Edexcel IGCSE Physics 5(c) Change of State — Notes
Written for Pearson Edexcel International GCSE Physics (4PH1), linear and modular. Everything on this page is Physics-only content, examined on Paper 2P.
Which parts do I need?
This whole sub-topic is Physics only (4PH1 Paper 2P; modular 4XPH1 Unit 1). Science Single Award (4SS0) and Double Award (4SD0, 4XSD1) candidates can skip this page entirely and go straight to 5(d) Ideal gas molecules.
What this page covers
| Spec | Modular no. | You need to be able to… | Route |
|---|---|---|---|
| 5.8P | 4.8P | explain why heating increases the energy stored in a system, either raising its temperature or changing its state | Physics only · 2P |
| 5.9P | 4.9P | describe what happens during melting, evaporation and boiling | Physics only · 2P |
| 5.10P | 4.10P | describe how particles are arranged and how they move in solids, liquids and gases | Physics only · 2P |
| 5.11P | 4.11P | obtain a temperature–time graph that shows the temperature staying constant while the state changes (practical) | Physics only · 2P |
| 5.12P | 4.12P | know what specific heat capacity means and its unit, J/kg °C | Physics only · 2P |
| 5.13P | 4.13P | use ΔQ = m × c × ΔT | Physics only · 2P |
| 5.14P | 4.14P | measure the specific heat capacity of water and of solids (practical) | Physics only · 2P |
Modular candidates: these are statements 4.8P–4.14P of 4XPH1 Unit 1 (4WPH1), under “Solids, liquids and gases: Part 1”. The unit J/kg °C (5.2P / 4.2P) is on the 5(b) page.
In one minute
- Solids: particles touching in a regular pattern, vibrating about fixed positions. Liquids: touching, no regular pattern, moving around each other. Gases: far apart, moving quickly and randomly.
- Heating a system increases the energy stored in its particles. Either the particles move faster (the temperature rises) or the energy separates the particles (the state changes).
- While a substance melts or boils, its temperature stays constant — the flat sections of a heating or cooling curve.
- Specific heat capacity, \(c\), is the energy needed to raise the temperature of 1 kg of a material by 1 °C, in J/kg °C.
- \(\Delta Q=m\times c\times\Delta T\): the energy needed depends on the mass, the material and the temperature rise.
How are the particles arranged in solids, liquids and gases? Physics only · 2P
| State | Arrangement | Motion | Forces between particles |
|---|---|---|---|
| solid | close together, touching, in a regular pattern (lattice) | vibrate about fixed positions | strong — hold particles in place, so a solid keeps its shape |
| liquid | close together, touching, irregular arrangement | move around and past each other | weaker — particles stay together but can flow, so a liquid takes the shape of its container |
| gas | far apart (spacing much larger than the particles) | move quickly in random directions in straight lines between collisions | very weak — a gas spreads out to fill its container |
- Liquid particles are only slightly further apart than solid particles, so the densities of a solid and its liquid are similar; a gas is far less dense.
- When you draw a liquid, make most particles touch — gaps big enough to fit another particle lose the mark.
What does heating do to the energy stored in a system? Physics only · 2P
The extra energy goes into the particles in one of two ways:
- The particles move faster (their kinetic energy increases). This is what a thermometer detects: the temperature rises.
- The particles are pulled further apart, overcoming the forces (bonds) holding them together. This increases the energy stored in the arrangement of the particles, not their speed, so the state changes while the temperature stays the same.
Which one happens depends on the situation: below the melting point, heating a solid raises its temperature; at the melting point, the energy is used to melt it; once it is all liquid, the temperature rises again.
What happens when a solid melts and when a liquid evaporates or boils? Physics only · 2P
| Change | What happens to the particles | Temperature |
|---|---|---|
| melting (solid → liquid) | particles gain enough energy to break free of their fixed positions; the regular pattern breaks down and they start moving around each other | stays at the melting point until all the solid has melted |
| evaporation (liquid → gas, at the surface) | the fastest-moving particles near the surface escape from the liquid | happens at any temperature; the liquid left behind cools because its average particle speed falls |
| boiling (liquid → gas, throughout) | particles everywhere in the liquid gain enough energy to separate completely; bubbles of gas form inside the liquid | happens only at the boiling point and stays there until all the liquid has boiled |
The reverse changes are freezing (liquid → solid) and condensing (gas → liquid). During these, the substance transfers energy to the surroundings, and again the temperature stays constant while the state changes.
Evaporation compared with boiling
- Evaporation takes place only at the surface and at any temperature; boiling takes place throughout the liquid and only at the boiling point.
- Evaporation cools a liquid: the particles that escape are the fastest ones, so the average kinetic energy of the particles left behind falls. This is why sweat evaporating cools your skin.
- Evaporation is faster when the liquid is warmer, has a larger surface area, or has air moving across its surface.
What does a heating curve show? Physics only · 2P
If a substance is heated at a steady rate and its temperature is plotted against time, the graph has sloping sections (temperature rising) and flat sections (state changing).
The graph shows how the temperature of some ice changes when it is heated steadily from −20 °C. (a) State what is happening in sections B and D. (b) Explain why the temperature does not change in section B, even though energy is still being supplied.
(a) B: the ice is melting at 0 °C. D: the water is boiling at 100 °C.
(b) The energy supplied is used to break the bonds between the particles / to overcome the forces holding them in a fixed pattern (1). It increases the energy stored in the arrangement of the particles, not their kinetic energy, so the particles do not move faster (1) and the temperature stays the same (1).
Section D is longer than B: much more energy is needed to separate the particles completely (boiling) than to free them from their fixed positions (melting). Section C is less steep than A because liquid water has a higher specific heat capacity than ice, so each degree of temperature rise needs more energy.
Practical: how do you obtain a temperature–time graph for a change of state? Physics only · 2P
The clearest results come from a substance that solidifies just above room temperature, such as stearic acid (melting point about 69 °C). You melt it, then record its temperature as it cools and solidifies.
Obtaining a cooling curve for stearic acid
- Aim: show that the temperature of a substance stays constant while it changes state.
- Apparatus: boiling tube of solid stearic acid, beaker of hot water (water bath) or kettle water, thermometer or temperature probe and data logger, stopwatch, clamp and stand.
- Independent variable: time. Dependent variable: temperature of the stearic acid.
- Control variables: the same mass of stearic acid; the tube cools in the same place, away from draughts.
Method
- Stand the boiling tube in the hot water until all the stearic acid has melted. Put the thermometer in the liquid.
- When the temperature is about 90 °C, lift the tube out of the water, clamp it in air and start the stopwatch.
- Record the temperature every 30 s until it has fallen to about 50 °C, noting when the solid starts to form and when it is all solid.
- Plot temperature (vertical axis) against time (horizontal axis) and draw a smooth curve of best fit, including the flat section.
Processing, errors and improvements
- The flat section shows the change of state; its temperature is the melting (freezing) point.
- A temperature probe and data logger records more often than a person can and avoids reading errors.
- Gently stir the liquid with the thermometer while it is still liquid so that the reading is the temperature of all of it, not just one part.
- The same experiment can be done by heating crushed ice in a beaker: the temperature stays close to 0 °C until all the ice has melted.
- Safety: heat stearic acid only in a water bath, never directly in a flame; wear eye protection; take care with hot water.
What is specific heat capacity and how do you use \(\Delta Q=mc\Delta T\)? Physics only · 2P
- \(\Delta Q\) = energy transferred in J, \(m\) = mass in kg, \(c\) = specific heat capacity in J/kg °C, \(\Delta T\) = temperature change in °C.
- Water has a high specific heat capacity, about 4200 J/kg °C: 4200 J raises 1 kg of water by only 1 °C. Metals have much lower values (copper about 385 J/kg °C), so they heat up and cool down quickly.
- \(\Delta T\) is a change: final temperature − starting temperature. A rise from 15 °C to 65 °C is \(\Delta T=50\ ^\circ\text{C}\).
- The equation applies only while the material stays in the same state — not during melting or boiling.
Calculate the energy needed to heat 2.0 kg of water from 15 °C to 65 °C. The specific heat capacity of water is 4200 J/kg °C.
\(\Delta T=65-15=50\ ^\circ\text{C}\)
\(\Delta Q=m\times c\times\Delta T=2.0\times4200\times50=\) 420 000 J (4.2 × 10⁵ J)
An electric kettle of power 2400 W contains 1.2 kg of water at 20 °C. The specific heat capacity of water is 4200 J/kg °C.
(a) Calculate the energy needed to bring the water to 100 °C. (b) Calculate the shortest time this could take. (c) In practice it takes 190 s. Suggest why.
(a) \(\Delta Q=1.2\times4200\times(100-20)=\) 403 200 J (4.0 × 10⁵ J)
(b) power = energy ÷ time, so time = 403 200 ÷ 2400 = 168 s (170 s)
(c) Some of the energy supplied heats the kettle itself and some is transferred to the surroundings, so more than 403 200 J must be supplied and it takes longer.
9600 J of energy is transferred to 0.50 kg of copper and, separately, to 0.50 kg of water. Calculate the temperature rise of each, and explain the difference. (c for copper = 385 J/kg °C; c for water = 4200 J/kg °C.)
Rearrange: \(\Delta T=\dfrac{\Delta Q}{m\times c}\)
copper: \(\Delta T=\dfrac{9600}{0.50\times385}=49.9=\) 50 °C; water: \(\Delta T=\dfrac{9600}{0.50\times4200}=4.57=\) 4.6 °C
Water has a specific heat capacity about 11 times larger, so each kilogram needs about 11 times as much energy for each degree. The same energy gives copper a temperature rise about 11 times larger.
Practical: how do you measure specific heat capacity? Physics only · 2P
You supply a measured amount of energy to a known mass of material and measure its temperature rise. Then \(c=\dfrac{\Delta Q}{m\times\Delta T}\).
Investigating the specific heat capacity of a solid and of water
- Aim: determine the specific heat capacity of a metal block and of water.
- Apparatus: metal block (with holes for a heater and a thermometer), immersion heater, low-voltage supply, joulemeter (or ammeter, voltmeter and stopwatch), thermometer, balance, insulation; for water, a beaker of water on an insulated base.
- Measured quantities: mass of the material, energy supplied and the temperature rise.
- Control variables: keep the insulation the same for every material tested, and use the same heater.
Method
- Measure the mass of the block on a balance (or of the water, as the mass of the beaker with water minus the empty beaker).
- Put the heater and the thermometer in the holes; add a drop of oil in the thermometer hole for good thermal contact. Wrap the block in insulation.
- Record the starting temperature. Switch on the heater and reset the joulemeter (or start the stopwatch and record the current and voltage).
- After a set time (for example 10 minutes) switch off. Keep watching the thermometer and record the highest temperature reached.
- Energy supplied = joulemeter reading, or energy = current × voltage × time (or power × time). Calculate \(c=\dfrac{\Delta Q}{m\times\Delta T}\).
- For water, stir it before reading the temperature so that the reading is the temperature of all the water.
Errors and improvements
- Some energy is transferred to the surroundings and to the heater, so the temperature rise is smaller than it should be and the calculated value of \(c\) is too high. Insulation (and a lid for water) reduces this.
- The block keeps warming for a short time after the heater is switched off, because the heater is still hotter than the block — so wait for the highest reading.
- A larger temperature rise (heating for longer) reduces the effect of the thermometer's reading uncertainty.
- Safety: the heater and block become hot — do not touch them until they have cooled; keep water away from the electrical supply.
In the experiment shown, a 48 W heater is switched on for 300 s. The temperature of the 1.0 kg block rises from 20.0 °C to 36.0 °C. Calculate the specific heat capacity of the metal.
Energy supplied: \(\Delta Q=P\times t=48\times300=14\,400\ \text{J}\) (this matches the joulemeter reading)
\(c=\dfrac{\Delta Q}{m\times\Delta T}=\dfrac{14\,400}{1.0\times(36.0-20.0)}=\dfrac{14\,400}{16}=\) 900 J/kg °C
Common error: using the power (48 W) as the energy, without multiplying by the time.
Exam tips for 5(c)
In specific heat capacity questions the heater is often given as a power in watts. Multiply by the heating time in seconds to get the energy in joules before using \(\Delta Q=mc\Delta T\).
For “explain why the temperature stays constant”, say where the energy goes (breaking bonds / overcoming forces between particles) and that the kinetic energy of the particles does not increase. “The energy is used to change state” on its own is not enough for full marks.
Label the axes with units, show a horizontal section at the correct melting or boiling point, and do not add extra flat sections for changes of state that do not happen. Sloping lines, not curves, are fine for a sketch.
A full-mark method names every measuring instrument (balance, thermometer, joulemeter or ammeter + voltmeter + stopwatch), says what is calculated from the readings, and includes at least one detail that improves accuracy — insulation, stirring, or waiting for the highest temperature.
Give a contrast in each part of your answer: surface versus throughout the liquid; any temperature versus only at the boiling point. Two separate descriptions without a comparison may not score.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Heating always makes the temperature rise. | During a change of state the energy separates the particles instead. | The temperature is constant while a substance melts or boils. |
| Particles in a liquid are far apart. | Liquids are almost as dense as solids. | Liquid particles touch; they are just not in a regular pattern. |
| Particles themselves expand or melt when heated. | The particles do not change; their motion and spacing do. | Heating makes particles move faster or move further apart. |
| Evaporation only happens at 100 °C. | Puddles dry up on cool days. | Evaporation happens at any temperature, from the surface only. |
| A material with a high specific heat capacity heats up quickly. | It needs more energy for each degree of temperature rise. | High \(c\) means a slow temperature rise for the same energy input. |
Key equations
| Equation | Use it to find | Route | On the Equation List? |
|---|---|---|---|
| change in thermal energy = mass × specific heat capacity × change in temperature, \(\Delta Q=m\times c\times\Delta T\) | energy needed for a temperature change, or \(c\), \(m\) or \(\Delta T\) | Physics only · 2P | yes — Physics 2P only |
| power = work done ÷ time taken, so energy = power × time (from 4(c)) | the energy supplied by a heater | All | yes — all papers |
The \(\Delta Q=mc\Delta T\) equation is on the additional Physics-only page of the Equation List, used for Paper 2P.
Check your understanding
1. Describe the arrangement and motion of the particles in a gas. Physics only · 2P
Far apart (spacing much larger than the particles), randomly arranged (1); moving quickly in random directions (1).
2. Water is boiling in a pan on a hob that is still switched on. Explain why the temperature of the water stays at 100 °C. Physics only · 2P
The energy supplied is used to separate the particles / overcome the forces between them (1), so the kinetic energy of the particles does not increase (1) and the temperature does not rise (1).
3. Give two differences between evaporation and boiling. Physics only · 2P
Evaporation happens only at the surface; boiling happens throughout the liquid (bubbles form inside). Evaporation happens at any temperature; boiling happens only at the boiling point.
4. Calculate the energy needed to raise the temperature of 0.25 kg of cooking oil by 30 °C. (c = 2000 J/kg °C) Physics only · 2P
\(\Delta Q=0.25\times2000\times30=\) 15 000 J
5. 33 600 J of energy is supplied to 0.40 kg of water at 18 °C. Calculate its final temperature. (c = 4200 J/kg °C) Physics only · 2P
\(\Delta T=\dfrac{33\,600}{0.40\times4200}=20\ ^\circ\text{C}\), so final temperature = 18 + 20 = 38 °C. (Common error: giving 20 °C as the answer.)
6. A student heats 0.50 kg of a liquid with a heater (current 5.0 A, voltage 12 V) for 600 s. The temperature rises from 20 °C to 62 °C. Calculate the specific heat capacity, and state whether the true value is likely to be higher or lower. Physics only · 2P
Energy = 5.0 × 12 × 600 = 36 000 J; \(c=\dfrac{36\,000}{0.50\times42}=1714\approx\) 1700 J/kg °C. The true value is likely to be lower: some energy was transferred to the surroundings, so less than 36 000 J actually heated the liquid.
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