AQA A-Level Physics: Thermal Physics Complete Guide
AQA A-Level Physics: Thermal Physics Complete Guide
Thermal Physics is the second half of Year 2 Section 3.6 of the AQA A-Level Physics specification. On paper it looks short — internal energy, the gas laws, kinetic theory and a single required practical — but it routinely surfaces some of the trickiest questions in Paper 2 because the everyday language of "heat" and "temperature" works against you. To do well in thermal physics you have to be willing to abandon your intuitive vocabulary and adopt a precise statistical-mechanics one. This guide takes you through the whole of Section 3.6.2 in the order an exam paper is likely to test it.
The single biggest conceptual hurdle is the distinction between heat, internal energy and temperature. Examiners reward candidates who never confuse the three. The mathematical hurdle is the derivation of the kinetic theory equation pV = (1/3)Nm⟨c²⟩ and its consequences for the relationship between temperature and average molecular kinetic energy. Once those two ideas are secure, the rest of the topic — specific heat capacity, latent heat, the ideal gas equation, Brownian motion, the modes of energy transfer — is largely an exercise in applying definitions carefully.
Internal Energy: Heat vs Temperature vs Internal Energy
The AQA specification defines internal energy as the sum of the randomly distributed kinetic and potential energies of all the particles in a substance. Three distinctions matter:
- Internal energy is a property of the substance — a state function. A given block of copper at 25 °C has a definite internal energy.
- Heat (or heating, Q) is the energy transferred from one body to another because of a temperature difference. It is not stored in the body; it flows across the boundary.
- Temperature is a measure of the average random kinetic energy per particle. Heat does not equal internal energy, and temperature is not a kind of energy.
A common Paper 2 vignette: ice at 0 °C is heated until it becomes water at 0 °C. The temperature is unchanged, but the internal energy has increased — the heat supplied went into raising the molecular potential energies as the lattice broke apart. This is the textbook case where temperature, internal energy and heat must be discussed as three separate things, and the answer is worth full marks only if all three are distinguished.
You should also be ready to discuss the absolute zero definition: 0 K is the temperature at which all molecular kinetic energy is at its minimum quantum value. Temperatures in equations must always be in kelvin; using degrees Celsius in pV = nRT is the single most common Year 2 error.
Specific Heat Capacity and Specific Latent Heat
The specific heat capacity c is the energy required to raise the temperature of one kilogram of a substance by one kelvin, in J kg⁻¹ K⁻¹:
ΔQ = mcΔθ
For water, c ≈ 4180 J kg⁻¹ K⁻¹ — one of the highest values for any common substance, which is why coastal climates are mild and why water makes such a good coolant.
The specific latent heat L is the energy required to change the state of one kilogram of a substance without changing its temperature, in J kg⁻¹:
ΔQ = mL
Two types matter: fusion (solid ↔ liquid) and vaporisation (liquid ↔ gas). Vaporisation always requires more energy than fusion for the same substance, because the molecules have to be separated completely against the intermolecular forces, not merely loosened from the lattice. For water, the specific latent heat of fusion is about 334 kJ kg⁻¹ and of vaporisation is about 2260 kJ kg⁻¹ — roughly seven times larger.
Required Practical 8: Specific Heat Capacity
Required Practical 8 asks you to measure the specific heat capacity of a metal block (typically aluminium or copper) and/or a liquid (typically water). The standard procedure is:
- Place a known mass m of material on a top-pan balance, with an immersion heater and a thermometer in the holes drilled into a metal block (or directly into a beaker of liquid).
- Insulate the block to minimise heat loss.
- Connect the heater to a low-voltage supply, switch on, and start a stopwatch. Record V (across the heater) and I (through the heater) at regular intervals.
- Record the temperature at fixed time intervals until a measurable rise has occurred.
- Energy supplied = VIt; energy absorbed = mcΔθ. Equating and rearranging gives c = VIt / (mΔθ).
The main systematic error is heat loss to the surroundings, which causes measured values of c to be too high — energy you assumed went into the block actually escaped, so you need more energy than expected per kelvin of rise. The fix is to insulate well, to use as small a temperature rise as is practical (so heat-loss rate is low), and to apply a cooling correction by allowing the block to cool naturally after the heater is switched off.
A useful exam phrase: "Use a thin layer of oil in the thermometer hole to improve thermal contact between the thermometer bulb and the block." Examiners reward candidates who can identify specific improvements rather than vague calls for "better insulation."
The Gas Laws: Boyle, Charles and Gay-Lussac
Three empirical gas laws describe how a fixed amount of gas behaves when one variable is held constant:
- Boyle's law (constant temperature): pV = constant. So p ∝ 1/V at fixed T.
- Charles's law (constant pressure): V / T = constant. So V ∝ T at fixed p.
- Gay-Lussac's pressure law (constant volume): p / T = constant. So p ∝ T at fixed V.
In all three, T must be in kelvin. Plotting Boyle's law as p against 1/V gives a straight line through the origin; plotting Charles's law as V against T (in K) gives the same. Extrapolating either back to V = 0 or p = 0 historically gave one of the first estimates of absolute zero, around −273 °C.
A subtle point that comes up almost every year: the gas laws hold only for a fixed amount of gas. If you change the number of moles (for example by letting some escape), you must use the full ideal gas equation rather than ratioing initial and final values.
The Ideal Gas Equation: pV = nRT
The three gas laws combine into the ideal gas equation:
pV = nRT
where p is pressure in pascals, V is volume in cubic metres, n is the number of moles, R is the molar gas constant (8.31 J mol⁻¹ K⁻¹) and T is the absolute temperature in kelvin.
An equivalent form uses the number of molecules N and the Boltzmann constant k = R / N_A:
pV = NkT
with k = 1.38 × 10⁻²³ J K⁻¹ and N_A = 6.02 × 10²³ mol⁻¹. You should be fluent in both forms, because exam questions sometimes give n and sometimes N. A common slip is to convert moles to molecules and then keep using R rather than switching to k; pick one form at the start of the question and stick to it.
An ideal gas is one that perfectly obeys pV = nRT at all pressures and temperatures. Real gases approximate ideal behaviour at low pressures (so intermolecular forces are negligible) and high temperatures (so kinetic energy swamps any potential energy of interaction). Examiners often ask candidates to state these conditions explicitly.
Molecular Kinetic Theory
The kinetic theory model derives the macroscopic gas equation from the microscopic motion of molecules. Its assumptions are stark and idealised, and you must be able to state them:
- A gas consists of a very large number of molecules in continual random motion.
- The volume of the molecules themselves is negligible compared with the volume of the container.
- Collisions between molecules and with the walls are perfectly elastic.
- The time between collisions is much greater than the time taken for a collision.
- There are no intermolecular forces between molecules except during collisions.
Newton's laws of motion apply to every collision.
From these assumptions the derivation considers a single molecule of mass m bouncing back and forth between two opposite walls of a cubic container of side L. The change of momentum at each collision is 2mu_x (where u_x is the component of velocity perpendicular to the wall), the time between collisions is 2L / u_x, and so the rate of change of momentum — and hence the force on the wall — is mu_x² / L. Summing over all N molecules and recognising that ⟨u_x²⟩ = (1/3)⟨c²⟩ in three dimensions gives:
pV = (1/3) N m ⟨c²⟩
You must be able to reproduce this derivation or at least its outline. Many AQA Paper 2 mark schemes reserve up to six marks for the steps. Common slip: writing ⟨c⟩² instead of ⟨c²⟩. The mean of the squares is not the same as the square of the mean.
Kinetic Energy and Temperature
Combining pV = (1/3)Nm⟨c²⟩ with pV = NkT gives the central result of A-Level kinetic theory:
(1/2) m ⟨c²⟩ = (3/2) kT
The average translational kinetic energy of a gas molecule is proportional to the absolute temperature. This is the deepest single statement in the topic, and it is the bridge between the everyday concept of temperature and the microscopic motion of particles.
A useful consequence: the root-mean-square speed c_rms = √⟨c²⟩ scales as √T. Doubling the absolute temperature increases c_rms by a factor of √2, not 2. Examiners exploit this regularly.
Brownian Motion: Evidence for Kinetic Theory
In 1827 Robert Brown observed that pollen grains suspended in water moved in jerky, random paths. The accepted explanation, developed by Einstein and others, is that the grains are constantly being bombarded by water molecules. Because the bombardment is random, occasionally more molecules hit one side of a grain than the other, producing a net impulse that the grain — being relatively heavy — responds to with a small visible displacement.
You can observe an analogous effect with smoke particles in air illuminated under a microscope: the smoke particles dance about because air molecules are bombarding them from all sides. The fact that the dance is random is direct evidence for the random motion assumption of kinetic theory.
Examiners often ask "explain how Brownian motion supports the kinetic theory of gases." The answer must mention both the random direction of the motion (consistent with random molecular motion) and the discreteness of the displacements (consistent with discrete molecular impacts). A common one-mark omission is failing to state that the air molecules must themselves be moving randomly, not just present in large numbers.
Energy Transfer: Conduction, Convection, Radiation
Although the AQA Year 2 specification is brief on the three modes of energy transfer, expect them to feature in describe-and-explain questions.
- Conduction. Energy transferred through a material by collisions between particles and (in metals) by mobile electrons. Solids conduct best because their particles are close together; gases conduct poorly because their particles are far apart.
- Convection. Energy transferred through a fluid by bulk motion of the fluid. Hotter fluid expands, becomes less dense, rises, and carries energy with it.
- Radiation. Energy transferred by electromagnetic waves and requires no medium. All bodies above 0 K emit thermal radiation; the rate depends on the fourth power of temperature (Stefan's law) — a useful link to astrophysics if you study Option A.
Synoptic Links Forward
Thermal Physics underpins several Year 2 topics:
- Gravitational and electric fields reuse the kT bookkeeping when treating thermal velocities of charged particles.
- Nuclear physics uses the kT / particle energy comparison when discussing the conditions required for nuclear fusion (≈10⁷ K for D-T fusion).
- Astrophysics uses kinetic theory implicitly in stellar atmospheres and the Sun's core, and explicitly in calculations of the rms speed of hydrogen in different stellar regions.
How to Study This Topic
Examiners reward careful definitions in thermal physics more than in any other topic. Build your revision plan around them.
- Make a one-page glossary distinguishing internal energy, heat and temperature; rewrite it from memory weekly.
- Memorise the four core equations: ΔQ = mcΔθ, ΔQ = mL, pV = nRT (and pV = NkT), pV = (1/3)Nm⟨c²⟩, and (1/2)m⟨c²⟩ = (3/2)kT.
- Be able to reproduce the kinetic theory derivation in outline — examiners reward correct structure even when arithmetic slips occur.
- Practise Required Practical 8 with a fake dataset so you can describe the method, identify three error sources, and propose three improvements without seeing the equipment.
- Drill unit care, especially K vs °C and Pa vs kPa.
Related LearningBro Courses
LearningBro's AQA A-Level Physics: Thermal Physics course covers all of the above in ten lessons:
- Thermal Energy and Internal Energy — the heat / temperature / internal energy distinction.
- Specific Heat Capacity — ΔQ = mcΔθ and worked examples.
- Specific Latent Heat — ΔQ = mL, fusion vs vaporisation.
- Gas Laws: Experimental Foundations — Boyle, Charles, Gay-Lussac and the route to absolute zero.
- The Ideal Gas Equation — pV = nRT and pV = NkT.
- Molecular Kinetic Theory — assumptions and derivation of pV = (1/3)Nm⟨c²⟩.
- Kinetic Energy and Temperature — (1/2)m⟨c²⟩ = (3/2)kT and c_rms scaling.
- Brownian Motion Evidence — random motion and the molecular impact picture.
- Energy Transfer Mechanisms — conduction, convection, radiation.
- Required Practicals: Thermal Physics — the RP8 method, error analysis, and graph extraction of c.
Common Exam Pitfalls
Thermal physics is a unit where careless unit handling, sign conventions and conceptual sloppiness account for most lost marks. Defending against them takes only a few minutes of revision.
- Kelvin, not Celsius. Every gas-law and kinetic-theory equation requires absolute temperature. A common error is to plug 25 in for T in pV = nRT and get a pressure off by a factor of about twelve. Always rewrite the temperatures in kelvin before substituting.
- Heat versus internal energy versus temperature. Internal energy U is a property of the system; heat Q is a transfer process; temperature T is what you measure with a thermometer. Phase changes happen at constant temperature but with non-zero heat flow and a large change in internal energy. A question asking what happens to the temperature during boiling has the answer "stays constant", and what happens to the internal energy has the answer "increases".
- Specific heat capacity in cooling-correction questions. When using ΔQ = m c Δθ on an experimental dataset, the energy supplied by the heater is never the energy that actually raised the temperature of the sample. Heat losses to the surroundings make the measured c too high. The standard fix is to extrapolate the cooling curve back to the moment the heater was switched off, which gives the true temperature rise.
- rms and mean square confused. ⟨c^2⟩ is the mean of the squares; c_rms is the square root of that. Plugging c_rms^2 into (1/2) m ⟨c^2⟩ is correct; plugging ⟨c⟩^2 is not. The two differ because the distribution of molecular speeds has a non-zero spread.
- Latent heat treated as continuing across a phase change. A heating curve has flat plateaus at the melting and boiling points where temperature does not change. During those plateaus the specific heat capacity equation does not apply; you must use Q = m L instead, with the appropriate L_f or L_v.
- Wrong gas constant in pV = nRT. R = 8.31 J K^-1 mol^-1 only works if n is in moles, p in pascals, V in m^3 and T in kelvin. Litres and atmospheres lead to wrong answers. Switching to pV = N k_B T (where N is the number of molecules) needs k_B = 1.38 x 10^-23 J K^-1.
- Kinetic-theory derivation shortcuts. A "show that pV = (1/3) N m ⟨c^2⟩" question is worth 5-6 marks. Examiners want the assumptions listed (point particles, elastic collisions, no intermolecular forces, random motion, negligible volume of molecules), the time-between-collisions argument, the change-in-momentum-per-collision step, and the factor-of-three split between the three orthogonal directions. Skipping any of these costs marks even when you write the right final expression.
Worked Example: Cooling a Drink with Ice
A 250 g glass of water at 20 degrees C has a 30 g ice cube at -10 degrees C dropped into it. The drink is in a well-insulated cup. Find the final temperature, assuming all the ice melts. Take c_water = 4180 J kg^-1 K^-1, c_ice = 2100 J kg^-1 K^-1 and L_f(water) = 3.34 x 10^5 J kg^-1.
Heat flows from the water to the ice. There are three stages absorbing energy from the water:
- Warm the ice from -10 to 0 degrees C: Q_1 = m_ice c_ice ΔT = 0.030 x 2100 x 10 = 630 J.
- Melt the ice at 0 degrees C: Q_2 = m_ice L_f = 0.030 x 3.34 x 10^5 = 10 020 J.
- Warm the melted ice (now liquid water at 0 degrees C) up to the final temperature T_f: Q_3 = 0.030 x 4180 x T_f = 125.4 T_f J (with T_f in degrees C).
The water cools from 20 degrees C to T_f, releasing Q_4 = m_water c_water (20 - T_f) = 0.250 x 4180 x (20 - T_f) = 1045 (20 - T_f) J.
Conservation of energy (no heat lost to surroundings) gives Q_1 + Q_2 + Q_3 = Q_4:
630 + 10 020 + 125.4 T_f = 1045 (20 - T_f) 10 650 + 125.4 T_f = 20 900 - 1045 T_f 1170.4 T_f = 10 250 T_f ~ 8.8 degrees C.
So the drink ends up at about 9 degrees C. Notice how the latent-heat term dominates: melting the ice takes more than ten kilojoules, whereas warming the ice from -10 to 0 takes only 630 J. This is why ice is such an effective coolant — most of the cooling comes from the phase change, not from the temperature rise of the cold ice.