Thermal Physics — OCR A Level Physics Revision
Module 5 · Newtonian World and Astrophysics

Thermal Physics

Specification: OCR A H556  |  Section: 5.1  |  Teaching time: ~10 hours

By the end of this topic you should be able to…
Part 1

Thermal Concepts

Temperature

Temperature is a measure of how hot or cold an object is. On a microscopic level, it is related to the average kinetic energy of the particles in a substance. When two objects are at the same temperature, there is no net energy transfer between them — they are in thermal equilibrium.

💡 Key Point

Temperature is not the same as heat. Heat is energy transferred due to a temperature difference. Temperature is a measure of the average kinetic energy of particles. A spark at 1000 °C has a very high temperature but very little internal energy (few particles); a bath of water at 40 °C has much more internal energy despite a lower temperature.

The Kelvin Scale

The absolute temperature scale (kelvin, K) is the SI unit of temperature. It starts at absolute zero (0 K), the lowest possible temperature at which particles have minimum kinetic energy.

Celsius to Kelvin Conversion $$T\text{ (K)} = \theta\text{ (°C)} + 273$$

A change of 1 K is exactly equal to a change of 1 °C — the scales have the same interval size, just different starting points. At absolute zero (0 K = −273 °C), the particles in a substance have their minimum possible kinetic energy (but not zero — quantum zero-point energy remains).

⚠️ Common Error

Never write "degrees kelvin" or use the ° symbol with kelvin. Write 300 K, not 300 °K. Similarly, when using temperature differences in equations, $\Delta\theta$ in °C is numerically equal to $\Delta T$ in K — so you can use either.

Thermal Equilibrium

When two objects are placed in contact, energy transfers from the hotter object to the cooler one until they reach the same temperature. At this point they are in thermal equilibrium — no net energy transfer occurs.

This principle underpins the zeroth law of thermodynamics: if A is in thermal equilibrium with B, and B is in thermal equilibrium with C, then A is in thermal equilibrium with C. This is why thermometers work — the thermometer reaches thermal equilibrium with the object being measured.

Part 2

Internal Energy

Internal energy is the sum of the randomly distributed potential and kinetic energies of all molecules (or atoms) in a system. It is defined as:

Internal Energy $$U = \sum E_k\text{ (random)} + \sum E_p\text{ (random)}$$

The kinetic energy component is due to the random motion and vibration of molecules (translational, rotational, and vibrational). The potential energy component is due to the intermolecular forces between molecules.

The word randomly is important — internal energy does not include ordered kinetic energy (like a moving object) or ordered potential energy (like a raised object in a gravitational field). It only includes the disordered, random energy at the molecular level.

StateKinetic EnergyPotential EnergyInternal Energy
SolidLow (vibration only)Large and negative (strong bonds)Lowest
LiquidMediumMedium (weaker bonds)Intermediate
GasHigh (free motion)≈ 0 (negligible forces)Highest
💡 Key Point

For an ideal gas, the potential energy is zero because there are no intermolecular forces. This means the internal energy of an ideal gas is entirely kinetic, and is therefore directly proportional to its absolute temperature.

Changes of State

During a change of state (melting, boiling, freezing, condensing), the temperature stays constant even though energy is still being transferred. This is because the energy goes into breaking or forming intermolecular bonds (changing potential energy) rather than increasing kinetic energy.

Heating a substance from solid to gas produces a characteristic heating curve with flat regions during phase changes:

  • Solid — temperature rises as KE increases
  • Melting — temperature constant, PE increases (bonds breaking)
  • Liquid — temperature rises as KE increases
  • Boiling — temperature constant, PE increases (bonds breaking)
  • Gas — temperature rises as KE increases

Brownian Motion

Brownian motion is the random, jittery movement of smoke or pollen particles suspended in a fluid (gas or liquid). It occurs because the much smaller, fast-moving molecules of the fluid collide randomly with the larger particles.

⚡ Evidence for the Particle Model

Brownian motion provides direct evidence for the kinetic particle model of matter. The random motion of visible particles can only be explained if they are being bombarded by even smaller, invisible molecules moving randomly. Robert Brown first observed this in 1827 using pollen grains in water. The smaller the suspended particle, the greater the observed Brownian motion.

Check your understanding

Knowledge Check

1
Convert −40 °C to kelvin. Explain why −40 °F also equals −40 °C.
3 marks
  • $T = -40 + 273 = 233$ K ✓
  • The Fahrenheit and Celsius scales intersect at −40° (i.e. −40 °F = −40 °C is a coincidence of their definitions) ✓
2
Explain why the temperature of a substance remains constant while it is melting, even though energy is still being supplied.
3 marks
  • During melting, energy is used to break intermolecular bonds rather than increase kinetic energy ✓
  • This increases the potential energy component of internal energy ✓
  • Since temperature is related to average kinetic energy, and KE doesn't change, temperature stays constant ✓
3
Explain what is meant by Brownian motion and describe how it provides evidence for the kinetic particle model.
4 marks
  • Brownian motion is the random, erratic movement of visible particles suspended in a fluid ✓
  • Caused by bombardment by much smaller, fast-moving fluid molecules ✓
  • Large numbers of collisions from all directions lead to a resultant force that constantly changes in magnitude and direction ✓
  • This is evidence for the existence of molecules and their random motion, supporting the kinetic particle model ✓
Part 3

Specific Heat Capacity

The specific heat capacity $c$ of a substance is the energy required per unit mass to raise the temperature by 1 K (or 1 °C) without a change of state.

Specific Heat Capacity $$Q = mc\Delta\theta$$

Where $Q$ is the energy transferred (J), $m$ is mass (kg), $c$ is specific heat capacity (J kg⁻¹ K⁻¹), and $\Delta\theta$ is the temperature change (K or °C).

SubstanceSpecific Heat Capacity / J kg⁻¹ K⁻¹
Water4180
Aluminium902
Iron449
Copper385
Lead128
💡 Key Point

Water has a very high specific heat capacity (4180 J kg⁻¹ K⁻¹). This means it can absorb a lot of energy for a small temperature rise — which is why it's used in cooling systems and why coastal climates are more stable than inland ones.

Worked Example
An electric heater supplies 84 000 J of energy to 2.0 kg of water initially at 20 °C. Calculate the final temperature of the water. ($c_{\text{water}} = 4180$ J kg⁻¹ K⁻¹)
$\Delta\theta = Q / mc = 84\,000 / (2.0 \times 4180)$
$\Delta\theta = 84\,000 / 8360 = 10.05$ K $\approx 10$ K
Final temperature $= 20 + 10 = 30$ °C

Method of Mixtures

The method of mixtures is a technique for finding an unknown specific heat capacity. A known mass of a hot substance is mixed with a known mass of cold water in an insulated calorimeter. At thermal equilibrium:

Conservation of Energy (Method of Mixtures) $$m_1 c_1 \Delta\theta_1 = m_2 c_2 \Delta\theta_2 + m_c c_c \Delta\theta_c$$
⚠️ Common Error

In the method of mixtures, remember to include the calorimeter's heat capacity if it's not negligible. Also, some energy is always lost to the surroundings — this is the main source of systematic error and leads to a final temperature that is lower than predicted.

Part 4

Specific Latent Heat

When a substance changes state, energy is transferred without a change in temperature. The specific latent heat $l$ of a substance is the energy required per unit mass to change its state without a temperature change.

Specific Latent Heat $$Q = ml$$

Where $Q$ is the energy (J), $m$ is mass (kg), and $l$ is specific latent heat (J kg⁻¹).

There are two types:

  • Specific latent heat of fusion ($l_f$) — energy per unit mass to change from solid to liquid at the melting point
  • Specific latent heat of vaporisation ($l_v$) — energy per unit mass to change from liquid to gas at the boiling point
Substance$l_f$ / J kg⁻¹$l_v$ / J kg⁻¹
Water$3.34 \times 10^5$$2.26 \times 10^6$
Ethanol$1.05 \times 10^5$$8.55 \times 10^5$
Aluminium$3.95 \times 10^5$$1.05 \times 10^7$
💡 Key Point

The latent heat of vaporisation is always much larger than the latent heat of fusion. This is because vaporisation requires all intermolecular bonds to be completely broken (particles become free), whereas melting only weakens them enough to allow sliding past each other.

Worked Example
A kettle rated at 2.2 kW contains 0.50 kg of water at 100 °C. How long does it take to completely boil away the water? ($l_v = 2.26 \times 10^6$ J kg⁻¹)
$Q = ml = 0.50 \times 2.26 \times 10^6 = 1.13 \times 10^6$ J
$P = Q/t$, so $t = Q/P = 1.13 \times 10^6 / 2200$
$t = 514$ s $\approx 8.6$ minutes
Worked Example — Combined Heating
Calculate the total energy needed to convert 0.20 kg of ice at −20 °C to steam at 120 °C. ($c_{\text{ice}} = 2100$, $c_{\text{water}} = 4180$, $c_{\text{steam}} = 2010$ J kg⁻¹ K⁻¹; $l_f = 3.34 \times 10^5$, $l_v = 2.26 \times 10^6$ J kg⁻¹)
Heat ice from −20 °C to 0 °C: $Q_1 = 0.20 \times 2100 \times 20 = 8400$ J
Melt ice at 0 °C: $Q_2 = 0.20 \times 3.34 \times 10^5 = 66\,800$ J
Heat water from 0 °C to 100 °C: $Q_3 = 0.20 \times 4180 \times 100 = 83\,600$ J
Boil water at 100 °C: $Q_4 = 0.20 \times 2.26 \times 10^6 = 452\,000$ J
Heat steam from 100 °C to 120 °C: $Q_5 = 0.20 \times 2010 \times 20 = 8040$ J
$Q_{\text{total}} = 8400 + 66\,800 + 83\,600 + 452\,000 + 8040 = 618\,840$ J $\approx 619$ kJ
Check your understanding

Energy Transfer Check

1
A 0.50 kg block of aluminium at 200 °C is dropped into 1.0 kg of water at 20 °C inside an insulated calorimeter of heat capacity 50 J K⁻¹. Calculate the final temperature. ($c_{\text{Al}} = 902$, $c_{\text{water}} = 4180$ J kg⁻¹ K⁻¹)
5 marks
  • Energy lost by Al = Energy gained by water + calorimeter ✓
  • $0.50 \times 902 \times (200 - T) = 1.0 \times 4180 \times (T - 20) + 50 \times (T - 20)$ ✓
  • $451(200 - T) = 4230(T - 20)$ ✓
  • $90\,200 - 451T = 4230T - 84\,600$ ✓
  • $174\,800 = 4681T$, so $T = 37.3$ °C ✓
2
Explain why steam at 100 °C causes a more severe burn than water at 100 °C.
2 marks
  • Steam condenses on the skin, releasing its latent heat of vaporisation ✓
  • This transfers additional energy ($2.26 \times 10^6$ J kg⁻¹) beyond just the heat content of the water at 100 °C ✓
Part 5

Ideal Gas Laws

An ideal gas is a theoretical model that makes these assumptions:

  • The gas consists of many small particles moving in rapid, random motion
  • Collisions between particles and between particles and walls are perfectly elastic
  • The volume of each particle is negligible compared to the volume of the container
  • Intermolecular forces are negligible except during collisions
  • The duration of collisions is negligible compared to the time between collisions
💡 Key Point

Real gases behave like ideal gases at low pressures and high temperatures (well above their boiling point). Under these conditions, particles are far apart and moving fast, so intermolecular forces are negligible.

Boyle's Law

At constant temperature, the pressure of a fixed mass of gas is inversely proportional to its volume:

Boyle's Law (constant T) $$pV = \text{constant} \quad \text{or} \quad p_1V_1 = p_2V_2$$

A Boyle's law graph of $p$ against $1/V$ gives a straight line through the origin.

Charles's Law

At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature:

Charles's Law (constant p) $$\frac{V}{T} = \text{constant} \quad \text{or} \quad \frac{V_1}{T_1} = \frac{V_2}{T_2}$$

Temperature must be in kelvin. A graph of $V$ against $T$ (K) is a straight line through the origin. Extrapolating back to $V = 0$ gives $T = 0$ K — a theoretical justification for absolute zero.

The Pressure Law

At constant volume, the pressure of a fixed mass of gas is directly proportional to its absolute temperature:

Pressure Law (constant V) $$\frac{p}{T} = \text{constant} \quad \text{or} \quad \frac{p_1}{T_1} = \frac{p_2}{T_2}$$

The Ideal Gas Equation

Combining all three gas laws into one unified equation:

Ideal Gas Equation $$pV = nRT$$

Where $p$ is pressure (Pa), $V$ is volume (m³), $n$ is number of moles (mol), $R$ is the molar gas constant ($8.31$ J mol⁻¹ K⁻¹), and $T$ is absolute temperature (K).

Ideal Gas Equation (molecular form) $$pV = NkT$$

Where $N$ is the total number of molecules and $k$ is the Boltzmann constant ($1.38 \times 10^{-23}$ J K⁻¹). Since $n = N/N_A$, we have $R = N_A \times k$.

⚠️ Unit Traps
  • Pressure must be in pascals (Pa). If given in kPa or atm, convert first ($1$ atm $= 101\,325$ Pa).
  • Volume must be in m³. If given in cm³, divide by $10^6$. If in litres, divide by 1000.
  • Temperature must be in kelvin. Always add 273.
Worked Example
A gas cylinder of volume 0.040 m³ contains oxygen at a pressure of $2.5 \times 10^6$ Pa and temperature 280 K. Calculate the number of moles of oxygen and the number of molecules present.
$n = pV/RT = (2.5 \times 10^6 \times 0.040) / (8.31 \times 280)$
$n = 100\,000 / 2326.8 = 42.97 \approx 43$ mol
$N = n \times N_A = 42.97 \times 6.02 \times 10^{23} = 2.59 \times 10^{25}$ molecules
Part 6

Molecular Kinetic Theory

The gas laws can be derived from the behaviour of individual molecules. The kinetic theory model links macroscopic properties (pressure, temperature) to microscopic ones (molecular speed, mass).

Pressure from Molecular Collisions

When a molecule bounces off a wall, it reverses its momentum. By Newton's second and third laws, this produces a force on the wall. The total pressure on the wall is the sum of forces from all molecular collisions per unit area.

📐 Derivation of $pV = \tfrac{1}{3}Nm\langle c^2 \rangle$

Consider $N$ molecules in a cubic box of side $L$, each of mass $m$. One molecule moves with velocity components $(u_x, u_y, u_z)$.

When it collides with the wall perpendicular to $x$, momentum change $= 2mu_x$. Time between collisions with this wall $= 2L/u_x$.

Force on wall from one molecule: $F = \Delta p/\Delta t = 2mu_x / (2L/u_x) = mu_x^2/L$

Total force from $N$ molecules: $F = (m/L) \times \sum u_x^2 = mN\langle u_x^2 \rangle/L$ (using average)

Pressure $p = F/L^2 = mN\langle u_x^2 \rangle/L^3 = mN\langle u_x^2 \rangle/V$

Since $\langle c^2 \rangle = \langle u_x^2 \rangle + \langle u_y^2 \rangle + \langle u_z^2 \rangle$ and by symmetry $\langle u_x^2 \rangle = \langle u_y^2 \rangle = \langle u_z^2 \rangle = \langle c^2 \rangle/3$:

$pV = \tfrac{1}{3}Nm\langle c^2 \rangle$

Root Mean Square Speed

The root mean square speed $c_{\text{rms}}$ is a measure of the average molecular speed:

Root Mean Square Speed $$c_{\text{rms}} = \sqrt{\langle c^2 \rangle} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3RT}{M}}$$

Where $M$ is the molar mass (kg mol⁻¹). Note that $M$ must be in kg, not g — so for O₂, $M = 0.032$ kg mol⁻¹.

Worked Example
Calculate the root mean square speed of nitrogen molecules (N₂, $M = 0.028$ kg mol⁻¹) at 300 K.
$c_{\text{rms}} = \sqrt{3RT/M} = \sqrt{3 \times 8.31 \times 300 / 0.028}$
$c_{\text{rms}} = \sqrt{7479 / 0.028} = \sqrt{267\,100}$
$c_{\text{rms}} = 517$ m s⁻¹

Average Kinetic Energy

From $pV = \tfrac{1}{3}Nm\langle c^2 \rangle$ and $pV = NkT$, we can equate:

📐 Derivation of $E_k = \tfrac{3}{2}kT$

$pV = \tfrac{1}{3}Nm\langle c^2 \rangle$ and $pV = NkT$

Therefore: $\tfrac{1}{3}Nm\langle c^2 \rangle = NkT$

$\tfrac{1}{2}m\langle c^2 \rangle = \tfrac{3}{2}kT$

So the average translational kinetic energy per molecule:

$E_k = \tfrac{3}{2}kT$

💡 Key Point

The average kinetic energy of a gas molecule depends only on temperature, not on the mass of the molecule or the type of gas. At the same temperature, helium and xenon molecules have the same average kinetic energy — but helium moves much faster because it's lighter.

Interactive

Gas Laws Simulator

Ideal Gas: $pV = NkT$
Temperature 300 K
Volume 0.025 m³
Molecules 50
Gas type
Pressure $p$—
$c_{\text{rms}}$—
Avg KE per molecule—
Moles $n$—
Live Calculation
—
—
—

Adjust the sliders to explore how $T$, $V$, and $N$ affect pressure via $pV = NkT$. Compare $c_{\text{rms}}$ for different gases at the same temperature — lighter gases move faster. The live calculation panel shows the actual numbers plugged into the equation.

Practical

Investigating Gas Laws

Boyle's Law Experiment

A fixed mass of gas is trapped in a sealed syringe or column of oil. By changing the volume (using the syringe plunger or varying the pressure with a pump), and measuring the corresponding pressure with a Bourdon gauge:

  • Plot a graph of $p$ against $1/V$ — should be a straight line through the origin
  • Or plot $pV$ against $p$ — should be approximately constant (horizontal line)

Specific Heat Capacity Experiment

A known mass of a solid is heated by an electric heater. The energy supplied is $E = VIt$ (voltage × current × time). Measuring the temperature rise gives:

Determining $c$ from an electrical method $$c = \frac{VIt}{m\Delta\theta}$$
⚠️ Sources of Error
  • Energy losses to the surroundings — reduce with insulation
  • Non-uniform heating — stir liquid, or use a thermometer at the centre
  • The heater itself has heat capacity — include it in the calculation or pre-heat
Exam Practice

Exam-Style Questions

1
A student carries out an experiment to determine the specific latent heat of fusion of ice using an electrical heater.

(a) Describe the experimental procedure. [4 marks]
(b) The heater supplies 25 W for 4.0 minutes and completely melts 7.2 g of ice at 0 °C. Calculate the specific latent heat of fusion. [3 marks]
(c) The accepted value is $3.34 \times 10^5$ J kg⁻¹. Suggest a reason why the student's value may differ from the accepted value. [2 marks]
9 marks

(a) Procedure [4]:

  • Place ice in a funnel above a measuring cylinder / calorimeter ✓
  • Use an electrical heater of known power ($P = VI$) immersed in the ice ✓
  • Measure the mass of melted water collected over a measured time ✓
  • Ensure all energy goes to melting (no temperature rise, insulate the apparatus) ✓

(b) Calculation [3]:

  • Energy supplied: $Q = Pt = 25 \times 240 = 6000$ J ✓
  • $l = Q/m = 6000 / 0.0072$ ✓
  • $l = 8.33 \times 10^5$ J kg⁻¹ ✓

(c) Error [2]:

  • Energy lost to the surroundings from the heater ✓
  • This means not all electrical energy goes into melting the ice, so the calculated $l$ is too high ✓
2
A balloon contains 0.030 m³ of helium gas at a temperature of 290 K and pressure $1.1 \times 10^5$ Pa. ($M_{\text{He}} = 0.004$ kg mol⁻¹)

(a) Calculate the number of moles of helium in the balloon. [2 marks]
(b) Calculate the root mean square speed of the helium atoms. [3 marks]
(c) The balloon rises to an altitude where the temperature is 250 K and the pressure is $6.0 \times 10^4$ Pa. The balloon expands. Calculate the new volume. [3 marks]
(d) State and explain whether the average kinetic energy of the helium atoms has increased, decreased, or stayed the same. [2 marks]
10 marks

(a) [2]:

  • $n = pV/RT = (1.1 \times 10^5 \times 0.030) / (8.31 \times 290)$ ✓
  • $n = 3300 / 2409.9 = 1.37$ mol ✓

(b) [3]:

  • $c_{\text{rms}} = \sqrt{3RT/M}$ ✓
  • $c_{\text{rms}} = \sqrt{3 \times 8.31 \times 290 / 0.004}$ ✓
  • $c_{\text{rms}} = \sqrt{1\,807\,350} = 1344$ m s⁻¹ ✓

(c) [3]:

  • $p_1V_1/T_1 = p_2V_2/T_2$ ✓
  • $V_2 = p_1V_1T_2/(p_2T_1) = (1.1 \times 10^5 \times 0.030 \times 250)/(6.0 \times 10^4 \times 290)$ ✓
  • $V_2 = 825\,000 / 17\,400\,000 = 0.047$ m³ ✓

(d) [2]:

  • Average KE has decreased ✓
  • Because $E_k = \tfrac{3}{2}kT$, and temperature has decreased from 290 K to 250 K ✓
3
(a) State two assumptions of the kinetic theory of gases. [2 marks]
(b) Use the kinetic theory to explain why the pressure of a gas in a sealed container increases when the temperature is increased at constant volume. [4 marks]
(c) Calculate the average kinetic energy of a gas molecule at 350 K. [2 marks]
8 marks

(a) Any two [2]:

  • Molecules are point masses (negligible volume) ✓
  • Collisions are perfectly elastic ✓
  • No intermolecular forces except during collisions ✓
  • Molecules move in rapid random motion ✓

(b) Explanation [4]:

  • Increasing temperature increases the average speed of molecules ✓
  • Molecules collide with the walls more frequently (higher speed, shorter time between collisions) ✓
  • Each collision involves a greater change in momentum (molecules hit harder) ✓
  • Greater force per unit area on the walls → higher pressure ✓

(c) [2]:

  • $E_k = \tfrac{3}{2}kT = 1.5 \times 1.38 \times 10^{-23} \times 350$ ✓
  • $E_k = 7.25 \times 10^{-21}$ J ✓
4
A student investigates Boyle's law using a sealed syringe connected to a pressure sensor. The student records the following data:

Volume / cm³Pressure / kPa
20250
25198
30166
40124
50100
(a) Explain why the student should wait a few seconds after changing the volume before taking a reading. [2 marks]
(b) Plot a graph to verify Boyle's law. State what graph you would plot and what you would expect to see. [3 marks]
(c) Calculate $pV$ for each data point and comment on the results. [3 marks]
8 marks

(a) [2]:

  • Allow time for thermal equilibrium to be reached ✓
  • Compressing the gas does work on it, temporarily raising temperature and pressure above the Boyle's law value ✓

(b) [3]:

  • Plot pressure (y-axis) against $1/V$ (x-axis) ✓
  • Convert $V$ to m³ first (divide cm³ by $10^6$) ✓
  • Should give a straight line through the origin ✓

(c) [3]:

  • $pV$ values: $20 \times 250 = 5000$, $25 \times 198 = 4950$, $30 \times 166 = 4980$, $40 \times 124 = 4960$, $50 \times 100 = 5000$ ✓
  • $pV$ is approximately constant ($\approx 5000$ kPa cm³) ✓
  • Small variations due to experimental error, confirming Boyle's law ✓

Topic Summary

Temperature & Internal Energy

Temperature (K) measures average KE. Internal energy = sum of randomly distributed KE + PE of all molecules. For an ideal gas, PE = 0, so U is purely kinetic.

Heat Capacity & Latent Heat

$Q = mc\Delta\theta$ for temperature changes. $Q = ml$ for phase changes. $l_v > l_f$ always.

Gas Laws

Boyle: $pV = \text{const}$. Charles: $V/T = \text{const}$. Pressure law: $p/T = \text{const}$. Combined: $pV = nRT = NkT$.

Kinetic Theory

$pV = \tfrac{1}{3}Nm\langle c^2 \rangle$. $E_k = \tfrac{3}{2}kT$. $c_{\text{rms}} = \sqrt{3RT/M}$. Average KE depends only on $T$.

Equations to Know

$T\text{(K)} = \theta\text{(°C)} + 273$
$Q = mc\Delta\theta$
$Q = ml$
$p_1V_1 = p_2V_2$
$pV = nRT$
$pV = NkT$
$pV = \tfrac{1}{3}Nm\langle c^2 \rangle$
$E_k = \tfrac{3}{2}kT$
$c_{\text{rms}} = \sqrt{3RT/M}$