Internal Energy

Spec mapping: OCR H556 Module 5.1 — Thermal physics (internal energy as the sum of the randomly distributed kinetic and potential energies of all particles within a system; dependence on temperature and state; the ideal-gas special case where $U$U depends only on $T$T). Refer to the official OCR H556 specification document for exact wording.

In lesson 1 we introduced temperature as a measure of the average kinetic energy of the particles in a substance. We now move from average to total. A body contains a huge number of atoms or molecules, each in ceaseless motion and each interacting with its neighbours. The total amount of energy stored in those microscopic motions and interactions is called the internal energy of the body. This is the key quantity of thermal physics — it is what changes when a body is heated, cooled, compressed or allowed to expand. Module 5.1 of the OCR A-Level Physics A specification (H556) requires you to understand what internal energy is, what contributes to it, and how it responds to changes in temperature and state.

The leap from GCSE to A-Level is substantial. At GCSE you may have spoken loosely about a body containing "heat energy". At A-Level you must replace that language with two distinct concepts: internal energy (energy stored in the body, in the random motions and configurations of its particles) and heat (energy in transit between bodies at different temperatures). This lesson sets out the rigorous A-Level picture and unpacks the special case — the ideal gas — where the potential component vanishes and $U$U depends only on $T$T.

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Definition of Internal Energy

Internal energy $U$U is the sum of the randomly distributed kinetic and potential energies of the particles within a system.

Two points matter in this definition:

1. "Randomly distributed" — ordinary macroscopic kinetic energy (e.g. a football flying through the air) does not count. The whole football has a lot of kinetic energy, but all the atoms of the football are moving together in the same direction. That is bulk or ordered kinetic energy, not random, and it is not part of the internal energy. If you warm the football with your hands while holding it still, on the other hand, the random motion of its atoms increases, and you raise its internal energy.
2. "Kinetic and potential energies" — both contribute. The kinetic part comes from the motion of the particles (translation, rotation and vibration). The potential part comes from the forces the particles exert on one another through their separations.

Internal energy is symbolised by $U$U and measured in joules (J), like any other energy.

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The Kinetic Component

Every particle in the body is in motion. In a gas, the motion is free translation through empty space, punctuated by occasional collisions. In a liquid, the motion is hindered by near neighbours, so particles jostle about in a confined space rather than flying freely. In a solid, the particles are held at fixed lattice sites and can only vibrate about them.

In every case, the kinetic energy of each particle is $\tfrac{1}{2}mv^2$21​mv2 where $v$v is that particle's instantaneous speed. Since every particle has a different $v$v at any given moment, the total translational kinetic energy of the body is

$KE_{\text{total}} = \sum_i \tfrac{1}{2} m_i v_i^{2}$KEtotal​=∑i​21​mi​vi2​

summed over all $N$N particles.

At higher temperatures, the average kinetic energy per particle is larger. A later kinetic-theory lesson will show precisely that for an ideal monatomic gas, each particle has an average translational kinetic energy of $\langle E_k \rangle = \tfrac{3}{2} k T$⟨Ek​⟩=23​kT, where $k = 1.38 \times 10^{-23}$k=1.38×10−23 J K$^{-1}$−1 is the Boltzmann constant. Increasing the temperature therefore increases the kinetic component of the internal energy directly.

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The Potential Component

The potential component arises because the particles are not independent: they attract or repel each other. The forces between molecules — the intermolecular forces — give the particles potential energy that depends on their separations.

In a solid

In a solid, particles are bound tightly to their equilibrium positions in a lattice. If you pull two neighbouring atoms apart, you do work against the attractive force between them, raising their potential energy. This stored energy is part of the internal energy of the solid.

In a liquid

In a liquid, particles are still in close contact with their neighbours, but they are not fixed in a lattice. They slide past one another. There is still significant intermolecular potential energy (which is why the density of a liquid is similar to that of a solid), but the arrangement of particles is constantly changing.

In a gas

In a gas (especially at low density), particles are widely separated. Except during brief collisions, they barely interact, and the potential component of the internal energy is therefore very small. In an ideal gas — which will be our main model from lesson 5 onward — we assume there are no intermolecular forces, so the internal energy consists of kinetic energy alone.

flowchart LR
    U["Internal energy U"] --> KE["Kinetic component"]
    U --> PE["Potential component"]
    KE --> K1["Translation"]
    KE --> K2["Rotation"]
    KE --> K3["Vibration"]
    PE --> P1["Intermolecular forces"]
    PE --> P2["Intramolecular bonds"]

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Internal Energy and Temperature

Since increasing the temperature increases the average kinetic energy of the particles, it follows that increasing the temperature increases the internal energy (provided the number of particles does not change). For an ideal monatomic gas at fixed $N$N,

$U = \tfrac{3}{2} N k T,$U=23​NkT,

so the internal energy is directly proportional to the absolute temperature. Doubling $T$T doubles $U$U.

But be careful: for a real substance — a solid, a liquid, or a non-ideal gas — the relationship is more complicated because the potential part of the energy also changes with temperature (atoms vibrate with larger amplitude, occupying larger PE values on average). And most importantly, during a change of state the temperature does not change at all, yet the internal energy does. We will return to this in lesson 4 when we discuss latent heat.

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Changes in Internal Energy

You can change the internal energy of a body in two ways:

1. By heating it — supplying thermal energy from a hotter body. The added energy increases the random motion of the particles.
2. By doing work on it — for example, compressing a gas with a piston. The work done on the gas increases its internal energy.

This is the content of the first law of thermodynamics, which you will meet formally only if you study physics beyond A-Level. For A-Level you simply need to know that any input of energy to the system — whether as heat or as work — increases its internal energy by the same amount.

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Worked Example 1 — Which motion counts as internal?

A 0.50 kg lump of metal is moving through the air at $v = 10$v=10 m s$^{-1}$−1. Its temperature is $25\,^{\circ}$25∘C. Which of the following contributes to its internal energy?

1. Its bulk kinetic energy of $\tfrac{1}{2}(0.50)(10)^2 = 25$21​(0.50)(10)2=25 J as it flies through the air.
2. The vibrational energy of its atoms about their equilibrium positions.
3. The potential energy of the interatomic bonds in the lattice.

The answer: only 2 and 3. The 25 J of bulk kinetic energy is ordered motion of the whole lump, not random motion of the particles. It does not count as internal energy, even though it is certainly energy of the lump. Internal energy is strictly the random, microscopic part.

If you were to catch the lump and bring it to rest, that 25 J of ordered kinetic energy would be converted (by friction, air resistance or impact) into additional random motion — that would increase the internal energy of the lump and whatever it collides with.

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Worked Example 2 — Heating a solid

A copper block of mass 0.40 kg is heated from $20\,^{\circ}$20∘C to $70\,^{\circ}$70∘C. The specific heat capacity of copper is $c = 385$c=385 J kg$^{-1}$−1 K$^{-1}$−1. Assuming no change of state, by how much does the internal energy of the block change?

The energy supplied, and therefore the increase in internal energy, is

$\Delta U = mc \Delta \theta = (0.40)(385)(70 - 20) = (0.40)(385)(50) = 7700 \text{ J} = 7.7 \text{ kJ}.$ΔU=mcΔθ=(0.40)(385)(70−20)=(0.40)(385)(50)=7700 J=7.7 kJ.

The block now has 7.7 kJ more internal energy than before. The increase is stored partly as an increase in the vibrational kinetic energy of the atoms about their lattice sites, and partly as an increase in the potential energy of the lattice (the atoms vibrate with larger amplitude and spend more time further from equilibrium).

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Worked Example 3 — Doing work on a gas

A gas is compressed by a piston. The piston does 120 J of work on the gas, and no heat is allowed to flow in or out of the gas (the walls are insulated). By how much does the internal energy of the gas change?

Because no heat is exchanged, all the work done on the gas goes into increasing its internal energy:

$\Delta U = W_{\text{on}} = 120 \text{ J}.$ΔU=Won​=120 J.

The gas becomes hotter, even though it has not been "heated" in the everyday sense. This is exactly what happens in a bicycle pump when you compress air rapidly — the pump and the air become noticeably warm.

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Worked Example 4 — Phase change with no temperature rise

A 0.10 kg sample of ice at exactly 0 $^{\circ}$∘C is supplied with 33.4 kJ of energy from an electric heater. The accepted value of the specific latent heat of fusion of ice is $L_f = 334$Lf​=334 kJ kg$^{-1}$−1. Find (a) the new mass of liquid water produced, (b) the change in temperature, and (c) the change in internal energy of the sample.

(a) Mass of ice melted: $m_{\text{melted}} = E/L_f = 33\,400 / (334 \times 10^3) = 0.100$mmelted​=E/Lf​=33400/(334×103)=0.100 kg. So all the ice has melted; the sample is now 0.10 kg of water at 0 $^{\circ}$∘C.

(b) $\Delta T = 0$ΔT=0 K. The temperature has not changed.

(c) $\Delta U = +33.4$ΔU=+33.4 kJ. The internal energy has increased by exactly the energy supplied (no heat lost, no work done by or on the sample). But because $\Delta T = 0$ΔT=0, the average random kinetic energy of the molecules has not changed; the kinetic component of $U$U is unchanged. The entire 33.4 kJ has gone into the potential component of $U$U, breaking the rigid hydrogen-bond lattice of ice and allowing molecules to slide freely in the liquid.

This is the cleanest possible illustration of why internal energy is not simply "the energy of temperature". Phase changes redistribute $U$U between its kinetic and potential parts without changing $T$T.

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States of Matter and Internal Energy

The three familiar states of matter — solid, liquid, gas — correspond to qualitatively different arrangements of particles and therefore different balances between the kinetic and potential parts of the internal energy.

State	Particle motion	Particle separation	Forces	Typical density
Solid	Vibrational about fixed positions	Small (close-packed)	Strong	High (few g cm⁻³)
Liquid	Free but hindered; particles jostle	Small (close-packed)	Moderate	High, similar to solid
Gas	Free translational flight	Large (many molecular diameters)	Very weak (or zero, for an ideal gas)	Low (~10⁻³ g cm⁻³)

As you add energy to a solid, its internal energy rises and its temperature rises — until you reach the melting point. At that point, all the energy you add goes into overcoming the intermolecular forces that hold the lattice together; the kinetic energy (and therefore the temperature) does not increase at all. This is why a pan of melting ice stays at 0 $^{\circ}$∘C. The energy of fusion raises the potential part of the internal energy without changing the kinetic part. We will study this in detail in lesson 4.

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Ideal Gas: Internal Energy Depends Only on Temperature

A key idealisation in this course is the ideal gas. For an ideal gas:

• There are no intermolecular forces (except during collisions).
• The potential component of the internal energy is zero.
• The internal energy consists entirely of the random kinetic energy of translation.

Therefore, for an ideal gas,

$U = U(T) \text{ only — not a function of } p \text{ or } V.$U=U(T) only — not a function of p or V.

This is a very powerful result. If you compress an ideal gas isothermally (keeping $T$T constant), its internal energy does not change, even though you did work on it, because heat must flow out of the gas to keep $T$T constant. If you double the volume of an ideal gas without changing $T$T, the internal energy is the same as before.

For real gases at normal laboratory pressures and temperatures, this approximation is excellent. It only begins to fail near the condensation point or at very high pressure, where intermolecular forces become significant and a small additional PE contribution appears.

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Exam Tip: Spot the Ordered Energy

Exam Tip: OCR love to ask questions that require you to distinguish between ordered and disordered energy. Watch out for phrases like "a moving ball", "a flowing liquid", "a rotating wheel". All of these involve ordered kinetic energy that is not part of the internal energy. The internal energy refers only to the random, disordered microscopic motions of the particles relative to the centre of mass of the body.

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The Microscopic Versus the Macroscopic

It is worth stepping back to appreciate the conceptual shift that thermal physics demands. In mechanics you are used to thinking of a body as a single point with a single velocity. In thermal physics you have to think of it as $\sim 10^{23}$∼1023 particles, each with its own velocity and its own kinetic energy, all jostling and colliding. The macroscopic quantities — temperature, pressure, internal energy — are averages or sums over this microscopic chaos.

The beauty of thermal physics is that the averages obey clean laws, even though the individual motions are hopelessly complicated. Temperature is an average; internal energy is a sum. In a later kinetic-theory lesson the derivation will give us the sharpest possible version of this connection: $pV = \tfrac{1}{3} N m \langle c^{2} \rangle$pV=31​Nm⟨c2⟩, with the macroscopic quantities on the left and the microscopic ones on the right.

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