What are the thermal properties of materials in the context of Cambridge IGCSE Physics?

In Cambridge IGCSE Physics, the thermal properties of materials refer to how substances respond to heat. This includes specific heat capacity, thermal conductivity, and thermal expansion. Specific heat capacity is the amount of heat required to change the temperature of a unit mass by one degree Celsius. Thermal conductivity measures how well a material conducts heat, while thermal expansion describes how a material's dimensions change with temperature.

How does temperature affect the state of matter according to IGCSE Thermal Physics?

In IGCSE Thermal Physics, temperature is a key factor that influences the state of matter. As temperature increases, solids may melt into liquids, and liquids may evaporate into gases. Conversely, decreasing temperature can cause gases to condense into liquids and liquids to freeze into solids. These changes occur because temperature affects the kinetic energy of particles, influencing their movement and the forces between them.

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

Specific heat capacity is a fundamental concept in Thermal Physics, defined as the amount of heat energy required to raise the temperature of one kilogram of a substance by one degree Celsius. It is important because it helps predict how substances will respond to heat changes, which is crucial in applications like heating systems, cooking, and climate science. Different materials have different specific heat capacities, affecting how they absorb and release heat.

Why is thermal expansion significant in everyday applications?

Thermal expansion is significant because it affects how materials behave when subjected to temperature changes. In everyday applications, this is crucial for designing structures like bridges and railways, which expand and contract with temperature fluctuations. Understanding thermal expansion helps engineers prevent structural damage and ensure safety by allowing for these changes in their designs.

How is thermal conductivity relevant in the study of Thermal Physics?

Thermal conductivity is a measure of a material's ability to conduct heat and is a key concept in Thermal Physics. It is relevant because it determines how quickly heat can transfer through a material, influencing insulation, heating, and cooling processes. Materials with high thermal conductivity, like metals, transfer heat efficiently, while those with low thermal conductivity, like wood or foam, are used as insulators to reduce heat loss.

Why is "Including $\Delta\theta$ during a phase change (note: the $E = mL$ part is beyond the 0625 syllabus)" a common mistake in Thermal Properties and Temperature?

Why it happens: Forgetting that temperature stays constant during melting/boiling. How to avoid it: On 0625, just know qualitatively that stays constant while ice melts. The quantitative fix — use E = mL for the phase change — is enrichment beyond 0625 (A-level / other-board IGCSE).

Why is "Using $L_{f}$ when $L_{v}$ is needed, or vice versa (beyond the 0625 syllabus)" a common mistake in Thermal Properties and Temperature?

Why it happens: Mislabelling the change. How to avoid it: Fusion = solid↔liquid. Vaporisation = liquid↔gas. Note: latent-heat calculations are not examined on 0625 — included only as enrichment.

Why is "Mixing J/(g·°C) with J/(kg·K)" a common mistake in Thermal Properties and Temperature?

Why it happens: Different textbooks use different units. How to avoid it: Stick with SI: c in J/(kg·K), m in kg.

Why is "Saying particles get bigger when heated" a common mistake in Thermal Properties and Temperature?

Why it happens: Confusing expansion with particle size. How to avoid it: Particle SIZE doesn't change. Only the SPACING between them grows.

Thermal PhysicsCambridge IGCSE Physics (0625)

Thermal Properties and Temperature

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Thermal Properties and Temperature — Cambridge IGCSE 0625 Physics Extended (2026)

Thermal expansion, specific heat capacity, changes of state, and evaporation. The energy bookkeeping for heated solids and fluids. On 0625, phase changes are treated qualitatively — specific latent heat ( $E = mL$ ) is beyond the syllabus and is flagged as enrichment where it appears.

What you’ll learn

Mapped to the Cambridge IGCSE 0625 syllabus (2026-2028).

2.2 — Describe thermal expansion of solids, liquids and gases.
2.2 — Recall and use $E = mc\Delta T$ to calculate energy in heating.
2.2 — Describe melting and boiling in terms of energy input without a temperature change.
2.2 — Distinguish between boiling and evaporation.

Thermal expansion

Heat → particles vibrate more → spacing increases → solid expands. Gas expands the most.

Why? Heating gives particles more KE, so they vibrate further from each other on average → material expands.

Order of expansion: gases (most) > liquids > solids (least). Gases have weak forces and free particles, so expansion is dramatic.

Practical consequences.

Bridges: have expansion gaps to allow concrete to expand on hot days without buckling.
Bimetallic strip: two metals with different expansion rates bonded together. On heating, the strip BENDS (towards the slower-expanding metal). Used in thermostats.
Thermometers (liquid-in-glass): mercury or coloured alcohol expands up a thin tube. Read off the temperature from the column height.

The metal that expands more ends up on the outside of the curve — this bending switches a thermostat.

Worked qualitative. Why do tightly closed glass bottles sometimes crack on heating? Liquid expands more than glass. The trapped liquid pushes outward → glass cracks.

Gas > liquid > solid in expansion.
Bimetallic strip bends — used in thermostats.
Bridges have expansion gaps.
Liquid-in-glass thermometers exploit liquid expansion.

Specific heat capacity

$E = mc\Delta T$ . Different materials need different amounts of energy to heat by the same temperature.

Specific heat capacity $c$ = the energy required to raise the temperature of $1\,\text{kg}$ of a substance by $1\degree\text{C}$ (or $1\,\text{K}$ ). Units: $\text{J/(kg}\,\degree\text{C)}$ .

Formula. $E = mc\Delta T,$ where $E$ is energy ( $\text{J}$ ), $m$ is mass ( $\text{kg}$ ), $\Delta T$ is temperature change.

Worked. Heating $0.5\,\text{kg}$ of water (c = $4200\,\text{J/(kg}\,\degree\text{C)}$ ) from $20\degree\text{C}$ to $80\degree\text{C}$ .

$E = 0.5 \times 4200 \times 60 = 126{,}000\,\text{J} = 126\,\text{kJ}$ .

Common values to remember.

Water: $\sim 4200\,\text{J/(kg}\,\degree\text{C)}$ — VERY high.
Aluminium: $\sim 900$ .
Copper: $\sim 380$ .
Iron: $\sim 460$ .

Why water's $c$ is so high. Hydrogen bonds need extra energy to break/loosen. Useful for car coolant and heating systems — water carries lots of heat per kg.

$E = mc\Delta T$ .
Water $c \approx 4200$ — very high.
More $c$ → more energy needed for same temperature rise.
Same $\Delta T$ , different materials, different energies.

Changes of state — melting and boiling

During melting and boiling, temperature stays constant — energy breaks bonds rather than raising $T$ .

Melting and boiling are changes of state. On 0625 you describe them qualitatively: energy is supplied (or removed) at a constant temperature while the state changes.

Melting: solid → liquid, at the melting point. Freezing/solidification is the reverse.
Boiling: liquid → gas throughout the liquid, at the boiling point. Condensation is the reverse.
For water at standard atmospheric pressure: melting point $0\degree\text{C}$ , boiling point $100\degree\text{C}$ .

Why temperature stays constant. The supplied energy goes into separating particles / breaking the bonds between them, not into increasing their average kinetic energy — so the temperature does not change while the state is changing.

Heating curve for water. $T$ against energy added:

Solid (ice) heats up — temperature rises ( $E = mc\Delta T$ , $c_{\text{ice}}$ ).
Melting at $0\degree\text{C}$ — temperature constant (a flat plateau); energy goes into breaking the lattice.
Liquid water heats up to $100\degree\text{C}$ .
Boiling at $100\degree\text{C}$ — temperature constant (a second plateau); energy escapes as steam.
Steam continues heating.

Sloped parts: temperature rises (E = mcΔT). Flat parts: energy breaks bonds at constant temperature (E = mL).

At each phase change, temperature is constant — that's the flat part of the curve. The energy still flows in, just not into raising $T$ .

Melting/boiling happen at constant temperature.
Phase change: energy breaks bonds, not raise $T$ .
Heating curve: alternating slopes and plateaux.
Water: melts at $0\degree\text{C}$ , boils at $100\degree\text{C}$ .

Going deeper for A*

Beyond the 0625 syllabus — the amount of energy for a phase change can be calculated with the specific latent heat $L$ (energy to change the state of $1\,\text{kg}$ at constant temperature, units $\text{J/kg}$ ) using $E = mL$ . There are two values: specific latent heat of fusion (melting/freezing) and specific latent heat of vaporisation (boiling/condensing), and $L_{\text{vap}} > L_{\text{fus}}$ because all the inter-particle bonds must be broken to form a gas. For example, melting $0.2\,\text{kg}$ of ice needs $E = mL_{\text{f}} = 0.2 \times 334{,}000 = 66{,}800\,\text{J}$ . This equation and the term specific latent heat are NOT in the Cambridge IGCSE 0625 syllabus — §2.2.3 treats melting, boiling and evaporation qualitatively only. They belong to A-level (and some other-board IGCSE) physics and will not be examined on 0625.

Boiling vs evaporation

Boiling: bulk, at fixed $T$ . Evaporation: surface, at any $T$ . Both cool what's left.

Evaporation. Liquid molecules near the surface, with above-average KE, escape into the air. Happens at ANY temperature — even ice can sublimate.

Boiling. Bulk liquid turns to vapour at the boiling temperature. Bubbles form throughout, not just the surface.

The most energetic molecules leave, so the average kinetic energy — and temperature — of what remains drops.

Why evaporation cools. Only the FASTEST molecules escape. The remaining molecules have lower average KE → lower temperature. That's why sweat cools you — energy taken from your skin to break the water-water bonds.

Factors that increase evaporation rate.

Higher temperature: more molecules with enough KE.
Larger surface area: more molecules near the surface.
Lower humidity / wind: removes water vapour, prevents saturation.

Cambridge tip. When asked "explain why evaporation cools the liquid", give the FULL chain: faster molecules escape → average KE of remaining molecules drops → temperature is lower.

Evaporation: surface, any $T$ .
Boiling: bulk, at boiling point.
Evaporation cools — fastest molecules leave.
Increase: higher $T$ , more area, drier air, wind.

How it’s examined

Thermal properties appear every Paper 4 (6-8 marks): a heating curve question and $E = mc\Delta T$ for a heater. Examiner reports flag two errors: (i) treating phase changes as gradual (they're at constant $T$ ), (ii) confusing evaporation with boiling — they're different. Note: latent-heat calculations ( $E = mL$ ) are beyond the 0625 syllabus and are not examined.

Sources: Cambridge IGCSE Physics 0625 syllabus 2026-2028 (2.2); 0625/42 Oct/Nov 2024 — Q12 (heating curve, specific heat capacity); 0625 Examiner Reports 2022-2024. Last reviewed 2026-05-06.

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

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

1Specific heat capacity
Extended• Adapted from 0625/42 May/Jun 2024 Q10• SHC
Question
$2\,\text{kg}$ of water is heated from $20\degree\text{C}$ to $80\degree\text{C}$ . The specific heat capacity of water is $4200\,\text{J/(kg \cdot K)}$ . Find the energy required.
Step-by-step solution
1. Step 1
  $E = mc\Delta\theta$ .
  $E = 2 \times 4200 \times 60 = 504{,}000\,\text{J}$
Answer
$504{,}000\,\text{J} = 504\,\text{kJ}$
2Mixing two substances at different temperatures
Extended• mixing
Question
$0.5\,\text{kg}$ of water at $80\degree\text{C}$ is mixed with $0.5\,\text{kg}$ of water at $20\degree\text{C}$ . Find the final temperature, ignoring losses.
Step-by-step solution
1. Step 1
  Heat lost = heat gained.
  $0.5 \times c \times (80 - T) = 0.5 \times c \times (T - 20)$
2. Step 2
  Cancel and solve.
  $80 - T = T - 20 \Rightarrow T = 50\degree\text{C}$
Answer
$50\degree\text{C}$
3Beyond 0625 — specific latent heat of fusion
Challenge• latent heat, beyond-0625
Question
Find the energy needed to melt $0.4\,\text{kg}$ of ice at $0\degree\text{C}$ . $L_{f} = 334{,}000\,\text{J/kg}$ .
Step-by-step solution
1. Step 1
  $E = mL$ .
  $E = 0.4 \times 334{,}000 = 133{,}600\,\text{J}$
Answer
$\approx 1.34 \times 10^{5}\,\text{J}$
Examiner tip
Enrichment beyond the Cambridge IGCSE 0625 syllabus. Specific latent heat and the equation $E = mL$ are not in 0625 — §2.2.3 treats melting and boiling qualitatively (energy input at constant temperature) without a quantitative formula. This calculation belongs to A-level (and other-board IGCSE) physics. The qualitative point still applies on 0625: during a phase change the temperature does not rise because the energy goes into breaking bonds.
4Why solids expand on heating
Core• expansion
Question
Explain why a solid expands when heated, using the kinetic model.
Step-by-step solution
1. Step 1
  Particles vibrate FASTER and through a LARGER amplitude.
2. Step 2
  Average separation increases → solid expands.
Answer
Greater vibration amplitude increases the average particle separation.
5Find specific heat capacity from an experiment
Extended• Adapted from 0625/42 Oct/Nov 2023 Q11• SHC, experiment
Question
A $0.50\,\text{kg}$ block of aluminium is heated by a $50\,\text{W}$ immersion heater for $3.0$ minutes. The temperature rises from $20\degree\text{C}$ to $53\degree\text{C}$ . Find the specific heat capacity of aluminium, ignoring losses.
Step-by-step solution
1. Step 1
  Energy supplied by the heater.
  $E = P t = 50 \times (3 \times 60) = 9000\,\text{J}$
2. Step 2
  Apply $E = mc\Delta\theta$ and rearrange for $c$ .
  $c = \dfrac{E}{m\,\Delta\theta} = \dfrac{9000}{0.50 \times 33}$
3. Step 3
  Compute.
  $c \approx 545\,\text{J/(kg \cdot K)}$
Answer
$c \approx 545\,\text{J/(kg\cdot K)}$ (book value $\approx 900$ — losses explain the low value).
Examiner tip
The examiner report flags candidates often forget to convert minutes to seconds before computing $E = Pt$ . Energy must be in joules, time in seconds.
6Heat water in a copper kettle
Extended• SHC, multi-step
Question
$1.5\,\text{kg}$ of water inside a $0.40\,\text{kg}$ copper kettle is heated from $20\degree\text{C}$ to $100\degree\text{C}$ . $c_{\text{water}} = 4200\,\text{J/(kg\cdot K)}$ , $c_{\text{copper}} = 390\,\text{J/(kg\cdot K)}$ . Find the total energy supplied (no losses).
Step-by-step solution
1. Step 1
  Energy to heat the water.
  $E_{w} = 1.5 \times 4200 \times 80 = 504{,}000\,\text{J}$
2. Step 2
  Energy to heat the kettle.
  $E_{k} = 0.40 \times 390 \times 80 = 12{,}480\,\text{J}$
3. Step 3
  Total.
  $E = E_{w} + E_{k} = 5.16 \times 10^{5}\,\text{J}$
Answer
$\approx 5.2 \times 10^{5}\,\text{J} = 520\,\text{kJ}$
7Beyond 0625 — specific latent heat of vaporisation
Challenge• latent heat, beyond-0625
Question
An electric kettle takes $480\,\text{s}$ at $2.0\,\text{kW}$ to boil away $0.42\,\text{kg}$ of water already at $100\degree\text{C}$ . Find the specific latent heat of vaporisation of water, ignoring losses.
Step-by-step solution
1. Step 1
  Energy supplied.
  $E = P t = 2000 \times 480 = 9.6 \times 10^{5}\,\text{J}$
2. Step 2
  Apply $E = m L_{v}$ .
  $L_{v} = \dfrac{E}{m} = \dfrac{9.6 \times 10^{5}}{0.42}$
3. Step 3
  Compute.
  $L_{v} \approx 2.29 \times 10^{6}\,\text{J/kg}$
Answer
$\approx 2.3 \times 10^{6}\,\text{J/kg}$ (close to data-book value).
Examiner tip
Enrichment beyond the Cambridge IGCSE 0625 syllabus. Specific latent heat of vaporisation and the equation $E = mL_{v}$ are not in 0625 — §2.2.3 treats boiling qualitatively only. This belongs to A-level (and other-board IGCSE) physics and is not examined on 0625.
8Cold metal block dropped into hot water
Extended• Adapted from 0625/42 May/Jun 2023 Q11• mixing, energy conservation
Question
A $0.20\,\text{kg}$ aluminium block at $20\degree\text{C}$ is dropped into $0.30\,\text{kg}$ of water at $90\degree\text{C}$ . $c_{\text{Al}} = 900\,\text{J/(kg\cdot K)}$ , $c_{\text{w}} = 4200\,\text{J/(kg\cdot K)}$ . Find the final temperature (no losses).
Step-by-step solution
1. Step 1
  Heat lost by water = heat gained by aluminium.
  $m_{w} c_{w} (90 - T) = m_{Al} c_{Al} (T - 20)$
2. Step 2
  Substitute values.
  $0.30 \times 4200 \times (90 - T) = 0.20 \times 900 \times (T - 20)$
3. Step 3
  Simplify.
  $1260(90 - T) = 180(T - 20)$
4. Step 4
  Solve.
  $113{,}400 - 1260T = 180T - 3600 \Rightarrow T \approx 81.3\degree\text{C}$
Answer
$T \approx 81\degree\text{C}$
Examiner tip
The examiner report flags candidates often forget the directionality of $\Delta\theta$ — the substance that COOLS appears as (hot − $T$ ), the one that WARMS as ( $T$ − cold).
9Linear thermal expansion of a metal rod
Extended• expansion
Question
A steel railway rail of length $20.0\,\text{m}$ at $5\degree\text{C}$ expands by $4.8\,\text{mm}$ when the temperature rises to $25\degree\text{C}$ . (a) State the temperature change. (b) Explain why short gaps are left between rails.
Step-by-step solution
1. Step 1
  Temperature change.
  $\Delta\theta = 25 - 5 = 20\,\text{K}$
2. Step 2
  The rail's particles vibrate more vigorously when heated → average separation grows → length increases by $4.8\,\text{mm}$ .
3. Step 3
  If rails butt directly together, expansion has nowhere to go and the rails buckle. Gaps allow free expansion in summer.
Answer
(a) $\Delta\theta = 20\,\text{K}$ . (b) Gaps accommodate thermal expansion and prevent buckling.
10A* — Read a cooling curve with plateaus
Challenge• Adapted from 0625/42 Oct/Nov 2024 Q10• graph, phase change, synoptic
Question
A liquid wax cools steadily and the temperature against time graph shows: a steady fall from $120\degree\text{C}$ to $80\degree\text{C}$ , a horizontal plateau at $80\degree\text{C}$ for $5$ minutes, then a fall to room temperature. Identify (a) what is happening on the plateau and why temperature stays constant, (b) what the wax's melting point is, (c) which section involves the largest energy release per unit time.
Step-by-step solution
1. Step 1
  (a) On the plateau the wax is changing state (freezing — liquid → solid). All the energy released is latent heat, used as bonds form. Average particle KE (i.e. temperature) is unchanged.
2. Step 2
  (b) The plateau temperature is the melting/freezing point.
  $T_{m} = 80\degree\text{C}$
3. Step 3
  (c) The plateau: a lot of energy is released in a short interval while $\theta$ does not change. So energy per unit time is greater than the surrounding sloping sections (where only mc $\Delta\theta$ is being released).
Answer
(a) Freezing/solidification — latent heat released, no change in average KE. (b) $80\degree\text{C}$ . (c) The plateau section.
Examiner tip
The examiner report flags candidates often say 'the wax is at its hottest' on the plateau instead of identifying the phase change. State explicitly that bonds are forming and latent heat is being released.
11Bimetallic strip as a thermometer
Extended• expansion, application
Question
A bimetallic strip is made by riveting together a strip of brass and a strip of iron. Brass expands more than iron for the same temperature rise. (a) Describe what happens to the strip when it is heated. (b) Explain in particle terms why brass expands more than iron.
Step-by-step solution
1. Step 1
  (a) Both metals get longer when heated, but brass expands more. Because they are rigidly joined, the only way to accommodate the different lengths is for the strip to BEND, with brass on the outside (longer) curve.
2. Step 2
  (b) Heating increases the amplitude of atomic vibration in both metals. Brass atoms have weaker average bonding forces than iron atoms, so the same energy produces a larger increase in average separation in brass.
3. Step 3
  Use: as a temperature-sensitive switch in thermostats — the strip bends to break or complete a circuit at a chosen temperature.
Answer
(a) The strip bends, with brass on the outside (longer side). (b) Weaker bonding in brass gives a larger increase in particle separation for the same temperature rise.
12Beyond 0625 — energy to take ice at $-10\degree\text{C}$ to steam at $100\degree\text{C}$
Challenge• synoptic, multi-step, beyond-0625
Question
Find the total energy needed to convert $0.10\,\text{kg}$ of ice at $-10\degree\text{C}$ to steam at $100\degree\text{C}$ . Data: $c_{\text{ice}} = 2100\,\text{J/(kg\cdot K)}$ , $L_{f} = 3.34 \times 10^{5}\,\text{J/kg}$ , $c_{\text{w}} = 4200\,\text{J/(kg\cdot K)}$ , $L_{v} = 2.26 \times 10^{6}\,\text{J/kg}$ .
Step-by-step solution
1. Step 1
  Warm ice from $-10$ to $0\degree\text{C}$ .
  $E_{1} = 0.10 \times 2100 \times 10 = 2100\,\text{J}$
2. Step 2
  Melt ice at $0\degree\text{C}$ .
  $E_{2} = 0.10 \times 3.34 \times 10^{5} = 33{,}400\,\text{J}$
3. Step 3
  Warm water from $0$ to $100\degree\text{C}$ .
  $E_{3} = 0.10 \times 4200 \times 100 = 42{,}000\,\text{J}$
4. Step 4
  Boil water at $100\degree\text{C}$ .
  $E_{4} = 0.10 \times 2.26 \times 10^{6} = 226{,}000\,\text{J}$
5. Step 5
  Sum.
  $E_{\text{total}} = 2100 + 33{,}400 + 42{,}000 + 226{,}000 \approx 3.0 \times 10^{5}\,\text{J}$
Answer
$\approx 3.0 \times 10^{5}\,\text{J} = 304\,\text{kJ}$ . Note vaporisation dominates (≈74% of total).
Examiner tip
Enrichment beyond the Cambridge IGCSE 0625 syllabus. The melting and boiling stages here use specific latent heat ( $E = mL$ ), which is not in 0625 — §2.2.3 treats phase changes qualitatively only. The specific-heat-capacity stages ( $E = mc\Delta\theta$ ) are in scope, but the full ice-to-steam calculation belongs to A-level (and other-board IGCSE) physics and is not examined on 0625.

Key Formulae — Thermal Properties and Temperature

The formulae you need to memorise for thermal properties and temperature on the Cambridge IGCSE 0625 paper, with every variable defined in plain English and a note on when to use it.

Specific heat capacity
$E = mc\Delta\theta$
$m$
mass (kg)
$c$
specific heat capacity (J/(kg·K))
$\Delta\theta$
temperature change (K or °C)
When to use
Heating or cooling without a phase change.
Latent heat (beyond the 0625 syllabus)
$E = mL$
$L$
specific latent heat (J/kg)
When to use
Energy required for a phase change at constant temperature. NOTE: this equation is NOT in the Cambridge IGCSE 0625 syllabus — §2.2.3 treats melting and boiling qualitatively only. It belongs to A-level (and other-board IGCSE) physics. Shown here as enrichment.

Key Definitions and Keywords — Thermal Properties and Temperature

Definitions to memorise and the exact keywords mark schemes credit for thermal properties and temperature answers — sharpened from recent examiner reports for the 2026 0625 sitting.

Temperature
Examiner keyword
A measure of the average kinetic energy of the particles.
Specific heat capacity
Examiner keyword
Energy required per unit mass per unit rise in temperature. SI unit J/(kg·K).
Specific latent heat (beyond the 0625 syllabus)
Energy required per unit mass to change state at constant temperature. Not examined on Cambridge IGCSE 0625 — §2.2.3 covers phase changes qualitatively only; this term belongs to A-level (and other-board IGCSE) physics.
Latent heat of fusion / vaporisation (beyond the 0625 syllabus)
Fusion: solid ↔ liquid. Vaporisation: liquid ↔ gas. Use the right one for the change. Not examined on Cambridge IGCSE 0625 — included as enrichment beyond the syllabus.
Thermal expansion
Examiner keyword
Increase in size when temperature rises, caused by greater amplitude of particle vibration.

Common Mistakes and Misconceptions — Thermal Properties and Temperature

The traps other students keep falling into on thermal properties and temperature questions — taken from recent Cambridge IGCSE 0625 examiner reports and mark schemes — and how to avoid them.

✕Including $\Delta\theta$ during a phase change (note: the $E = mL$ part is beyond the 0625 syllabus)
Why it happens
Forgetting that temperature stays constant during melting/boiling.
How to avoid it
On 0625, just know qualitatively that $\theta$ stays constant while ice melts. The quantitative fix — use $E = mL$ for the phase change — is enrichment beyond 0625 (A-level / other-board IGCSE).
✕Using $L_{f}$ when $L_{v}$ is needed, or vice versa (beyond the 0625 syllabus)
Why it happens
Mislabelling the change.
How to avoid it
Fusion = solid↔liquid. Vaporisation = liquid↔gas. Note: latent-heat calculations are not examined on 0625 — included only as enrichment.
✕Mixing J/(g·°C) with J/(kg·K)
Why it happens
Different textbooks use different units.
How to avoid it
Stick with SI: c in J/(kg·K), m in kg.
✕Saying particles get bigger when heated
Why it happens
Confusing expansion with particle size.
How to avoid it
Particle SIZE doesn't change. Only the SPACING between them grows.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

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Short walkthrough of the concepts students most often get stuck on.

Thermal Properties and Temperature — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Thermal Properties and Temperature

Short Study Notes in the form of Slides

Detailed Notes

What you’ll learn

How it’s examined

1Specific heat capacity

2Mixing two substances at different temperatures

3Beyond 0625 — specific latent heat of fusion

4Why solids expand on heating

5Find specific heat capacity from an experiment

6Heat water in a copper kettle

7Beyond 0625 — specific latent heat of vaporisation

8Cold metal block dropped into hot water

9Linear thermal expansion of a metal rod

10A* — Read a cooling curve with plateaus

11Bimetallic strip as a thermometer

12Beyond 0625 — energy to take ice at −10°C-10\degree\text{C}−10°C to steam at 100°C100\degree\text{C}100°C

Specific heat capacity

Latent heat (beyond the 0625 syllabus)

Temperature

Specific heat capacity

Specific latent heat (beyond the 0625 syllabus)

Latent heat of fusion / vaporisation (beyond the 0625 syllabus)

Thermal expansion

✕Including Δθ\Delta\thetaΔθ during a phase change (note: the E=mLE = mLE=mL part is beyond the 0625 syllabus)

✕Using LfL_{f}Lf​ when LvL_{v}Lv​ is needed, or vice versa (beyond the 0625 syllabus)

✕Mixing J/(g·°C) with J/(kg·K)

✕Saying particles get bigger when heated

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Thermal Properties and Temperature — frequently asked questions

12Beyond 0625 — energy to take ice at $-10\degree\text{C}$ to steam at $100\degree\text{C}$

✕Including $\Delta\theta$ during a phase change (note: the $E = mL$ part is beyond the 0625 syllabus)

✕Using $L_{f}$ when $L_{v}$ is needed, or vice versa (beyond the 0625 syllabus)