AQA GCSE Combined Science (8464), Higher tier · Physics › Particle model of matter › Internal energy and energy transfers
Practise Changes of state and specific latent heat. 14 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Latent heat is the energy needed to change the state of a substance without changing its temperature. You need to define specific latent heat, use E = m L (on the equation sheet), tell latent heat of fusion from latent heat of vaporisation, and interpret heating and cooling graphs. Harder questions combine E = m L with ΔE = m c Δθ or with power and time.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
4
Know temperature stays constant during state changesWhile a substance melts, boils, freezes or condenses, its temperature does not change.
5
Define specific latent heatThe amount of energy needed to change the state of 1 kg of a substance with no change in temperature.
5
Use E = m LCalculate the energy for a change of state with E in J, m in kg and L in J/kg.
5
Find changes of state on heating graphsFlat sections show a change of state at the melting or boiling point; sloping sections show a temperature change.
6
Distinguish latent heat of fusion and vaporisationFusion: changing between solid and liquid. Vaporisation: changing between liquid and vapour (gas).
6
Tell specific heat capacity from specific latent heatSpecific heat capacity is for a temperature change with no change of state; specific latent heat is for a change of state with no temperature change.
7
Do multi-step heating and melting calculationsWork out each stage separately with ΔE = m c Δθ or E = m L, then add the energies.
8
Find specific latent heat from experimental dataUse E = P t for the energy supplied and the mass that changed state, then L = E ÷ m.
Notes
Latent heat
The energy needed for a substance to change state is called latent heat.
During a change of state, the energy supplied changes the internal energy (the potential energy of the particles) but not the temperature.
When a substance freezes or condenses, the same amount of energy per kilogram as was needed to melt or boil it is transferred to the surroundings, again at constant temperature.
Specific latent heat
The specific latent heat (L) of a substance is the amount of energy needed to change the state of 1 kg of it with no change in temperature.
thermal energy for a change of state = mass × specific latent heat: E = m L (on the equation sheet). E in J, m in kg, L in J/kg.
Specific latent heat of fusion: for a change between solid and liquid (melting, or freezing).
Specific latent heat of vaporisation: for a change between liquid and vapour (boiling, or condensing).
For water, the specific latent heat of vaporisation is several times bigger than that of fusion: to form a gas, the forces between the particles must be overcome so they can move far apart. grade 7+
Heating and cooling graphs
A heating graph (temperature against time) for a solid heated at a steady rate has sloping parts and flat parts.
Sloping parts: the temperature of one state is rising (energy calculations use ΔE = m c Δθ).
Flat parts: a change of state at constant temperature, at the melting point (first flat part) or the boiling point (second flat part). Energy calculations use E = m L.
With steady heating, a longer flat part means more energy was needed. For water, the boiling section is much longer than the melting section.
A cooling graph works the same way in reverse: its flat parts show condensing or freezing, where energy is released while the temperature stays constant.
Keep the two ideas separate: specific heat capacity is for a temperature change without a change of state; specific latent heat is for a change of state without a temperature change.
Cheatsheet
E = m L (on the equation sheet); E in J, m in kg, L in J/kg
Specific latent heat: energy to change the state of 1 kg with no change in temperature
During a change of state: internal energy changes, temperature does not
Heating graph: sloping part = temperature changing; flat part = state changing
Freezing and condensing release energy; E = m L gives how much
Specific heat capacity → temperature change. Specific latent heat → change of state
How to answer each type of question
Calculate the energy for a change of state
2 marks5
Check the question is about a change of state with no temperature change, then choose E = m L.
Choose the right L: fusion for melting or freezing, vaporisation for boiling or condensing.
Make sure the mass is in kg, substitute and give the unit (J).
Example. Calculate the energy needed to melt 0.25 kg of ice at 0 °C. specific latent heat of fusion of ice = 334 000 J/kg
Show the model answer
E = 0.25 × 334 000 (1) E = 83 500 J (1)
Interpret a heating or cooling graph
2 to 4 marks6
Find the flat sections: these are the changes of state. Read off their temperature (the melting or boiling point).
Sloping sections show one state warming up or cooling down.
Explain a flat section with internal energy: the energy changes the potential energy of the particles, not their kinetic energy, so the temperature stays constant.
Example. A student heats a solid substance at a steady rate and plots a graph of temperature against time. The temperature rises from 20 °C to 54 °C in the first 3 minutes, stays at 54 °C from 3 minutes to 8 minutes, then rises again. (a) State the melting point of the substance. (1 mark) (b) Explain why the temperature stays constant between 3 and 8 minutes, even though the substance is still being heated. (2 marks)
Show the model answer
(a) 54 °C (1) (b) The substance is melting (changing state) (1). The energy supplied increases the internal energy by increasing the potential energy of the particles, not their kinetic energy, so the temperature does not change (1).
Multi-step: change of state plus temperature change
4 to 5 marks7
Split the process into stages: each change of state and each temperature change.
Use E = m L for each change of state and ΔE = m c Δθ for each temperature change.
Add the energies for all the stages and give the unit.
Example. Calculate the energy needed to turn 0.50 kg of ice at 0 °C into water at 20 °C. specific latent heat of fusion of ice = 334 000 J/kg specific heat capacity of water = 4200 J/kg °C
Find the energy supplied with E = P t (power in W, time in s).
Find the mass that changed state (e.g. the mass of water boiled away) in kg.
Calculate L = E ÷ m and give the unit J/kg.
Example. A kettle with a power of 2.4 kW is kept boiling with its lid open. In 120 s, the mass of water in the kettle decreases by 0.125 kg. (a) Calculate the specific latent heat of vaporisation of water given by these results. Give your answer to 2 significant figures. (3 marks) (b) The true value is lower than the student's value. Suggest why. (1 mark)
Show the model answer
(a) E = 2400 × 120 = 288 000 J (1) L = 288 000 ÷ 0.125 (1) L = 2 304 000 = 2.3 × 106 J/kg (to 2 s.f.) (1) (b) Some of the energy from the kettle was transferred to the surroundings instead of boiling the water, so less than 288 000 J was used to boil the 0.125 kg (1).
Shortcuts and memory tricks
On a heating graph: Level line → Latent heat (E = m L); Climbing line → c (ΔE = m c Δθ).
There is no Δθ in E = m L. If the temperature doesn't change, you can't be using c.
Sense check for water: boiling away 1 kg (about 2.26 million J) takes far more energy than heating 1 kg from 0 °C to 100 °C (420 000 J).
Where marks are lost
Using the wrong specific latent heat: fusion when the question is about boiling, or the reverse.
Putting a temperature change into E = m L. The temperature does not change during a change of state.
Saying the energy supplied during melting is 'lost' or 'does nothing'. It increases the internal energy (the potential energy of the particles).
In multi-step questions, missing out a stage or leaving the mass in grams.
Writing the unit of specific latent heat as J/kg °C. It is J/kg.
Exam technique
Before calculating, decide: does the temperature change (use c) or does the state change (use L)? Some questions need both.
Set out multi-step answers stage by stage with a label (melting, warming, boiling) so each stage can earn its marks.
When explaining a flat section, use the words 'internal energy' and 'potential energy of the particles', and say the temperature does not change.
Read graph values carefully from the axes; a melting or boiling point is the temperature of a flat section.
Quick recall
Cover the answers and test yourself. The app has these as flashcards that come back just before you'd forget them.
Calculate the energy released when 0.40 kg of liquid stearic acid solidifies (freezes) at its melting point. Use the equation: thermal energy for a change of state = mass × specific latent heat
79 600 J
Sample questions
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy4 marks
(a) What is meant by the specific latent heat of a substance? Tick (✓) one box.[1]
The energy needed to change the state of 1 kg of the substance with no change in temperature
The energy needed to raise the temperature of 1 kg of the substance by 1 °C
The temperature at which the substance changes state
The total energy of all the particles in the substance
(b) What is the name of the specific latent heat for a change of state from liquid to vapour? Tick (✓) one box.[1]
Specific latent heat of condensation
Specific latent heat of fusion
Specific latent heat of vaporisation
Specific heat capacity
(c) Calculate the energy needed to melt 0.25 kg of ice at 0 °C. specific latent heat of fusion of ice = 334 000 J/kg Use the equation: thermal energy for a change of state = mass × specific latent heat[2]
Show the answer and mark scheme
(a)Answer: The energy needed to change the state of 1 kg of the substance with no change in temperature
(b)Answer: Specific latent heat of vaporisation
(c)Answer: 83 500 J
E = 0.25 × 334 000
83 500 (J)
Question 2Medium6 marks
Lead melts at 327 °C. 9200 J of energy is needed to melt a 0.40 kg block of lead that is already at 327 °C.
(a) Calculate the specific latent heat of fusion of lead. Use the Physics Equations Sheet.[2]
(b) Explain the difference between specific heat capacity and specific latent heat.[2]
(c) Before it melted, the block was heated from 20 °C to 327 °C. specific heat capacity of lead = 130 J/kg °C Calculate the energy needed to heat the block from 20 °C to 327 °C. Use the Physics Equations Sheet.[2]
Show the answer and mark scheme
(a)Answer: 23 000 J/kg
9200 = 0.40 × L
L = 23 000 (J/kg)
(b)Answer: Specific heat capacity is the energy needed to raise the temperature of 1 kg by 1 °C; specific latent heat is the energy needed to change the state of 1 kg without changing its temperature.
specific heat capacity is the energy needed to raise the temperature of 1 kg (of a substance) by 1 °C
specific latent heat is the energy needed to change the state of 1 kg (of a substance) with no change in temperature
(c)Answer: 16 000 J (15 964 J) J
ΔE = 0.40 × 130 × 307
15 964 (J) / 16 000 (J)
Question 3Hard8 marks
A student wants to determine the specific latent heat of fusion of ice. The student has:
two funnels, each above an empty beaker
crushed ice at 0 °C
an electric immersion heater connected to a joulemeter
a stopwatch
a top-pan balance
(a) Describe a method the student could use to determine the specific latent heat of fusion of ice. Your answer should include how the student would use the results.[6]
(b) Explain why the method uses a second funnel of ice that has no heater in it.[2]
Show the answer and mark scheme
(a)Answer: Two funnels of crushed ice, one with the heater. Collect meltwater from both for the same time while the heater runs; E from the joulemeter; m = heated mass − control mass; L = E ÷ m.
fill both funnels with crushed ice; put the immersion heater into the ice in one funnel, surrounded by ice; the other funnel has no heater (control)
wait until water drips from both funnels at a steady rate, then put empty (weighed) beakers under the funnels
switch on the heater, start the stopwatch and record the joulemeter reading
after a set time, e.g. 5 minutes, switch off the heater, remove the beakers and record the new joulemeter reading
energy transferred E = difference in joulemeter readings
measure the mass of water collected in each beaker using the balance
mass melted by the heater m = mass from the heated funnel − mass from the control funnel
calculate L = E ÷ m (in J/kg, with m in kg); repeat and calculate a mean
Marked with levels of response: the full level descriptors are in the app.
(b)Answer: Energy from the room melts some ice in both funnels; subtracting the control mass leaves just the mass melted by the heater.
some ice melts in both funnels because of energy transferred from the surroundings / the warm room
the mass melted in the funnel without the heater is subtracted, so that only the mass melted by the energy from the heater is used