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You have now met every idea in Topic P1: the particle model and the development of the atomic model, density, states of matter and changes of state, internal energy and specific heat capacity, specific latent heat, and particle motion and gas pressure. This final lesson pulls the topic together — gathering the key equations and rules in one place, working through multi-step problems that draw on several ideas at once, and warning you about the mistakes that most often cost marks in this topic. The aim is to help you move confidently between a density calculation, a heating-curve explanation, and an E=mcΔθ or E=mL calculation. This lesson is part of Topic P1 (Matter) of OCR Gateway Combined Science A.
By the end of this lesson you should be able to recall and select the P1 equations, carry out multi-step calculations that combine specific heat capacity and specific latent heat, explain the physics behind heating curves and gas pressure, recognise and avoid common P1 exam mistakes, and structure a longer answer.
This lesson exercises all three objectives together: AO1 to recall and select the right P1 equation, AO2 to carry out the multi-step E=mcΔθ and E=mL calculations and particle-model explanations, and AO3 to analyse heating curves and evaluate the reasoning in an extended answer.
Most of Topic P1 rests on three equations and a small set of particle-model ideas. Knowing exactly which equation to use, and when, is half the battle.
| Idea | The rule to remember |
|---|---|
| Density | ρ=Vm; unit kg/m3 or g/cm3; 1 g/cm3=1000 kg/m3 |
| Specific heat capacity | E=mcΔθ; c = energy to raise 1 kg by 1°C; changes the temperature |
| Specific latent heat | E=mL (no Δθ); L = energy to change the state of 1 kg; changes the state |
| States of matter | Solid (close, regular, vibrating); liquid (close, random, sliding); gas (far apart, fast) |
| Change of state | Physical, reversible; same particles rearranged; mass conserved |
| Internal energy | Total kinetic + potential energy of the particles |
| Gas pressure | Particles collide with the walls; hotter → faster → harder + more often → higher pressure |
| Kelvin scale | K=°C+273; absolute zero =−273°C=0 K |
| Atomic model | Dalton → Thomson (plum pudding) → Rutherford (nuclear) → Bohr (energy levels); atom ≈ 10−10 m |
Exam Tip: The single most useful skill in P1 is choosing between E=mcΔθ and E=mL. If the temperature changes, use E=mcΔθ; if the substance is changing state (melting, boiling) at constant temperature, use E=mL. A multi-step "heat the ice, melt it, heat the water, boil it" problem uses them in turn.
The commonest higher-mark P1 calculation asks for the total energy to take a substance through a temperature rise and a change of state. You handle it in stages, using E=mcΔθ for each temperature change and E=mL for each change of state, then add the energies together.
flowchart TD
A["Does the temperature change?"] -->|Yes| B["Use E = mcΔθ"]
A -->|"No — changing state<br/>(melting / boiling)"| C["Use E = mL"]
B --> D["Add the energies for<br/>each stage together"]
C --> D
How much energy is needed to take 0.5 kg of ice at −10°C and turn it all into water at 0°C? (Specific heat capacity of ice =2100 J/kg°C; specific latent heat of fusion of water =3.34×105 J/kg.)
Stage 1 — warm the ice from −10°C to 0°C (temperature change, so E=mcΔθ):
Δθ=0−(−10)=10°C E1=mcΔθ=0.5×2100×10=10500 J
Stage 2 — melt the ice at 0°C (change of state, so E=mL):
E2=mL=0.5×3.34×105=1.67×105 J=167000 J
Add the stages:
Etotal=10500+167000=177500 J
Answer: 177500 J (about 1.78×105 J). Notice that far more energy goes into the melting stage than the warming stage — the latent heat of fusion is large.
A rectangular gold bar measures 10 cm×5 cm×2 cm. Gold has a density of 19.3 g/cm3. Calculate the mass of the bar.
Step 1 — find the volume: V=10×5×2=100 cm3.
Step 2 — rearrange the density equation for mass: m=ρV.
Step 3 — substitute and calculate: m=19.3×100=1930 g.
Answer: the bar has a mass of 1930 g (which is 1.93 kg).
A kettle contains 0.4 kg of water at 80°C. Calculate the total energy needed to bring it to 100°C and then boil it all away. (Specific heat capacity of water =4200 J/kg°C; specific latent heat of vaporisation of water =2.26×106 J/kg.)
Stage 1 — warm the water from 80°C to 100°C (temperature change, so E=mcΔθ):
Δθ=100−80=20°C E1=mcΔθ=0.4×4200×20=33600 J
Stage 2 — boil the water at 100°C (change of state, so E=mL):
E2=mL=0.4×2.26×106=9.04×105 J=904000 J
Add the stages:
Etotal=33600+904000=937600 J
Answer: 937600 J (about 9.38×105 J). This shows starkly why a kettle takes far longer to boil dry than to reach 100°C — the boiling stage needs almost thirty times the energy of the final warming stage, because the latent heat of vaporisation is so large.
Exam Tip: In a multi-stage energy problem, label each stage and show its own equation, substitution and answer, then add. Never try to combine E=mcΔθ and E=mL into one line — keep them separate so the examiner can award each stage.
A heating curve has sloping sections (one state warming, temperature rising, E=mcΔθ at work) and flat plateaus (a change of state at constant temperature, E=mL at work). The plateau is flat because the energy is breaking the forces between the particles (raising their potential energy) rather than making them move faster (kinetic energy), so the temperature does not change.
On a heating curve for a pure substance, the first flat plateau is at 80°C and the second at 210°C. State the melting point and boiling point, and say what is happening to the particles during the second plateau.
Step 1 — the first plateau (lower temperature) is where the solid melts, so the melting point is 80°C.
Step 2 — the second plateau (higher temperature) is where the liquid boils, so the boiling point is 210°C.
Step 3 — during the second plateau the substance is boiling: the energy supplied is separating the particles completely (overcoming the forces between them), increasing their potential energy, so the temperature stays constant.
Answer: melting point 80°C, boiling point 210°C; during the second plateau the liquid is boiling as the particles are separated completely, with no temperature rise.
Exam Tip: On a curve, the lower plateau is always melting and the higher plateau is always boiling. If asked why a plateau is flat, always give the particle reason (energy breaks forces / raises potential energy, not kinetic energy).
Almost every explanation in P1 comes back to the arrangement, spacing and motion of the particles. It is worth seeing how one idea threads through the whole topic:
Exam Tip: When an "explain" question feels hard, fall back on the particles: describe their spacing, their motion and their energy, and link each to the property being asked about. This single habit answers the majority of P1 extended-response questions.
The extended-response questions in P1 (worth around six marks) reward answers that are organised, complete and use the right vocabulary. A few habits make the difference between a mid-band and a top-band response.
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