物质粒子模型

• $\rho$ density 密度 in kg/m³; $m$ mass in kg; $V$ volume in m³.
• Use consistent units. To express a result in kg/m³, convert g/cm³. $1\ \text{g/cm}^3 = 1000\ \text{kg/m}^3$ (multiply by 1000: a cm³ is a millionth of a m³ and a gram is a thousandth of a kg).

The particle model explains the states of matter:

State	Arrangement	Motion
solid	close, regular	vibrate about fixed positions
liquid	close, irregular	move past each other
gas	far apart	random; straight paths between collisions

• Solids and liquids have similar densities because their particles are similarly packed; a gas is mostly empty space.
• Ice is unusual: water expands on freezing, so ice is slightly less dense than water.

Worked example. A ring has mass 9.46 g and volume 0.44 cm³. Find its density in kg/m³.

• Convert both first: $m = 9.46\ \text{g} = 9.46\times10^{-3}\ \text{kg}$; $V = 0.44\ \text{cm}^3 = 4.4\times10^{-7}\ \text{m}^3$.
  $$\rho = \frac{m}{V} = \frac{9.46\times 10^{-3}\ \text{kg}}{4.4\times 10^{-7}\ \text{m}^3} = 21\,500\ \text{kg/m}^3$$

Actual exam demands: density

AQA June 2025 8463/1H Q01.3: 824000 kg of seawater passes a turbine each second; density is 1030 kg/m³. Choose $\rho=m/V$, then rearrange:

This is the volume passing in each second. AQA June 2024 Q07.5 reverses the ring example: given density 21500 kg/m³ and volume 0.44 cm³, calculate mass. Convert the volume, then use $m=\rho V=21500\ \mathrm{kg/m^3}\times4.4\times10^{-7}\ \mathrm{m^3}=0.00946\ \mathrm{kg}$.

Teacher-written liquid example. The empty cylinder is 42 g; cylinder plus 60 cm³ of liquid is 90 g. Subtract $m=90\ \mathrm{g}-42\ \mathrm{g}=48\ \mathrm{g}$, then $\rho=m/V=48\ \mathrm{g}/60\ \mathrm{cm^3}=0.80\ \mathrm{g/cm^3}$.

Required practical 5: density

• Regular solid: measure length, width and thickness with a ruler (or micrometer/Vernier callipers), multiply for $V$; find $m$ on a balance; $\rho = m/V$.
• Irregular solid: fill a displacement (eureka) can to the spout, wait for dripping to stop, lower the object in on thin string; the volume of water collected in a measuring cylinder equals the object's volume.
• Liquid: find the mass of an empty measuring cylinder, then the mass with a known volume inside; subtract for $m$.
• Accuracy points: read the measuring cylinder at eye level on a flat surface (avoid parallax); use thin string so it displaces almost no water; repeat and average.

AQA June 2022 8463/1H Q02.1–02.4: describe a complete rock-density method, then interpret $2.55\pm0.10$g/cm³ as the interval 2.45–2.65 g/cm³. Repeated readings allow a mean and reduce random-error effects; they do not remove a systematic calibration error. In a cylinder-displacement method, subtract initial volume from final volume, fully submerge the rock and avoid trapped bubbles.

When a substance melts, freezes, boils, evaporates, condenses or sublimates 升华:

• Mass is conserved in a closed system: the number of particles does not change. If vapour leaves an open container, the remaining material loses mass, but the total including the escaped vapour is conserved.
• Changes of state are physical changes 物理变化: reverse the change and the material recovers its original properties. (A chemical change makes new substances; melting does not.)

Internal energy 内能 is the total kinetic and potential energy of all the particles that make up a system. Heating a system increases the particles' energy, and that energy goes one of two ways:

1. it raises the temperature — the particles' kinetic energy grows;
2. it produces melting or boiling — the particles' potential energy increases as their arrangement changes. For a pure substance changing state at constant pressure, temperature stays constant. During freezing or condensation, energy is released and potential energy decreases.

While the temperature changes, the energy needed follows (also met in topic 1):

Specific heat capacity 比热容 $c$ (J/kg °C) is the energy needed to raise the temperature of one kilogram by one degree Celsius.

Worked example. 0.030 kg of olive oil ($c = 1800$ J/kg °C) warms from 21 °C to 96 °C.

• Temperature change first: $\Delta\theta = 96 - 21 = 75$ °C.
  $$\Delta E = mc\Delta\theta = 0.030 \times 1800 \times 75 = 4050\ \text{J}$$

Teacher-written heating-pad example. A 0.20 kg pad with $c=900\ \mathrm{J/(kg\,{}^{\circ}C)}$ warms from 22 °C to 46 °C. First find $\Delta\theta=46-22=24\,{}^{\circ}\mathrm{C}$, then:

The RP1 method, error analysis and percentage-difference work are covered on sheet 1.3 — the same equation, the same practical.

For a pure substance melting or boiling at constant pressure, temperature remains constant while energy enters. Freezing and condensation release energy at constant temperature under the same conditions. The energy needed is called latent heat 潜热:

• $E$ energy for the change of state in J; $m$ mass that changes state in kg; $L$ specific latent heat 比潜热 in J/kg.
• Specific latent heat is the energy needed to change the state of one kilogram of a substance with no change of temperature.
• Fusion 熔化: solid to liquid. Vaporisation 汽化: liquid to vapour. These are different changes, with different values of $L$. For water, the specific latent heat of vaporisation is much greater than that of fusion; use the value for the stated material and change.

Reading the graph:

• Rising sections: energy goes into kinetic energy — the temperature climbs ($\Delta E = mc\Delta\theta$).
• Flat sections: energy goes into potential energy — the state is changing ($E = mL$). For the same material, change of state and constant net heating power, a longer plateau means more mass changed state. If $L$ or heating power differs, time alone does not identify the mass.
• Cooling has the reverse sequence of state changes: flat while a pure substance freezes or condenses at constant pressure, releasing latent heat. Rates and durations need not mirror the heating graph.

Distinguishing the two: specific heat capacity involves a temperature change; specific latent heat involves a change of state at constant temperature.

Teacher-written worked example. A 30 W heater runs for 11 minutes and boils off $6.6\times10^{-3}$ kg of water already at its boiling point. Estimate $L$ assuming all heater energy reaches the boiling water, then explain the effect of heat loss.

• Convert: $E = Pt = 30\ \text{W} \times 660\ \text{s} = 19\,800$ J.
  $$L = \frac{E}{m} = \frac{19\,800\ \text{J}}{6.6\times 10^{-3}\ \text{kg}} = 3.0\times 10^6\ \text{J/kg}$$

Actual exam demands: boiling and energy accounting

AQA June 2025 8463/1H Q08.1–08.2: 9950 J boils 50 g of nitrogen at its boiling point. Convert $m=0.050\ \mathrm{kg}$, choose $E=mL$ and rearrange:

During boiling, potential energy increases while average kinetic energy and temperature remain constant; internal energy increases.

AQA June 2022 Q08.3–08.5: beaker-and-water mass falls from 0.080 kg to 0.071 kg while the heater transfers 25200 J. The evaporated mass is 0.009 kg, so $L=E/m=25200\ \mathrm{J}/0.009\ \mathrm{kg}=2.8\times10^6\ \mathrm{J/kg}$. Heat transferred to the surroundings makes the heater-energy estimate of $L$ too high. Conversely, including water lost before boiling overstates the mass associated with the measured boiling energy and makes the estimate too low. Identify which measured quantity is biased before predicting the result.