Matter: Synthesis and Exam Skills

You have now met every idea in Topic P1: density, the particle model, changes of state, internal energy, specific heat capacity, specific latent heat, gas pressure and temperature, and pressure in fluids. This final lesson pulls the topic together — collecting every equation in one place, recapping the two required practicals, working through a multi-step problem that draws on several ideas at once, and warning you about the mistakes that most often cost marks. The aim is to help you move confidently between the parts of P1 and to apply the right equation to an unfamiliar problem. This lesson is part of Topic P1 (Matter) of OCR Gateway Science A.

By the end of this lesson you should be able to select and use every P1 equation, recall the key required practicals, recognise and avoid common P1 exam mistakes, and work through a multi-step matter problem.

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Every P1 Equation in One Place

The whole of Topic P1 rests on a small set of equations. Knowing what each symbol means and which equation a question needs is half the battle.

Equation	What it finds	Symbols and units
$\rho = \dfrac{m}{V}$ρ=Vm​	Density	$\rho$ρ in $\text{kg/m}^3$kg/m3 (or $\text{g/cm}^3$g/cm3), $m$m mass, $V$V volume
$E = mc\Delta\theta$E=mcΔθ	Energy to change temperature	$E$E in J, $m$m in kg, $c$c in $\text{J/kg}\,°\text{C}$J/kg°C, $\Delta\theta$Δθ temperature change
$E = mL$E=mL	Energy to change state	$E$E in J, $m$m in kg, $L$L in $\text{J/kg}$J/kg
$p = h\rho g$p=hρg (Higher)	Pressure due to a liquid column	$p$p in Pa, $h$h in m, $\rho$ρ in $\text{kg/m}^3$kg/m3, $g = 9.8 \text{ N/kg}$g=9.8 N/kg
$\text{K} = °\text{C} + 273$K=°C+273	Convert to the kelvin scale	absolute zero $= -273\,°\text{C} = 0 \text{ K}$=−273°C=0 K

There is also the qualitative pressure–temperature rule for a fixed volume of gas (hotter → higher pressure) and, for Higher tier, the inverse pressure–volume rule ($pV = \text{constant}$pV=constant at constant temperature).

Exam Tip: The first job in any calculation is to choose the right equation by spotting the clue word: "density" → $\rho = \dfrac{m}{V}$ρ=Vm​; "temperature rise" → $E = mc\Delta\theta$E=mcΔθ; "melt/boil/change state" → $E = mL$E=mL; "depth in a liquid" → $p = h\rho g$p=hρg. Then write the equation, substitute, and give the answer with a unit.

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Telling $E = mc\Delta\theta$E=mcΔθ and $E = mL$E=mL Apart

The single most common source of confusion in P1 is choosing between the two energy equations. The rule is simple once you spot it:

• If the temperature is changing (a substance is warming up or cooling down within one state), use $E = mc\Delta\theta$E=mcΔθ — it has a $\Delta\theta$Δθ because the temperature changes.
• If the state is changing (melting, boiling, freezing or condensing) at constant temperature, use $E = mL$E=mL — it has no $\Delta\theta$Δθ because the temperature does not change.

A heating curve makes this visible: the sloping sections (temperature changing) use $E = mc\Delta\theta$E=mcΔθ, and the flat plateaus (state changing) use $E = mL$E=mL. To work out the total energy to take ice below freezing all the way to steam, you would add up several stages, using $mc\Delta\theta$mcΔθ for each slope and $mL$mL for each plateau.

Exam Tip: Ask yourself: "Is the temperature changing, or the state changing?" Temperature changing → $E = mc\Delta\theta$E=mcΔθ. State changing → $E = mL$E=mL. Getting this choice right is worth several marks across the paper.

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Required-Practical Recap

Two required practicals run through P1. Be ready to describe the apparatus, the measurements and the calculation for each.

Density of solids and liquids.

• Regular solid: measure mass (balance) and volume from dimensions ($V = l \times w \times h$V=l×w×h for a cuboid); then $\rho = \dfrac{m}{V}$ρ=Vm​.
• Irregular solid: measure mass (balance) and volume by displacement (eureka can or rise in a measuring cylinder); then $\rho = \dfrac{m}{V}$ρ=Vm​.
• Liquid: zero the balance with the empty measuring cylinder, add a known volume, read the mass; then $\rho = \dfrac{m}{V}$ρ=Vm​.

Specific heat capacity.

• Insulate a metal block of known mass; insert an immersion heater and a thermometer.
• Supply a measured energy (joulemeter, or $VIt$VIt) and record the temperature rise $\Delta\theta$Δθ.
• Calculate $c = \dfrac{E}{m\Delta\theta}$c=mΔθE​. The main error is heat loss to the surroundings, making $c$c come out too high; reduce it with insulation.

Exam Tip: For the density of a liquid, the marking point is to subtract the mass of the empty container (or zero the balance with it on). For specific heat capacity, it is to say the value is too high because of heat loss and to suggest insulation.

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A Multi-Step Worked Problem

Exam questions often combine several P1 ideas. Working through one carefully shows how the equations connect.

Worked Example — Heating water in a kettle

A kettle contains $0.8 \text{ kg}$0.8 kg of water at $20\,°\text{C}$20°C. (a) Calculate the energy needed to heat the water to its boiling point, $100\,°\text{C}$100°C. (b) The kettle continues to supply energy and boils away $0.1 \text{ kg}$0.1 kg of the water. Calculate the energy needed to do this. (c) State the total energy supplied. (Specific heat capacity of water $= 4200 \text{ J/kg}\,°\text{C}$=4200 J/kg°C; specific latent heat of vaporisation of water $= 2.26 \times 10^{6} \text{ J/kg}$=2.26×106 J/kg.)

(a) Heating the water (temperature changing → $E = mc\Delta\theta$E=mcΔθ):

Step 1 — find $\Delta\theta$Δθ: $100 - 20 = 80\,°\text{C}$100−20=80°C.

Step 2 — substitute: $E = mc\Delta\theta = 0.8 \times 4200 \times 80$E=mcΔθ=0.8×4200×80.

Step 3 — calculate: $E = 268\,800 \text{ J}$E=268800 J.

(b) Boiling some water away (state changing → $E = mL$E=mL):

Step 1 — use the mass that boils away, $0.1 \text{ kg}$0.1 kg: $E = mL = 0.1 \times 2.26 \times 10^{6}$E=mL=0.1×2.26×106.

Step 2 — calculate: $E = 226\,000 \text{ J}$E=226000 J.

(c) Total energy supplied:

$268\,800 + 226\,000 = 494\,800 \text{ J}$268800+226000=494800 J

Answer: (a) $268\,800 \text{ J}$268800 J; (b) $226\,000 \text{ J}$226000 J; (c) about $4.95 \times 10^{5} \text{ J}$4.95×105 J in total. Notice that boiling away just $0.1 \text{ kg}$0.1 kg takes nearly as much energy as heating the whole $0.8 \text{ kg}$0.8 kg from $20\,°\text{C}$20°C to $100\,°\text{C}$100°C — a direct consequence of the large latent heat of vaporisation.

Worked Example — Density and a floating check

A wooden block measures $0.1 \text{ m} \times 0.1 \text{ m} \times 0.05 \text{ m}$0.1 m×0.1 m×0.05 m and has a mass of $0.3 \text{ kg}$0.3 kg. (a) Calculate its density. (b) Will it float on water (density $1000 \text{ kg/m}^3$1000 kg/m3)?

Step 1 — volume: $V = 0.1 \times 0.1 \times 0.05 = 5 \times 10^{-4} \text{ m}^3$V=0.1×0.1×0.05=5×10−4 m3.

Step 2 — density: $\rho = \dfrac{m}{V} = \dfrac{0.3}{5 \times 10^{-4}} = 600 \text{ kg/m}^3$ρ=Vm​=5×10−40.3​=600 kg/m3.

Step 3 — compare with water: $600 \text{ kg/m}^3$600 kg/m3 is less than $1000 \text{ kg/m}^3$1000 kg/m3.

Answer: (a) $600 \text{ kg/m}^3$600 kg/m3; (b) it floats, because its density is less than that of water.

Worked Example — Pressure on a diver (Higher)

A diver descends to a depth of $25 \text{ m}$25 m in seawater of density $1030 \text{ kg/m}^3$1030 kg/m3. (a) Calculate the pressure on the diver due to the seawater. (b) The atmospheric pressure at the surface is $1.0 \times 10^{5} \text{ Pa}$1.0×105 Pa. State the total pressure on the diver. ($g = 9.8 \text{ N/kg}$g=9.8 N/kg.)

Step 1 — choose the equation for pressure due to a liquid column: $p = h\rho g$p=hρg.

Step 2 — substitute: $p = 25 \times 1030 \times 9.8$p=25×1030×9.8.

Step 3 — calculate: $p = 252\,350 \text{ Pa} \approx 2.52 \times 10^{5} \text{ Pa}$p=252350 Pa≈2.52×105 Pa.

Step 4 — add the atmospheric pressure pushing down on the surface: total $= 252\,350 + 100\,000 = 352\,350 \text{ Pa}$=252350+100000=352350 Pa.