OxfordAQA International AS and A-level Physics: First law of thermodynamics, heating and conduction (9630) – Revision Notes

These revision notes are aligned to the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, sections 3.11.1 (energy transfer by heating and doing work) and 3.11.2 (energy transfer by conduction). Section 3.11 is International A-level only. For full explanations and worked examples, read the study guide first.

Then test yourself with the practice questions. The course page, the checklist and the free diagnostics help you plan. You can also review work, power, efficiency and conservation of energy.

Key definitions

• Internal energy: the sum of the randomly distributed kinetic energies and potential energies of the particles in a body.
• Specific heat capacity, c: the energy needed to raise the temperature of 1 kg of a substance by 1 K. Unit J kg⁻¹ K⁻¹.
• Specific latent heat, l: the energy needed to change the state of 1 kg of a substance at constant temperature. Unit J kg⁻¹.
• Thermal conductivity, k: a property of the material that measures how well it conducts energy, as in rate = kAΔθ/L. Unit W m⁻¹ K⁻¹.
• U-value: U = k/L. Unit W m⁻² K⁻¹. Lower means better insulation.

Formulas

Quantity	Equation	Notes
First law	ΔU = Q + W	Q = energy input by heating. W = work done ON the system
Temperature change	Q = mcΔθ	No change of state. Δθ in °C or K
Change of state	Q = ml	Temperature stays constant
Continuous flow	P = (Δm/Δt) c Δθ	Steady flow. Power in W, flow rate in kg s⁻¹
Conduction	rate = kAΔθ/L	A at right angles to flow, L in metres
U-value rate	rate = UAΔθ	Parallel surfaces only. U = k/L

First law: signs

Situation	Sign
Energy flows into the system by heating	Q positive
Energy flows out of the system by heating	Q negative
Surroundings do work on the system (compression)	W positive
System does work on surroundings (expansion)	W negative

If ΔU is positive, the internal energy rises. If ΔU is zero, then Q = −W: the energy taken in by heating exactly equals the work done by the system.

Reminder: a gas is compressed. The surroundings do 210 J of work on it and 75 J leaves it by heating. ΔU = (−75) + 210 = +135 J.

Method: temperature change

1. Find the mass in kilograms.
2. Find Δθ as final minus initial.
3. Use Q = mcΔθ with the c value given.
4. Give the answer in joules, and check the number of significant figures against the data.

Reminder: 2.5 kg of water is heated from 20 °C to 100 °C. Take c = 4180 J kg⁻¹ K⁻¹. Q = 2.5 × 4180 × 80 = 836 000 J = 8.4 × 10⁵ J.

Method: change of state

1. Use Q = ml for each stage at constant temperature.
2. Use Q = mcΔθ for each stage where the temperature changes.
3. Add the stages.

Reminder: melting 0.60 kg of ice at 0 °C needs 0.60 × 3.34 × 10⁵ = 2.0 × 10⁵ J. A 2.5 kW heater would take 2.004 × 10⁵ / 2500 = 80 s, if all its energy went into the ice.

Particle explanation (must-know)

• On a heating curve, a sloping section means the average kinetic energy of the particles is rising.
• A flat section means a change of state. The supplied energy increases the potential energy of the particles as they move apart. The average kinetic energy, and so the temperature, does not change.
• Always name both kinds of energy. State which one changes.

Method: continuous flow

1. Write the power as energy per second delivered to the fluid.
2. Convert any volume flow rate to mass flow rate using the density.
3. Use P = (Δm/Δt) c Δθ.
4. Rearrange for whichever quantity is missing.

Reminder: a 6.0 kW heater warms water flowing at 0.032 kg s⁻¹. Taking all the power to reach the water, Δθ = 6000 / (0.032 × 4180) = 45 K. If only 85% of the power reaches the water, Δθ = 38 K.

Required practical 8: electrical method

Item	Detail
Aim	Determine the specific heat capacity of a material by an electrical method
Measurements	Mass m, potential difference V, current I, time t, temperature at the start and end (or at regular intervals)
Equation	VIt = mcΔθ, so c = VIt / (mΔθ)
Graph method	Plot temperature against time. c = VI / (m × gradient)
Main systematic error	Energy lost to the surroundings, so the temperature rise is too small and c is too large
Improvements	Lag the block, add a lid, use a little oil for thermal contact, stir a liquid, start below room temperature

Reminder: a 0.50 kg block, heater at 10.0 V and 3.0 A, temperature gradient 0.065 K s⁻¹. P = VI = 30 W. c = 30 / (0.50 × 0.065) = 923, so c = 9.2 × 10² J kg⁻¹ K⁻¹.

Conduction

• Rate = kAΔθ/L in watts.
• Rate is proportional to area and temperature difference. It is inversely proportional to thickness.
• A steady state is needed for the equation: the temperatures at each face are constant.
• Metals conduct well because free electrons carry energy as well as lattice vibrations.

Reminder: a copper bar has k = 390 W m⁻¹ K⁻¹, length 0.40 m and cross-sectional area 2.0 × 10⁻⁴ m². Its ends are at 100 °C and 30 °C. Rate = 390 × 2.0 × 10⁻⁴ × 70 / 0.40 = 14 W.

U-values

• U = k/L, so a wall with k = 0.025 W m⁻¹ K⁻¹ and L = 0.050 m has U = 0.50 W m⁻² K⁻¹.
• Rate = UAΔθ.
• For parallel surfaces, add the separate rates: total = (U₁A₁ + U₂A₂ + …) Δθ.
• The specification limits this to parallel surfaces only.

Reminder: a roof of area 12 m² has U = 0.25 W m⁻² K⁻¹. The temperature difference is 14 K. Rate = 0.25 × 12 × 14 = 42 W. In one hour this is 42 × 3600 = 1.5 × 10⁵ J.

Must-know distinctions

• Heating and work: both transfer energy and both change internal energy. Neither is a store.
• Temperature and internal energy: boiling water at 100 °C gains internal energy without any rise in temperature.
• k and U: k belongs to a material. U belongs to a particular thickness of that material.
• c and l: c is for a temperature change. l is for a change of state.

Quick self-test

1. Define internal energy.
2. A gas receives 55 J by heating and does 140 J of work on its surroundings. Find ΔU.
3. Calculate the energy needed to warm 0.250 kg of copper from 20 °C to 80 °C. Take c = 385 J kg⁻¹ K⁻¹.
4. Calculate the energy needed to melt 0.60 kg of ice at 0 °C. Take l = 3.34 × 10⁵ J kg⁻¹.
5. Explain why the temperature of boiling water stays constant while a heater is on.
6. Water flows at 0.032 kg s⁻¹ over a 6.0 kW heater. All the power reaches the water. Find the temperature rise. Take c = 4180 J kg⁻¹ K⁻¹.
7. State one reason why an electrical determination of c gives a value that is too high.
8. A polystyrene slab has k = 0.035 W m⁻¹ K⁻¹, area 0.50 m² and thickness 0.040 m. Its faces differ in temperature by 22 K. Find the rate of energy transfer.
9. Find U for a layer with k = 0.025 W m⁻¹ K⁻¹ and thickness 0.050 m.
10. A sheet of area 0.30 m² has faces 20 K apart. Its k is 0.015 W m⁻¹ K⁻¹ and it transfers 18 W. Find its thickness.

Answers

1. The sum of the randomly distributed kinetic energies and potential energies of the particles in the body.
2. Q = +55 J. W = −140 J. ΔU = 55 − 140 = −85 J.
3. Q = 0.250 × 385 × 60 = 5775 J = 5.8 × 10³ J.
4. Q = 0.60 × 3.34 × 10⁵ = 2.004 × 10⁵ J = 2.0 × 10⁵ J.
5. The supplied energy increases the potential energy of the molecules as they separate. Their average kinetic energy is unchanged, so the temperature is constant.
6. Δθ = 6000 / (0.032 × 4180) = 45 K.
7. Some energy is lost to the surroundings, so the measured temperature rise is too small and c comes out too large. Lagging reduces this.
8. Rate = 0.035 × 0.50 × 22 / 0.040 = 9.6 W.
9. U = 0.025 / 0.050 = 0.50 W m⁻² K⁻¹.
10. L = kAΔθ / rate = 0.015 × 0.30 × 20 / 18 = 0.0050 m (5.0 mm).

Where marks are usually lost

• Writing W with the wrong sign for an expanding gas.
• Defining internal energy without both kinetic and potential energies, or without “random”.
• Saying the kinetic energy increases at a boiling point.
• Leaving mass in grams, or thickness in centimetres or millimetres.
• Mixing up the formulas Q = mcΔθ and Q = ml in a question that has two stages.
• Giving a continuous flow answer in joules instead of a rate in watts or kg s⁻¹.
• Stating that lagging “stops” heat loss. Say it reduces the rate of loss.
• Adding thickness-based U-values for layers in series. This specification limits U-value work to parallel surfaces.
• Giving final answers to far more significant figures than the data supports.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams from May/June 2020, published by OxfordAQA. Sections 3.11.1 and 3.11.2.