HookThe fake gold bar that fooled a Manhattan dealer
In September 2012, Ibrahim Fadl — a veteran gold dealer in Manhattan's Diamond District — grew suspicious of a 10-ounce gold bar he had bought at full market price. The paperwork was immaculate: a genuine PAMP Suisse stamp, a legitimate serial number, the correct mass to the gram. So he clamped the bar and drilled into it. Out of the holes came curls of dull grey metal: tungsten. Only the shell was gold. The forgery worked because of a freak of nature — tungsten's density, 19.25 g/cm³, sits within half a per cent of gold's 19.32 g/cm³. A bar of the right mass was also, to any measurement a dealer could make by hand, a bar of the right size.
The check the forger defeated — weigh it, find its volume, divide — is the oldest measurement in physics. Vitruvius tells the story of Archimedes running it on King Hiero II's crown in Syracuse around 250 BC, and you will run it yourself on a pebble and a measuring cylinder in Required practical 17. P3 is the topic that explains what density actually is. Matter is made of particles: how tightly they pack sets density, how fast they move sets temperature, and how strongly they bind sets the price — in joules — of melting and boiling. By the end of this section you can put defensible numbers on all three, which is precisely what the exam pays for.
ModelDensity — the particle model's first number
The definition is one line: density = mass ÷ volume, \(\rho = m/V\), in kg/m³ or g/cm³ — and multiplying g/cm³ by 1,000 converts to kg/m³, because a cubic metre holds a million cubic centimetres while a kilogram holds only a thousand grams. Density is the particle model's first prediction. In a solid, particles pack touching, in a regular arrangement, vibrating about fixed positions. In a liquid they still touch but slide past one another — which is why melting barely changes density. In a gas the same particles are separated by around ten times their own diameter in every direction, so the same mass fills roughly a thousand times the volume: air manages about 1.2 kg/m³ against water's 1,000 kg/m³.
Density belongs to the substance, not the object. A gold ring and a gold ingot both come out at 19.32 g/cm³ — which is exactly why the measurement can identify an unknown material, and why the tungsten bar was such an elegant crime.
Run the forger's arithmetic. Ten troy ounces is 311 g. As pure gold the bar's volume must be \(V = m/\rho = 311 \div 19.32 = 16.10\ \text{cm}^3\). Cast in tungsten instead: \(311 \div 19.25 = 16.16\ \text{cm}^3\). The difference is 0.06 cm³ — about one drop of water, spread over an entire bar, far below anything callipers will catch. Now try the same fraud in lead: \(311 \div 11.3 = 27.5\ \text{cm}^3\), a bar 71% too big for its mass that any dealer would clock the moment it hit the pan. That is why fakers pay for tungsten and why bullion houses now test with ultrasound instead of scales. Finish in SI units: \(19.32\ \text{g/cm}^3 \times 1000 = 19{,}320\ \text{kg/m}^3\) — always show that conversion, because mixing g/cm³ with kg/m³ is the most common density error on the paper.
DataRequired practical 17 — three densities, one balance
RP17 asks for three measurements. Regular solid: measure the sides with a rule or vernier callipers, multiply for volume, find mass on a top-pan balance, divide. Irregular solid: geometry fails, so displace instead — lower the object on a thread into a part-filled measuring cylinder and take the rise in the reading as its volume, or use a eureka can and catch the overflow. Liquid: stand an empty measuring cylinder on the balance, zero (tare) it, pour in the liquid, and read mass and volume directly. In every case the independent quantity is the object or liquid chosen, and mass and volume are the measured variables feeding one calculated result.
The marks live in the error talk. Read the cylinder at eye level, at the bottom of the meniscus — reading from above is a parallax error that scatters results (random error). Forgetting to zero the balance, or letting water cling to the pebble and thread between measurements, shifts every reading the same way (systematic error). A school cylinder resolves ±0.5 cm³ at best, so a tiny object carries a huge percentage uncertainty — use the largest sample you can, repeat, discard anomalies, take a mean.
A pebble reads 84.0 g on the balance. The cylinder holds 50.0 cm³ of water before and 80.0 cm³ with the pebble submerged, so \(V = 80.0 - 50.0 = 30.0\ \text{cm}^3\) and \[\rho = \frac{m}{V} = \frac{84.0}{30.0} = 2.8\ \text{g/cm}^3 = 2{,}800\ \text{kg/m}^3.\] That lands exactly where rock should — granite is about 2.7 g/cm³ — so the answer passes the sanity check. And if the pebble had trapped air bubbles under its surface, the volume reading would be inflated and the calculated density dragged down. Naming an error and its direction is what separates top-band practical answers from the rest.
ModelChanges of state — same particles, new arrangement
Melting, freezing, boiling, evaporating, condensing, sublimating — all six are physical changes: the particles themselves are untouched; only their arrangement and energy change. Freeze the water back and you recover exactly the substance you started with, which is what separates a change of state from a chemical reaction. And because particles are rearranged but never created or destroyed, mass is conserved: 50 g of ice melts to precisely 50 g of water. A boiling kettle seems to lose mass only because the steam escapes the room's accounting — seal the system and the kilograms balance.
The spec also asks you to criticise the model that draws these pictures. The particle model uses hard, identical, inelastic spheres with nothing between them. Real particles are not solid spheres; they attract one another (otherwise nothing would ever condense); the spacing in diagrams is nowhere near to scale; and atoms of different substances are far from identical. Two limitations, stated crisply, is a standard 2-mark ask that catches out students who have only ever trusted the diagram.
ModelInternal energy — two stores in one total
Heat a system and where does the energy go? Into internal energy: the total kinetic energy plus potential energy of every particle in the system. That word 'total' hides a fork in the road, and the whole topic turns on it. If the supplied energy raises the particles' kinetic energy, they move faster and the temperature rises. If it raises their potential energy, it is prising particles apart against their mutual attractions — and the state changes while the temperature holds still.
Every flat section on a heating graph is the second branch in action: the thermometer parks at 0 °C while ice melts, and at 100 °C while water boils, not because energy stopped arriving but because it is being banked as potential energy rather than speed. Write the explanation in exactly those store terms — 'energy increases the potential energy of the particles, not their kinetic energy' — and the marks are yours.
MechanismSpecific heat capacity — the price of a degree
How much energy does a temperature change cost? \(\Delta E = mc\Delta\theta\): energy equals mass × specific heat capacity × temperature change. The specific heat capacity, c, is the price list — the energy needed to raise 1 kg of a substance by 1 °C. Water's is enormous: 4,200 J/kg°C, against roughly 900 for aluminium and about 450 for steel. That one number explains why coastal Britain stays milder than inland Europe in winter (the sea is a colossal thermal reservoir, slow to heat and slow to cool), why storage heaters bank cheap overnight electricity in dense blocks, and why engines are cooled with water jackets rather than air alone.
You first met the equation in P1 alongside Required practical 14; P3's job is to re-read it through the particle model. A high c simply means it takes a great deal of energy per kilogram to make the particles move faster by one degree's worth.
A 3 kW kettle holds 1.5 kg of water at 18 °C. Energy to reach the boil: \[\Delta E = mc\Delta\theta = 1.5 \times 4200 \times (100 - 18) = 1.5 \times 4200 \times 82 = 516{,}600\ \text{J}.\] At 3,000 joules per second that takes \(t = E \div P = 516{,}600 \div 3000 \approx 172\ \text{s}\) — a shade under three minutes, which is exactly what your kitchen kettle does. Two habits to copy: Δθ is the temperature change (82 °C here, never 100), and the cross-check against lived experience — 'does three minutes sound like a kettle?' — is free AO3 credit.
MechanismLatent heat — energy with no temperature rise
At the plateaus a different equation takes over: \(E = mL\), where L is the specific latent heat — the energy needed to change the state of 1 kg of a substance with no change in temperature. Water has two values. The latent heat of fusion (melting and freezing) is 334,000 J/kg; the latent heat of vaporisation (boiling and condensing) is 2,260,000 J/kg — nearly seven times larger, because boiling must separate the particles almost completely while melting merely loosens the structure.
The asymmetry runs both ways: freezing and condensing release exactly the same energy. Fruit growers exploit this on frost nights by spraying blossom with water — every kilogram that freezes hands back 334,000 J and holds the flower at 0 °C instead of a lethal few degrees below. And it is why steam at 100 °C scalds catastrophically worse than water at 100 °C: on your skin the steam first condenses, donating 2.26 million joules per kilogram before it has cooled by a single degree.
Take 0.50 kg of ice at 0 °C all the way to steam at 100 °C — three legs, two equations. Melt it: \(E = mL = 0.50 \times 334{,}000 = 167{,}000\ \text{J}\). Heat the water to the boil: \(E = mc\Delta\theta = 0.50 \times 4200 \times 100 = 210{,}000\ \text{J}\). Boil it all away: \(E = mL = 0.50 \times 2{,}260{,}000 = 1{,}130{,}000\ \text{J}\). The final leg costs three times the other two combined — which is why the boiling plateau is by far the longest stretch of a heating graph, and why a forgotten pan takes so long to boil dry. Total bill: roughly 1.5 MJ for half a kilogram of water.
MechanismGases — temperature, kinetic energy and pressure
Gas particles are in constant random motion, and the temperature of a gas is a measure of their average kinetic energy: hotter means faster, on average. Pressure is what their collisions do. Every impact on the container wall delivers a tiny outward force, and billions of impacts per second across every square centimetre add up to a steady force per unit area at right angles to the wall.
Hold the volume fixed and raise the temperature, and both collision factors climb at once: the particles strike the walls more often (they cross the container faster) and harder (each impact carries more energy). So the pressure rises. That is why tyre pressures are specified 'cold' — a motorway run warms the air inside and the gauge reading climbs; why a football left in a cold garage feels soft and leaden by morning; and why every aerosol can warns against leaving it in sunlight — the sealed volume cannot grow, so the pressure does, until the seams give way. The exam wants that two-part collision argument, frequency and energy of impacts, not the vague (and wrong) claim that 'the particles expand'.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Choose the equation with one question: is the temperature changing? Changing temperature → ΔE = mcΔθ. Changing state at constant temperature → E = mL. A journey that does both — ice to steam — is separate calculations added together, never one formula stretched across a plateau. Note where the memory work really lies: both of those equations are printed on the Physics Equations Sheet you get in the exam, but ρ = m/V is not — it must be recalled, along with what every symbol means and its SI unit.
Units are this topic's minefield, and AQA's mark schemes are unforgiving about them. Convert grams to kilograms and cm³ to m³ before substituting (1 m³ = 1,000,000 cm³), or work consistently in g/cm³ and convert the answer (×1,000 to reach kg/m³). Always write the substitution line — around 30% of the physics marks are mathematical, and method marks survive an arithmetic slip only if the examiner can see the method. On heating and cooling graphs, label each plateau with its state change and explain it in store language: energy is increasing the particles' potential energy while kinetic energy — and so temperature — stays constant. That phrasing is lifted from what level-3 answers look like, and 'the particles get hotter' is not it.
For Required practical 17, rehearse the six-marker as a sequence with accuracy points attached: zero the balance, measure the largest practical sample, read the meniscus at eye level, repeat and mean, and state the direction any named error pushes the result (trapped bubbles inflate volume, so density reads low). Two boundary notes for Trilogy students: pressure–volume calculations and the Kelvin temperature scale belong to the separate physics course, so your job here ends at the qualitative collision argument; and the AO balance (roughly 40% recall, 40% application, 20% analysis) means most P3 marks come from applying these few equations cleanly, not from exotic content.