Changes of state — OCR GCSE Physics
Test yourself on Changes of state with OCR GCSE practice questions.
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Changes of state explained
The equation ΔE = m × c × Δθ gives the thermal energy needed to raise a mass by a temperature change without a change of state.
Read the full explanation
Specific heat capacity c is the energy to raise 1 kg by 1 °C, in J/kg °C, so units cancel to leave joules. Worked method: heat 2.0 kg of water (c = 4200 J/kg °C) through 15 °C, giving ΔE = 2.0 × 4200 × 15 = 126 000 J, or 1.26 × 10⁵ J. Use the change in temperature, not the final temperature, and keep mass in kilograms. A larger c means more energy for the same rise, which is why water heats slowly. The equation applies while the substance stays in one state; melting or boiling needs latent heat instead.
apply: thermal energy for a change in state (J) = mass (kg) × specific latent heat (J/kg)
This equation gives the energy needed to change the state of a substance at constant temperature. Multiply the mass in kilograms by the specific latent heat in joules per kilogram, and the answer is in joules. For example, melting 0.50 kg of ice uses a specific latent heat of fusion of about 3.3 × 10⁵ J/kg, so the energy is 0.50 × 3.3 × 10⁵ = 1.65 × 10⁵ J. Use the latent heat of fusion for melting or freezing and the latent heat of vaporisation for boiling or condensing. Check units before substituting: convert grams to kilograms by dividing by 1000, and remember the temperature stays constant during the change of state.
describe how mass is conserved when substances melt, freeze, evaporate, condense or sublimate
A change of state rearranges particles but does not create or destroy them, so the total mass stays the same. Melting turns a solid into a liquid, freezing turns a liquid into a solid, evaporating turns a liquid into a gas, condensing turns a gas into a liquid, and sublimating turns a solid directly into a gas. In each case the same particles are present before and after, only their arrangement and energy change. For example, 100 g of ice melts to give 100 g of water, and the water vapour formed when it evaporates still has a mass of 100 g. In a sealed container the mass reading does not change during any of these processes.
describe that physical changes differ from chemical changes because the material recovers its original properties if the change is reversed
A physical change alters the form or arrangement of a substance without making a new substance, so reversing it restores the material's original properties. Melting ice is the standard example: solid ice becomes liquid water, and freezing that water returns ice with the same melting point, density and appearance as before. A chemical change produces a new substance with different properties, and reversing it is generally not possible by simple physical means. In a multiple-choice question, look for the option that names reversibility and recovery of original properties as the distinguishing feature, and reject options that describe only a change of appearance or a change of state without mentioning recovery.
describe how heating a system will change the energy stored within the system and raise its temperature or produce changes of state
Heating transfers energy to a system, increasing the energy stored in its thermal store. That extra energy can do one of two things. It can raise the temperature, because the particles gain kinetic energy and move or vibrate faster. Alternatively, during a change of state such as melting or boiling, the energy breaks the bonds or overcomes the forces between particles rather than raising the temperature, so the temperature stays constant while the change proceeds. A multiple-choice question may ask which statement correctly links heating to stored energy and to temperature or state change; choose the option that names energy transfer into the system and either a temperature rise or a change of state, and reject options claiming that heating always raises temperature.
define the term specific heat capacity and distinguish between it and the term specific latent heat (to include: specific latent heat of fusion and specific latent heat of vaporisation)
Specific heat capacity is the energy needed to raise the temperature of 1 kg of a substance by 1 °C (equivalently 1 K), with no change of state. For example, water needs about 4200 J to warm 1 kg by 1 °C, so its value is roughly 4200 J/(kg °C). Specific latent heat is different: it is the energy needed to change the state of 1 kg of a substance with no change in temperature. Melting or freezing uses the specific latent heat of fusion; boiling or condensing uses the specific latent heat of vaporisation. So a temperature change signals specific heat capacity, while a state change at constant temperature signals specific latent heat.
apply the relationship between change in internal energy of a material and its mass, specific heat capacity and temperature change to calculate the energy change involved
When a material is heated without changing state, its change in internal energy equals the energy supplied, given by ΔE = m × c × Δθ, where m is mass in kg, c is specific heat capacity in J/(kg °C) and Δθ is the temperature change in °C. For example, heating 2 kg of water (c ≈ 4200 J/(kg °C)) through 10 °C needs ΔE = 2 × 4200 × 10 = 84 000 J, or 84 kJ. Work out the temperature change by subtracting the initial temperature from the final temperature, keep mass in kilograms, and match the units of your answer to the options. The same relationship lets you find c or Δθ by rearranging.
apply the relationship between specific latent heat and mass to calculate the energy change involved in a change of state
Specific latent heat is the energy needed to change the state of 1 kg of a substance without changing its temperature. The relationship is E = m × L, where E is the energy change in joules (J), m is the mass in kilograms (kg) and L is the specific latent heat in joules per kilogram (J/kg). To apply it, identify the mass and the specific latent heat for the change happening (melting, freezing, boiling or condensing), then multiply. For example, melting 0.50 kg of ice with a specific latent heat of fusion of 3.34 × 10⁵ J/kg needs E = 0.50 × 3.34 × 10⁵ = 1.67 × 10⁵ J. Rearranged forms give m = E ÷ L and L = E ÷ m. Keep mass in kg and energy in J before calculating.
Your focus
- Apply ΔE = m × c × Δθ to calculate thermal energy change.
- Use consistent SI units of kg, J/kg °C and °C within the calculation.
- Explain how specific heat capacity affects the energy needed for a given temperature rise.
Show all 24 objectives
- Apply the equation thermal energy = mass × specific latent heat to a change of state
- Convert masses to kilograms and select the correct specific latent heat for fusion or vaporisation
- Calculate and state the thermal energy for a change of state in joules
- Describe how mass is conserved across melting, freezing, evaporation, condensation and sublimation
- Explain conservation of mass in terms of the same particles being rearranged
- Apply conservation of mass to a named change of state with a numerical example
- State that a physical change is reversible and restores the original material
- Explain why a chemical change is not usually reversed by physical means
- Apply the reversibility test to a familiar change of state
- Describe how heating increases the energy stored in a system
- Explain how added energy raises temperature or produces a change of state
- Distinguish between a temperature rise and a constant-temperature change of state
- Define specific heat capacity as the energy to raise 1 kg of a substance by 1 °C
- Define specific latent heat as the energy to change the state of 1 kg of a substance at constant temperature
- Distinguish specific heat capacity from specific latent heat, including fusion and vaporisation
- Apply ΔE = m × c × Δθ to calculate an energy change
- Calculate a temperature change from initial and final temperatures
- Substitute mass, specific heat capacity and temperature change with consistent units
- State the relationship E = m × L for a change of state
- Calculate the energy change for a given mass and specific latent heat
- Rearrange the relationship to find mass or specific latent heat
Changes of state exam tips
Marking Points
- Selects ΔE = m × c × Δθ and substitutes mass in kilograms, c in J/kg °C and temperature change in °C.
- Calculates the temperature change by subtracting the initial temperature from the final temperature.
- Computes a thermal energy value in joules and expresses it with a sensible power of ten.
- Explains that a substance with a larger specific heat capacity needs more energy for the same temperature rise.
- Recognises that the equation applies to a temperature change within one state rather than to melting or boiling.
- Selects the equation thermal energy = mass × specific latent heat and substitutes values in the correct units
- Converts a mass given in grams to kilograms by dividing by 1000 before multiplying
- Uses the latent heat of fusion for melting or freezing and the latent heat of vaporisation for boiling or condensing
- States the answer in joules and recognises that temperature remains constant during the change of state
- States that the total mass stays the same during melting, freezing, evaporation, condensation and sublimation
- Explains that the same particles are present before and after a change of state, only rearranged
- Applies conservation of mass to a named change of state such as ice melting to water of equal mass
- Recognises that mass is conserved because no particles are added or lost in a closed system
- Selects the option stating that a physical change is reversible and the material regains its original properties
- Rejects options that describe a chemical change as easily reversed by physical means
- Identifies melting and freezing as an example where the original substance is recovered
- Distinguishes physical change from chemical change by whether a new substance is formed
- States that heating transfers energy into the system and increases the energy stored in its thermal store
- Explains that the added energy raises temperature by increasing the kinetic energy of the particles
- Recognises that during a change of state the energy overcomes forces between particles and the temperature stays constant
- Selects the option that correctly links heating to either a temperature rise or a change of state
- States that specific heat capacity is the energy needed to raise the temperature of 1 kg of a substance by 1 °C without a change of state
- States that specific latent heat is the energy needed to change the state of 1 kg of a substance with no change in temperature
- Links specific latent heat of fusion to melting or freezing and specific latent heat of vaporisation to boiling or condensing
- Distinguishes the two by noting that specific heat capacity involves a temperature change while specific latent heat involves a change of state at constant temperature
- Selects the relationship ΔE = m × c × Δθ linking energy change to mass, specific heat capacity and temperature change
- Substitutes mass in kilograms, specific heat capacity in J/(kg °C) and the temperature change in °C into the relationship
- Calculates the temperature change by subtracting the initial temperature from the final temperature before multiplying
- Computes the energy change with correct arithmetic and consistent units, for example 2 × 4200 × 10 = 84 000 J
- Multiplies the mass in kilograms by the specific latent heat in joules per kilogram to obtain the energy in joules
- Selects the specific latent heat appropriate to the stated change of state rather than a value for a different change
- Converts a mass given in grams to kilograms by dividing by 1000 before substituting into the relationship
- Rearranges the relationship correctly to find mass or specific latent heat when the energy change is given
- Keeps units consistent, expressing energy in joules and specific latent heat in joules per kilogram
Examiner Tips
- 💡Write the equation, then fill in m, c and Δθ in a row before calculating.
- 💡Estimate the answer first, for example 2 × 4000 × 10 ≈ 80 000 J, to catch unit slips.
- 💡Check the sign and size of the temperature change and state the energy in joules with a power of ten.
- 💡Write down the equation, then list mass and specific latent heat with their units before substituting any numbers
- 💡Convert every mass to kilograms and check that the specific latent heat is in J/kg so the answer comes out in joules
- 💡Use the value of specific latent heat given in the question rather than a remembered figure, and keep powers of ten in standard form
- 💡Name the change of state in the question and state that the particles are the same before and after
- 💡Use a simple numerical example such as equal masses of ice and water to show conservation of mass
- 💡Mention that in a closed container the balance reading stays constant to support your answer
- 💡Read every option before choosing and eliminate those that ignore reversibility
- 💡Link the chosen option to a named example such as ice melting and refreezing
- 💡Check that the option mentions recovery of original properties, not just a change of state
- 💡Identify whether the option describes a temperature rise or a change of state and check it matches the physics
- 💡Use the phrase energy stored in the thermal store to justify your choice
- 💡Eliminate options that say temperature rises during melting or boiling
- 💡Read the question stem for the words 'temperature change' or 'change of state' before choosing, because that phrase decides which quantity is meant
- 💡Check that any definition you select includes both '1 kg' and the relevant condition (1 °C rise or change of state)
- 💡Eliminate options that give a total energy rather than an energy per kilogram, since both definitions are per unit mass
- 💡Write down m, c and Δθ separately before multiplying so a slip in one value is easy to spot
- 💡Convert any mass given in grams to kilograms, since c is quoted per kilogram
- 💡Estimate the answer roughly first (for example 2 × 4000 × 10 = 80 000 J) to reject implausible options quickly
- 💡Write down E = m × L and label each known quantity with its unit before substituting numbers
- 💡Check whether the mass is in grams and convert to kilograms first, since the specific latent heat is per kilogram
- 💡Sanity-check the answer: a few tenths of a kilogram typically needs tens or hundreds of thousands of joules to change state
Common Mistakes
- Substitutes the final temperature instead of the change in temperature; subtract the starting value first.
- Leaves mass in grams and multiplies directly; convert grams to kilograms by dividing by 1000.
- Uses the equation to find energy for melting ice at constant temperature; apply latent heat reasoning there instead.
- Mixes units so the answer is not in joules; keep c in J/kg °C with mass in kg and temperature in °C.
- Substitutes a mass in grams directly into the equation; convert grams to kilograms first by dividing by 1000
- Uses the specific latent heat of vaporisation when a substance is melting; select the latent heat of fusion for melting and freezing
- Adds a temperature change to the calculation; recognise that the temperature stays constant during a change of state
- Omits the unit or gives the energy in the wrong unit; state the answer in joules
- Claims that mass is lost when a liquid evaporates; recognise that the particles become a gas and the total mass is unchanged
- Assumes a gas has no mass because it is invisible; state that the gas particles still have mass
- Treats each change of state as a different process with different mass rules; apply the same conservation of mass to melting, freezing, evaporation, condensation and sublimation
- Forgets sublimation when listing changes of state; include solid-to-gas sublimation alongside the other four changes
- Treats any change of appearance as a chemical change; check instead whether a new substance forms and whether the original properties return on reversal
- Assumes all changes of state create new substances; recognise that melting and freezing recover the same material
- Confuses dissolving with a chemical reaction; note that evaporating the solvent can recover the dissolved solid
- Claims that heating always raises the temperature; recognise that during melting or boiling the temperature remains constant
- Says energy is destroyed during a change of state; state instead that it is used to overcome forces between particles
- Ignores the energy store and describes only the visible change; link heating to increased energy stored in the system
- Confuses the two quantities by describing specific heat capacity as the energy for a change of state; correct this by tying specific heat capacity to a 1 °C temperature rise of 1 kg
- Omits the per-kilogram basis and treats the values as totals for any amount; correct this by stating that both quantities are defined per 1 kg
- Attaches the latent heat terms to the wrong changes by swapping fusion and vaporisation; correct this by linking fusion to melting or freezing and vaporisation to boiling or condensing
- Uses the final temperature instead of the temperature change; correct this by subtracting the initial temperature from the final temperature first
- Leaves mass in grams and multiplies directly; correct this by converting grams to kilograms before substituting
- Mixes units so the answer is out by a factor of 1000; correct this by keeping c in J/(kg °C) with mass in kg and stating the answer in joules or kilojoules consistently
- Uses the specific heat capacity instead of the specific latent heat; select the latent heat value because the temperature stays constant during a change of state
- Forgets to convert grams to kilograms; divide the mass by 1000 so it is in kg before multiplying
- Adds the temperature change into the calculation; omit any temperature term because latent heat applies at constant temperature
- Mixes up the latent heat of fusion and the latent heat of vaporisation; match the value to the change of state described