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

This study guide is 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. It is not part of the shared International AS and A-level content in sections 3.1 to 3.5.

Use it with the revision notes and the practice questions. The course page and the checklist show where this unit sits. If you are unsure of your starting point, try the free diagnostics. Earlier energy ideas are in work, power, efficiency and conservation of energy.

What this unit covers

Section	What you must be able to do
3.11.1	Define internal energy as the sum of the randomly distributed kinetic and potential energies of the particles in a body
3.11.1	State when internal energy rises or falls: heating or work done on the system
3.11.1	Use the first law, ΔU = Q + W, with Q the energy input by heating and W the work done on the system
3.11.1	Explain that during a change of state the potential energies of the particles change but the kinetic energies do not
3.11.1	Calculate energy transfers, including continuous flow, using Q = mcΔθ and Q = ml
3.11.1	Required practical 8: determine specific heat capacity by an electrical method
3.11.2	Use rate of energy transfer by conduction = kAΔθ/L
3.11.2	Use U-values to calculate energy losses for parallel surfaces only, with rate = UAΔθ and U = k/L

Internal energy

A body is made of particles that move and attract or repel one another. Each particle has kinetic energy because it moves. Each pair of particles has potential energy because of the forces between them. Internal energy is the sum of the randomly distributed kinetic energies and potential energies of the particles in the body.

“Randomly distributed” means you do not count the kinetic energy of the whole body moving through the room. “Sum” means both kinds of energy are included. Writing “the kinetic energy of the molecules” alone gives only half the definition.

Heating and work are processes that transfer energy. Internal energy is the stored result, so a gas does not “contain 200 J of heat”.

The first law of thermodynamics

The internal energy of a system rises when energy is transferred to it by heating or when work is done on it. It falls in the reverse cases. In symbols:

ΔU = Q + W

• ΔU is the change in internal energy.
• Q is the energy input to the system by heating. It is negative if the system loses energy by heating.
• W is the work done ON the system. It is negative if the system does work on its surroundings.

Compressing a gas does work on it, so W is positive. A gas that expands and pushes a piston outwards does work on its surroundings, so W is negative.

Worked example 1

A sample of gas in a cylinder is compressed. The piston does 340 J of work on the gas. At the same time 120 J leaves the gas by heating through the cylinder walls. Find the change in internal energy.

Work done on the gas: W = +340 J. Energy leaves by heating: Q = −120 J. ΔU = Q + W = (−120) + 340 = +220 J

The internal energy of the gas rises by 220 J.

Heating a substance: Q = mcΔθ

When a substance is warmed without changing state, the energy needed is

Q = mcΔθ

where m is the mass, c is the specific heat capacity (the energy needed to raise the temperature of 1 kg of the substance by 1 K) and Δθ is the temperature change. A change of 1 °C equals a change of 1 K, so Δθ can be in either unit. The unit of c is J kg⁻¹ K⁻¹.

Worked example 2

An aluminium block of mass 0.80 kg is warmed from 18 °C to 64 °C. Take c = 900 J kg⁻¹ K⁻¹. Find the energy supplied.

Δθ = 64 − 18 = 46 K Q = mcΔθ = 0.80 × 900 × 46 = 33 120 J Q = 3.3 × 10⁴ J (2 significant figures, matching the data)

Change of state: Q = ml

During melting or boiling the temperature stays constant while energy is still being supplied. The energy goes into the potential energy of the particles as they move apart against the forces between them. Their average kinetic energy does not change, which is why the temperature does not change.

The energy to change the state of a mass m at constant temperature is

Q = ml

where l is the specific latent heat, the energy needed to change the state of 1 kg of the substance at constant temperature. The unit is J kg⁻¹. Fusion applies to melting and vaporisation to boiling.

On a temperature-time graph for steady heating, a sloping section shows the kinetic energy of the particles rising. A flat section shows the potential energy rising.

Worked example 3

How much energy is needed to melt 0.42 kg of ice at 0 °C? Take the specific latent heat of fusion of ice as l = 3.34 × 10⁵ J kg⁻¹.

Q = ml = 0.42 × 3.34 × 10⁵ = 1.403 × 10⁵ J Q = 1.4 × 10⁵ J (2 significant figures)

If a question includes both a temperature change and a change of state, split it into stages. Use mcΔθ for each stage where the temperature changes and ml for each stage where it does not. Then add the stages.

Continuous flow

In a continuous flow system a fluid passes steadily over a heater, as in a shower heater. You work with the mass that flows each second. The mass flow rate is Δm/Δt. Dividing Q = mcΔθ by time gives

P = (Δm/Δt) × c × Δθ

where P is the power transferred to the fluid. This applies when the fluid reaches a steady state, with a constant temperature rise between inlet and outlet.

If you are given a volume flow rate, convert it to a mass flow rate using the density first.

Worked example 4

An electric shower heater has a power of 9.0 kW. Water enters at 15 °C and leaves at 40 °C. Take c = 4180 J kg⁻¹ K⁻¹. Find the mass of water heated each second if all the electrical energy goes to the water.

Δθ = 40 − 15 = 25 K Δm/Δt = P / (cΔθ) = 9000 / (4180 × 25) = 0.0861 kg s⁻¹ Mass flow rate = 0.086 kg s⁻¹

If a question says only a percentage of the power reaches the water, multiply P by that fraction first.

Required practical 8: specific heat capacity by an electrical method

The specification requires you to determine specific heat capacity by an electrical method.

Method for a solid block. The block has two holes: an electric heater fits in one and a thermometer in the other, with a little oil for contact.

1. Measure the mass m of the block with a balance.
2. Record the starting temperature.
3. Switch on the heater and a stopwatch together. Record the potential difference V and the current I.
4. After a time t, record the final temperature.
5. Lag the block with insulating material to reduce losses.

With no energy lost, VIt = mcΔθ. So c = VIt / (mΔθ).

Better analysis uses a graph. Plot temperature against time at regular intervals. The gradient is Δθ/Δt = P/(mc), so c = P / (m × gradient), where P = VI. This reduces the effect of one poor reading.

Worked example 5

A 1.00 kg metal block is heated by a heater running at 12.0 V and 4.20 A for 420 s. The temperature rises by 24.0 K. Find c.

Energy supplied = VIt = 12.0 × 4.20 × 420 = 21 168 J c = 21 168 / (1.00 × 24.0) = 882 J kg⁻¹ K⁻¹ c = 882 J kg⁻¹ K⁻¹ (3 significant figures)

Sources of error. Some energy always warms the surroundings and the heater itself. The energy that actually reaches the block is smaller than VIt, so the temperature rise is smaller than it should be. Your value of c is then too large, a systematic error. Lagging, a lid on the block and starting a few degrees below room temperature all help.

Conduction

Conduction is the transfer of energy through a material without bulk movement of the material. Vibrating particles pass energy to their neighbours. In a metal, free electrons also carry energy, so metals conduct well.

In the steady state, the rate of energy transfer through a uniform slab is

rate = kAΔθ/L

• k is the thermal conductivity of the material, in W m⁻¹ K⁻¹.
• A is the cross-sectional area at right angles to the energy flow.
• Δθ is the temperature difference between the two faces.
• L is the thickness of the slab, or the length of the rod.

The rate is in watts. Doubling the thickness halves the rate. The ratio Δθ/L is the temperature gradient.

Worked example 6

A brick wall has area 14 m² and thickness 0.22 m. The thermal conductivity of the brick is 0.80 W m⁻¹ K⁻¹. The inside face is at 20 °C and the outside face is at 4 °C. Find the rate of energy transfer.

Δθ = 20 − 4 = 16 K rate = kAΔθ/L = (0.80 × 14 × 16) / 0.22 = 179.2 / 0.22 = 814.5 W rate = 8.1 × 10² W (2 significant figures)

U-values

The loss through a whole wall, roof or window is described by its U-value:

U = k/L

Substituting into the conduction equation gives

rate of energy transfer = UAΔθ

U has unit W m⁻² K⁻¹. A low U-value means a good insulator.

The specification restricts U-value calculations to parallel surfaces only. That means surfaces that are side by side in the same building envelope, such as a wall, a window and a roof. You find the total by adding the separate rates:

total rate = (U₁A₁ + U₂A₂ + …) × Δθ

You are not asked to combine layers placed one behind another in series.

Worked example 7

A room has a wall of area 18.5 m² with U = 0.35 W m⁻² K⁻¹ and a window of area 3.5 m² with U = 1.8 W m⁻² K⁻¹. The inside is at 21 °C and the outside at 6 °C. Find the total rate of energy loss, and the share lost through the window.

Δθ = 21 − 6 = 15 K Wall: 0.35 × 18.5 × 15 = 97.1 W Window: 1.8 × 3.5 × 15 = 94.5 W Total = 191.6 W = 1.9 × 10² W Share through the window = 94.5 / 191.6 = 49%

The window is under 16% of the area but loses nearly half the energy.

A slab 0.10 m thick made of material with k = 0.040 W m⁻¹ K⁻¹ has U = 0.040 / 0.10 = 0.40 W m⁻² K⁻¹.

Common errors

• Giving W the wrong sign. Work done by a gas on its surroundings is negative in ΔU = Q + W.
• Defining internal energy as kinetic energy only, or as “heat”.
• Saying kinetic energy rises during boiling. It is potential energy that rises.
• Forgetting to convert thickness to metres, or area to square metres, in conduction questions.
• Applying U = k/L to layers in series, which this specification does not ask for.
• Forgetting that heat losses in the heater experiment push the measured c upwards.

Next, try the practice questions.

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.