward.
2. Step 2
  (b) During heating only. Each joule of supplied heat increases the KINETIC ENERGY of the particles → temperature rises.
3. Step 3
  During the plateau (1 mark). Heat is STILL being supplied at the same rate. But the temperature does NOT rise.
4. Step 4
  (b) Why constant (1 mark). The supplied heat is used to BREAK the bonds holding particles in the solid lattice — increases POTENTIAL energy of particles, NOT kinetic. Since temperature is a measure of average KE, T stays constant.
5. Step 5
  Energy book-keeping (1 mark). The energy required is called the specific LATENT HEAT of fusion. Once all the solid has melted, supplied heat resumes increasing KE → temperature rises again.
Answer
Graph: rising line → horizontal plateau at melting point → rising line again. Constant T because supplied heat breaks bonds (PE↑), not KE (so T unchanged).
Examiner tip
Spec 5.11P is a REQUIRED PRACTICAL. Edexcel mark schemes reward: (1) correctly sketched three-segment graph; (2) labelled melting point; (3) explanation in terms of PE not KE. The phrase 'bonds breaking' is mark-worthy.
3Specific heat capacity calculation (4 marks, Paper 2)
Core• Adapted from 4PH1/2P May/Jun 2024• specific heat, spec-5.13P, Paper 2
Question
How much energy is required to raise the temperature of 0.50 kg of water from 20 °C to 80 °C? Specific heat capacity of water = 4200 J/(kg °C). (4 marks) [Paper 2 only]
Step-by-step solution
1. Step 1
  Write the relationship (spec 5.13P).
  $\Delta Q = m c \Delta T$
2. Step 2
  Substitute.
  $\Delta Q = 0.50 \times 4200 \times (80 - 20)$
3. Step 3
  Evaluate.
  $\Delta Q = 0.50 \times 4200 \times 60 = 126\,000\ \text{J}$
4. Step 4
  Quote answer. $\Delta Q = 1.26 \times 10^{5}$ J = 126 kJ.
Answer
$\Delta Q = 1.26 \times 10^{5}$ J (= 126 kJ).
Examiner tip
Edexcel: 1 mark formula, 1 substitution, 1 numerical evaluation, 1 final answer with unit. Water has an unusually HIGH $c$ (4200 J/(kg°C)) — much more than metals (~400-900 J/(kg°C)). This is why oceans moderate climate.
4Required practical: specific heat capacity of metal (6 marks, Paper 2)
Extended• Adapted from 4PH1/2P January 2024• required practical, spec-5.14P, Paper 2
Question
Describe a method to find the specific heat capacity of an aluminium block using an electric immersion heater, a thermometer and a stopwatch. State how the data should be processed. (6 marks) [Paper 2 only]
Step-by-step solution
1. Step 1
  Setup (1 mark). Insert immersion heater into a hole drilled in the aluminium block. Insert thermometer into a separate hole, with a drop of oil to improve thermal contact. INSULATE the block (wrap in cotton wool) to reduce heat loss.
2. Step 2
  Mass (1 mark). Use a balance to measure the mass $m$ of the block in kg.
3. Step 3
  Connect heater (1 mark). Connect the immersion heater (rated power $P$ in watts) to a joulemeter (or note the V and I from a voltmeter + ammeter). Record initial temperature $T_1$ .
4. Step 4
  Heat for measured time (1 mark). Turn on the heater; start a stopwatch. Heat for, say, 600 s (10 minutes), stirring or waiting for the temperature to plateau. Record final temperature $T_2$ .
5. Step 5
  Process — calculate energy in (1 mark). Energy supplied $\Delta Q = P \times t$ (joulemeter reading is a direct alternative).
6. Step 6
  Calculate $c$ (1 mark). Use $\Delta Q = mc\Delta T$ rearranged: $c = \Delta Q / (m \Delta T)$ . Repeat the experiment and take the mean.
Answer
Insulated aluminium block + immersion heater + thermometer + stopwatch. Measure mass, initial T, final T after measured heating time. $\Delta Q = Pt$ ; $c = \Delta Q/(m\Delta T)$ .
Examiner tip
Spec 5.14P is a REQUIRED PRACTICAL. Edexcel awards marks for: insulation (reduces heat loss → improves accuracy); recording mass, V, I (or P), time, initial and final T; rearranging $\Delta Q = mc\Delta T$ for $c$ ; identifying heat loss as a source of error (typically the experimental $c$ comes out LOWER than the textbook value because some energy heats the surroundings).

Key Formulae — Change of State

The formulae you need to memorise for change of state on the Pearson Edexcel IGCSE 4PH1 paper, with every variable defined in plain English and a note on when to use it.

Specific heat capacity (spec 5.13P)
$\Delta Q = m c \Delta T$
When to use
Paper 2 only. $\Delta Q$ = energy supplied (J), $m$ = mass (kg), $c$ = specific heat capacity (J/(kg °C)), $\Delta T$ = temperature change (°C or K). Water: $c$ = 4200 J/(kg °C); aluminium: 900; copper: 390.

Key Definitions and Keywords — Change of State

Definitions to memorise and the exact keywords mark schemes credit for change of state answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IGCSE 4PH1 sitting.

State of matter (spec 5.10P)
Solid, liquid or gas. Particles are arranged and move differently in each state. (Paper 2)
Melting point (spec 5.9P, 5.11P)
Examiner keyword
The temperature at which a solid changes to a liquid (and vice versa). During melting, the temperature stays CONSTANT even though energy is being supplied. (Paper 2)
Boiling point (spec 5.9P)
Examiner keyword
The temperature at which a liquid changes to a gas (and vice versa). During boiling, the temperature stays constant. For water at sea-level atmospheric pressure: 100 °C. (Paper 2)
Specific heat capacity (spec 5.12P)
Examiner keyword
The energy required to raise the temperature of 1 kg of a substance by 1 °C (or 1 K). Unit: J/(kg °C). Water has an unusually high value of 4200 J/(kg °C). (Paper 2)

Common Mistakes and Misconceptions — Change of State

The traps other students keep falling into on change of state questions — taken from recent Pearson Edexcel IGCSE 4PH1 examiner reports and mark schemes — and how to avoid them.

✕Saying the temperature rises while a solid is melting
4PH1/2P Examiner Reports 2022-2024
Why it happens
Intuitive — heat is being added, so temperature 'should' rise.
How to avoid it
During a change of state the temperature stays CONSTANT. The supplied heat goes into BREAKING BONDS (increasing potential energy), not increasing kinetic energy. Once all the solid has melted, T resumes rising.
✕Confusing 'closely packed' (liquid) with 'fixed positions' (solid)
Why it happens
Both look similar in a diagram.
How to avoid it
Solid = REGULAR, fixed positions, only vibrate. Liquid = close-packed but RANDOM arrangement, slide past each other.
✕Using $\Delta Q = mc\Delta T$ across a state change
Why it happens
Forgetting the latent-heat plateau.
How to avoid it
$mc\Delta T$ applies only when temperature is CHANGING. During melting/boiling, use latent heat (beyond 4PH1 scope but still worth knowing the principle). For multi-step problems: heat the solid up; melt; heat the liquid up; etc.
✕Quoting specific heat capacity in J/kg or J/°C instead of J/(kg °C)
4PH1/2P Examiner Reports 2022-2024
Why it happens
Compound unit confusion.
How to avoid it
$c$ has units of J per kilogram per degree. Always write J/(kg °C) — both 'per kg' AND 'per °C'.
✕Surprised when an experimental $c$ comes out lower than the textbook value
Why it happens
Not anticipating heat loss to surroundings.
How to avoid it
Heat losses → some of $\Delta Q$ goes to the surroundings rather than the block → experimental $\Delta T$ is smaller for a given $\Delta Q$ . Insulating reduces this; the spec 5.14P practical accepts a final $c$ slightly off from the standard value.

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Change of State — frequently asked questions

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Change of State

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Particle arrangement and motion (5 marks, Paper 2)

2Temperature-time graph showing change of state (6 marks, Paper 2)

3Specific heat capacity calculation (4 marks, Paper 2)

4Required practical: specific heat capacity of metal (6 marks, Paper 2)

Specific heat capacity (spec 5.13P)

State of matter (spec 5.10P)

Melting point (spec 5.9P, 5.11P)

Boiling point (spec 5.9P)

Specific heat capacity (spec 5.12P)

✕Saying the temperature rises while a solid is melting

✕Confusing 'closely packed' (liquid) with 'fixed positions' (solid)

✕Using ΔQ=mcΔT\Delta Q = mc\Delta TΔQ=mcΔT across a state change

✕Quoting specific heat capacity in J/kg or J/°C instead of J/(kg °C)

✕Surprised when an experimental ccc comes out lower than the textbook value

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Change of State — frequently asked questions

✕Using $\Delta Q = mc\Delta T$ across a state change

✕Surprised when an experimental $c$ comes out lower than the textbook value