What are the basic states of matter covered in the Cambridge IGCSE Coordinated Science syllabus?

In the Cambridge IGCSE Coordinated Science syllabus, the basic states of matter are solid, liquid, and gas. Each state has distinct characteristics: solids have a fixed shape and volume, liquids have a fixed volume but take the shape of their container, and gases have neither fixed shape nor volume, expanding to fill their container.

How does temperature affect the thermal properties of matter?

Temperature significantly affects the thermal properties of matter. As temperature increases, particles gain kinetic energy, causing solids to expand, liquids to become less viscous, and gases to increase in pressure if confined. This is a key concept in Thermal Physics, as it explains phenomena like thermal expansion and changes in state.

What is specific heat capacity and why is it important in Thermal Physics?

Specific heat capacity is the amount of heat energy required to raise the temperature of one kilogram of a substance by one degree Celsius. It is crucial in Thermal Physics because it helps predict how substances will respond to heat, influencing everything from climate science to engineering applications.

Can you explain the process of conduction and its relevance in the study of thermal properties?

Conduction is the process of heat transfer through a material without the movement of the material itself. It occurs when particles in a solid vibrate and pass on their energy to neighboring particles. This process is relevant in studying thermal properties as it explains how heat moves through materials, affecting their thermal conductivity.

What role does thermal expansion play in everyday applications?

Thermal expansion is the increase in volume of a substance due to an increase in temperature. It plays a crucial role in everyday applications such as the design of bridges, railways, and thermostats, where allowances must be made for expansion and contraction to prevent structural damage or malfunction.

Why is "Forgetting to include the latent heat stage when calculating total energy for heating through a state change" a common mistake in Matter and Thermal Properties ?

Why it happens: Students only apply Q = mc for the entire temperature range. How to avoid it: If the substance changes state, split into separate stages: (1) heating to transition temperature, (2) state change (Q = mL), (3) heating from transition temperature.

Why is "Saying evaporation and boiling are the same process" a common mistake in Matter and Thermal Properties ?

Why it happens: Both convert liquid to gas, so students treat them identically. How to avoid it: Evaporation: surface only, any temperature, cools liquid, no bubbles. Boiling: throughout liquid, specific temperature, bubbles form inside liquid.

Why is "Using the final temperature instead of the temperature change $\Delta\theta$ in $Q = mc\Delta\theta$" a common mistake in Matter and Thermal Properties ?

Why it happens: Students plug in the final temperature without subtracting the initial temperature. How to avoid it: = T_final - T_initial. Always subtract the initial temperature, not just use the final value.

Thermal PhysicsCambridge IGCSE Coordinated Sciences 0654

Matter and Thermal Properties

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Thermal Properties of Matter — Cambridge IGCSE Coordinated Science 0654

Changes of state involve latent heat (energy absorbed or released with no temperature change). Cambridge tests melting, boiling, evaporation, condensation, and the energy involved — along with specific heat capacity.

What you’ll learn

Mapped to the Cambridge IGCSE 0654 syllabus (2025-2027).

P3.3a — Define and use specific heat capacity (Q = mcΔT).
P3.3b — Define and use specific latent heat (Q = mL).
P3.3c — Explain changes of state in terms of particle energy.
P3.3d — Distinguish between evaporation and boiling.

Specific Heat Capacity

Q = mcΔT links heat energy, mass, material, and temperature change.

Specific heat capacity (c):

The energy required to raise the temperature of 1 kg of a substance by 1°C (or 1 K). Q = mcΔT

Q = heat energy (J), m = mass (kg), c = specific heat capacity (J/kg°C), ΔT = temperature change (°C or K)

Typical values:

Material	c (J/kg°C)
Water	4200
Aluminium	900
Iron	450
Copper	400

Why water has high c:

Water has strong hydrogen bonds that require much energy to break before temperature rises
High c makes water excellent for cooling systems and climate regulation

Example: How much energy is needed to heat 2 kg of water from 20°C to 100°C?

Q = 2 × 4200 × 80 = 672 000 J = 672 kJ

Experimental measurement:

Use electrical heater (power P, time t): energy = Pt
Measure temperature change with thermometer
c = Q/(mΔT) = Pt/(mΔT)

Q = mcΔT. c = specific heat capacity (J/kg°C).
Water: c = 4200 J/kg°C (highest of common substances).
Electrical heater: energy = power × time = Pt.

Specific Latent Heat and Changes of State

During a state change, energy is absorbed/released but temperature stays constant.

Specific latent heat (L):

The energy absorbed or released when 1 kg of a substance changes state at constant temperature. Q = mL

Q = heat energy (J), m = mass (kg), L = specific latent heat (J/kg)

Types:

Latent heat of fusion (Lf): melting (solid → liquid) or freezing (liquid → solid)
Latent heat of vaporisation (Lv): boiling (liquid → gas) or condensing (gas → liquid)
Lv > Lf for most substances (more energy needed to vaporise than melt)

Why temperature is constant during state change:

Energy input breaks intermolecular bonds (PE increases) rather than increasing KE
KE unchanged → temperature unchanged

Heating/cooling curve:

Flat section during melting (at melting point) and boiling (at boiling point)
Sloped sections when temperature is changing within a single state

Evaporation vs Boiling:

Feature	Evaporation	Boiling
Temperature	Below boiling point	At boiling point only
Location	Surface only	Throughout the liquid
Cooling effect	Yes (liquid cools)	No net cooling of remaining liquid
Rate affected by	Temperature, surface area, wind	Fixed by boiling point

Cooling by evaporation:

Fastest molecules escape from surface → average KE of remaining molecules decreases → liquid cools
Used in: sweating, refrigerators, air conditioning

Q = mL for state change. Temperature constant during melting/boiling.
Latent heat vaporisation > latent heat fusion (more energy to vaporise).
Evaporation: surface, any temperature, cools liquid. Boiling: throughout, at boiling point.

How it’s examined

Paper 4: 'Calculate the energy to melt 0.5 kg of ice. Latent heat of fusion = 336 000 J/kg' (2 marks — Q = 0.5 × 336 000 = 168 000 J). 'Explain why sweating cools the body' (2 marks — fast-moving water molecules evaporate from skin surface; remaining molecules have lower average KE → temperature falls). 'A heating curve shows a horizontal section. Explain what is happening' (2 marks — state change; energy absorbed breaking intermolecular bonds without raising temperature). 'Calculate Q to heat 3 kg of water by 40°C' (2 marks — Q = 3 × 4200 × 40 = 504 000 J).

Sources: Cambridge IGCSE Coordinated Sciences 0654 syllabus 2025-2027 (P3); 0654 Examiner Reports 2022-2024. Last reviewed 2026-05-14.

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Step-by-step worked examples — Matter and Thermal Properties

Step-by-step solutions to past-paper-style questions on matter and thermal properties , written exactly the way a tutor would explain them at the board.

1Specific heat capacity calculation
Core• specific heat capacity, Q=mcΔθ
Question
Calculate the energy required to heat $2.0\,\text{kg}$ of water from $20\,°\text{C}$ to $80\,°\text{C}$ . The specific heat capacity of water is $4200\,\text{J/(kg\,°C)}$ .
Step-by-step solution
1. Step 1
  Identify variables: $m = 2.0\,\text{kg}$ ; $\Delta\theta = 80 - 20 = 60\,°\text{C}$ ; $c = 4200\,\text{J/(kg\,°C)}$ .
2. Step 2
  Apply $Q = mc\Delta\theta$ .
  $Q = 2.0 \times 4200 \times 60 = 504\,000\,\text{J} = 504\,\text{kJ}$
Answer
$Q = 504\,\text{kJ}$
2Specific latent heat — energy during melting
Extended• Adapted from 0654/42 Oct/Nov 2022 Q5• latent heat, melting, Q=mL
Question
Ice at $0\,°\text{C}$ is melted and then heated to $100\,°\text{C}$ . The mass of ice is $0.50\,\text{kg}$ . Calculate the total energy required. Specific latent heat of fusion of ice $= 336\,000\,\text{J/kg}$ ; specific heat capacity of water $= 4200\,\text{J/(kg\,°C)}$ .
Step-by-step solution
1. Step 1
  Energy to melt the ice (latent heat).
  $Q_{1} = mL = 0.50 \times 336\,000 = 168\,000\,\text{J}$
2. Step 2
  Energy to heat water from $0$ to $100\,°\text{C}$ .
  $Q_{2} = mc\Delta\theta = 0.50 \times 4200 \times 100 = 210\,000\,\text{J}$
3. Step 3
  Total energy.
  $Q = Q_{1} + Q_{2} = 168\,000 + 210\,000 = 378\,000\,\text{J}$
Answer
$Q = 378\,000\,\text{J} = 378\,\text{kJ}$
Examiner tip
Two separate stages must be calculated and added. Do not forget the latent heat stage — temperature does not change during melting.
3Explaining a heating curve
Challenge• heating curve, latent heat, plateau
Question
Sketch and annotate a heating curve for a substance heated at constant rate from below its melting point to above its boiling point. Explain what is happening at each stage in terms of particle energy.
Step-by-step solution
1. Step 1
  Stage 1 (solid heating): Temperature rises as particles gain kinetic energy, vibrating more vigorously.
2. Step 2
  Stage 2 (melting plateau at $T_{\text{melt}}$ ): Temperature stays constant. Energy input goes into breaking intermolecular bonds (increasing potential energy), not kinetic energy. The graph is horizontal.
3. Step 3
  Stage 3 (liquid heating): Temperature rises again as particles gain kinetic energy.
4. Step 4
  Stage 4 (boiling plateau at $T_{\text{boil}}$ ): Temperature stays constant. Energy goes into completely separating particles (liquid → gas). The graph is horizontal again.
5. Step 5
  Stage 5 (gas heating): Temperature rises as gas particles speed up.
Answer
Heating curve has two horizontal plateaux at melting point and boiling point, where energy is used to increase particle potential energy (break bonds) rather than kinetic energy (temperature).

Key Formulae — Matter and Thermal Properties

The formulae you need to memorise for matter and thermal properties on the Cambridge IGCSE 0654 paper, with every variable defined in plain English and a note on when to use it.

Specific heat capacity
$Q = mc\Delta\theta$
$Q$
heat energy transferred (J)
$m$
mass (kg)
$c$
specific heat capacity (J/(kg °C) or J/(kg K))
$\Delta\theta$
temperature change (°C or K)
When to use
Calculating energy transferred to change an object's temperature (no state change).
Specific latent heat
$Q = mL$
$Q$
heat energy transferred (J)
$m$
mass (kg)
$L$
specific latent heat (J/kg)
When to use
Calculating energy transferred during a change of state (temperature constant).

Key Definitions and Keywords — Matter and Thermal Properties

Definitions to memorise and the exact keywords mark schemes credit for matter and thermal properties answers — sharpened from recent examiner reports for the 2026 0654 sitting.

Specific heat capacity
Examiner keyword
The energy required to raise the temperature of $1\,\text{kg}$ of a substance by $1\,°\text{C}$ (or $1\,\text{K}$ ). SI unit: $\text{J/(kg\,°C)}$ or $\text{J/(kg\,K)}$ .
Specific latent heat
Examiner keyword
The energy required to change the state of $1\,\text{kg}$ of a substance without changing its temperature. Fusion (melting/freezing) or vaporisation (boiling/condensing).
Evaporation
Examiner keyword
The change of state from liquid to gas occurring at the surface of a liquid at any temperature, as high-energy molecules escape. It causes cooling of the remaining liquid.
Related: boiling
Boiling
Examiner keyword
The change of state from liquid to gas throughout the bulk of the liquid at a fixed temperature (the boiling point), where vapour bubbles form within the liquid.
Related: evaporation

Common Mistakes and Misconceptions — Matter and Thermal Properties

The traps other students keep falling into on matter and thermal properties questions — taken from recent Cambridge IGCSE 0654 examiner reports and mark schemes — and how to avoid them.

✕Forgetting to include the latent heat stage when calculating total energy for heating through a state change
Why it happens
Students only apply $Q = mc\Delta\theta$ for the entire temperature range.
How to avoid it
If the substance changes state, split into separate stages: (1) heating to transition temperature, (2) state change ( $Q = mL$ ), (3) heating from transition temperature.
✕Saying evaporation and boiling are the same process
Why it happens
Both convert liquid to gas, so students treat them identically.
How to avoid it
Evaporation: surface only, any temperature, cools liquid, no bubbles. Boiling: throughout liquid, specific temperature, bubbles form inside liquid.
✕Using the final temperature instead of the temperature change $\Delta\theta$ in $Q = mc\Delta\theta$
Why it happens
Students plug in the final temperature without subtracting the initial temperature.
How to avoid it
$\Delta\theta = T_{\text{final}} - T_{\text{initial}}$ . Always subtract the initial temperature, not just use the final value.

Practice questions

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