Thermal energy transfers

All matter is made of particles (atoms or molecules) that are in constant random motion and exert forces on one another. The three states of matter differ in how tightly the particles are held:

State	Arrangement	Movement	Intermolecular forces
Solid	Regular, fixed lattice; particles touching	Vibrate about fixed positions	Strong
Liquid	Close together but disordered	Move/slide past one another	Moderate
Gas	Far apart, random	Move freely at high speed	Very weak (almost none)

Temperature is a measure of the average random kinetic energy of the particles. The hotter the substance, the faster its particles move (on average). This is why we use an absolute temperature scale, the Kelvin scale:

$T(\text{K}) = \theta(°\text{C}) + 273$

At absolute zero ($0\,\text{K} \approx -273\,°\text{C}$) the particles have the minimum possible kinetic energy. Because the kelvin scale starts at this natural zero, the absolute temperature is directly proportional to the average kinetic energy of the particles — which is exactly why the Stefan–Boltzmann and Wien laws later on demand kelvin, not Celsius.

Internal energy is the total energy of all the particles:

$\text{internal energy} = \underbrace{\text{total random kinetic energy}}_{\text{depends on temperature}} + \underbrace{\text{total intermolecular potential energy}}_{\text{depends on separation / state}}$

• The kinetic part depends on temperature (how fast the particles move).
• The potential part depends on the spacing of the particles, i.e. the state — it changes when bonds are broken or formed (during a phase change).

When you supply thermal energy to a substance without changing its state, its temperature rises. The energy needed is:

$Q = mc\Delta T$

where $c$ is the specific heat capacity — the energy needed to raise the temperature of 1 kg of the substance by 1 K (or 1 °C). Its unit is $\text{J kg}^{-1}\,\text{K}^{-1}$.

• Water has an unusually large specific heat capacity ($c \approx 4180\,\text{J kg}^{-1}\,\text{K}^{-1}$), which is why it takes a lot of energy to heat and why it is used as a coolant.
• Metals have small specific heat capacities (copper ≈ 385, aluminium ≈ 900), so they heat up (and cool down) quickly.

Heat capacity vs specific heat capacity — a distinction examiners love. The heat capacity of a whole object is $C = mc$ (unit $\text{J K}^{-1}$) — the energy to raise that object's temperature by 1 K. The specific heat capacity is per kilogram and is a property of the material. If your equation contains a mass $m$, you need the specific value.

Since $\Delta T$ is a temperature difference, it is the same number in kelvin and in Celsius — a rise of 60 °C is a rise of 60 K. So you do not need to convert to kelvin in $Q = mc\Delta T$ (unlike the star laws later). What you must never do is mix a mass in grams with a $c$ quoted per kilogram.

During a change of state (melting, boiling, and their reverses) the temperature stays constant even though energy is still being transferred. The energy involved is the latent heat:

$Q = mL$

• Specific latent heat of fusion $L_f$: energy per kilogram to melt a solid into a liquid (for ice, $\approx 3.34\times10^{5}\,\text{J kg}^{-1}$).
• Specific latent heat of vaporisation $L_v$: energy per kilogram to boil a liquid into a gas (for water, $\approx 2.26\times10^{6}\,\text{J kg}^{-1}$).

Why is the temperature constant? Temperature measures average kinetic energy. During a phase change the supplied energy goes into breaking the intermolecular bonds — it increases the particles' potential energy, not their kinetic energy. The average kinetic energy (and so the temperature) is unchanged until the phase change is complete. This is the single most important idea in the whole subtopic.

Why is $L_v$ so much bigger than $L_f$? To melt a solid you only need to loosen the bonds enough for the particles to slide past each other. To boil a liquid you must completely separate the particles against the intermolecular forces and do work pushing back the atmosphere — far more energy per kilogram.

A heating curve (temperature against time, at constant power) makes this visible: sloping sections are single states warming up ($Q = mc\Delta T$), and the flat plateaus are phase changes ($Q = mL$).

Thermal energy always flows from hotter to cooler regions, by three mechanisms:

1. Conduction — energy passes through a material as particles vibrate more vigorously and collide with their neighbours, handing on kinetic energy. The particles themselves stay put. In metals, free (delocalised) electrons also carry energy, which is why metals are the best conductors. Conduction dominates in solids.

2. Convection — in a fluid (liquid or gas), a warmer region expands, becomes less dense and rises, while cooler, denser fluid sinks to take its place. This circulating convection current physically carries energy from place to place. Convection cannot occur in a solid (the particles cannot flow).

3. Thermal radiation — all objects emit electromagnetic waves (mainly infrared at everyday temperatures) because of the motion of their charged particles. Crucially, this needs no medium, so radiation is the only mechanism that can transfer energy through a vacuum. This is how the Sun's energy reaches Earth across empty space.

Mechanism	Needs a medium?	Where it dominates
Conduction	Yes (particles)	Solids, especially metals
Convection	Yes (a fluid)	Liquids and gases
Thermal radiation	No — works in a vacuum	Across space; all hot objects

The hotter an object, the more energy it radiates per second (this is quantified for a black body by the Stefan–Boltzmann law in the next section).

A black body is an idealised object that absorbs all radiation falling on it and, at a given temperature, emits the maximum possible radiation at every wavelength. Stars are modelled as black bodies. The graph of emitted intensity against wavelength is the black-body (emission) spectrum, and it has two key features you must know.

1. The hotter the body, the more total power it radiates — Stefan–Boltzmann law. The luminosity $L$ (total power radiated, in watts) of a spherical black body is:

$L = 4\pi r^2 \sigma T^4$

where $r$ is the radius, $T$ is the surface temperature in kelvin, and $\sigma = 5.67\times10^{-8}\,\text{W m}^{-2}\,\text{K}^{-4}$ (the Stefan–Boltzmann constant, given in the data booklet). Note the fourth power of temperature: doubling $T$ multiplies the luminosity by $2^4 = 16$.

2. The hotter the body, the shorter its peak wavelength — Wien's displacement law. The wavelength at which the spectrum peaks obeys:

$\lambda_{\text{max}} = \frac{b}{T}, \qquad b = 2.9\times10^{-3}\,\text{m K}$

So a hot, blue star peaks at a short wavelength, while a cool, red star peaks at a longer wavelength. Measuring $\lambda_{\text{max}}$ from a star's spectrum immediately gives its surface temperature.

Both laws demand kelvin. Because $T$ is raised to the fourth power in Stefan–Boltzmann and sits in a ratio with $b$ in Wien, using Celsius gives a wildly wrong answer. Always convert first.

Luminosity $L$ is the total power a star radiates in every direction. But an observer on Earth does not receive all of it — only the tiny fraction that lands on each square metre where they are. That received power per unit area is the apparent brightness $b$, and we can derive its formula from first principles.

Step 1 — the power spreads out over a sphere. Imagine the star at the centre of a giant imaginary sphere of radius $d$ (your distance from the star). By the time the light reaches you, the star's total power $L$ has spread evenly over the whole surface of that sphere (assuming nothing absorbs it on the way).

Step 2 — divide the power by the sphere's area. The surface area of a sphere of radius $d$ is $4\pi d^2$. So the power arriving on each square metre is the total power divided by that area:

$b = \frac{L}{4\pi d^2}$

Why this matters. The $d^2$ in the denominator means apparent brightness obeys an inverse-square law: move twice as far away and the star looks four times fainter. Two stars can look equally bright either because they have the same luminosity at the same distance, or because a very luminous distant star and a modest nearby star happen to deliver the same power per square metre. Combining this relationship with the Stefan–Boltzmann law (to get $L$ from $r$ and $T$) is a classic multi-step Paper 2 question.