Thermal Physics — IB Physics SL Study Guide

For: IB Physics SL candidates sitting IB Physics SL — Theme B (The particulate nature of matter).

Covers: Temperature and absolute Kelvin scale, internal energy, specific heat capacity, phase changes and latent heat, ideal gas equation, kinetic theory of gases, first law of thermodynamics, and heat transfer mechanisms.

You should already know: Basic algebra, units (kelvin, joule), particle model of matter.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Physics SL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Is Thermal Physics?

Thermal physics is the study of energy transfer between systems and the resulting changes in temperature, phase, or microscopic particle behavior, forming the core of Theme B (the particulate nature of matter) in the 2024 IB Physics SL syllabus. It bridges observable macroscopic properties (pressure, temperature, volume) to the unobserved motion and interactions of atoms and molecules, with applications ranging from energy production to astrophysics.

2. Temperature and absolute (Kelvin) scale

Temperature is a scalar quantity that measures the average translational kinetic energy of particles in a system, and determines the direction of spontaneous heat flow (always from higher to lower temperature systems). The Celsius scale is calibrated to the freezing (0°C) and boiling (100°C) point of pure water at 1 atm pressure, but it is a relative scale that can take negative values.

The absolute Kelvin scale is defined relative to absolute zero (0 K), the theoretical temperature at which all random particle motion stops and a system has the minimum possible internal energy. No negative values exist on the Kelvin scale, and it is required for all gas law and thermodynamics calculations in the exam. The conversion formula is: $$T_K=T_C+273$$ Note that a change in temperature measured in Celsius is identical to the change measured in Kelvin: $\Delta T_K=\Delta T_C$, so you do not need to add 273 when calculating temperature differences.

Worked example: A sample of gas is cooled from 127°C to 27°C. What is the initial temperature in Kelvin, and the magnitude of the temperature change?

• Initial T: $T_K=127+273=400$ K
• Temperature change: $\Delta T=127-27=100$ °C = 100 K

Exam tip: Examiners explicitly award marks for converting temperature to Kelvin in gas law questions, so always show this step in your working.

3. Internal energy

Internal energy ($U$) is the total sum of all microscopic kinetic energy (translational, rotational, vibrational motion of particles) and intermolecular potential energy (from attractive/repulsive forces between particles) of all particles in a system. Crucially, internal energy does not include the macroscopic kinetic or potential energy of the system as a whole: for example, the kinetic energy of a moving hot drink, or the gravitational potential energy of a water tank on a roof, are not counted in the internal energy of the water.

Internal energy changes in three scenarios:

1. Heat is transferred to or from the system
2. Work is done on or by the system
3. The system undergoes a phase change (even if temperature stays constant, potential energy changes as intermolecular bonds are broken or formed)

Example: When 1 kg of ice melts to liquid water at 0°C, its internal energy increases by 334 kJ, even though its temperature (and average kinetic energy) stays the same. All added energy goes to breaking the rigid intermolecular bonds in ice, increasing the potential energy of the water molecules.

4. Specific heat capacity

When heat is added to a substance that does not change phase, its temperature increases. The specific heat capacity ($c$) of a substance is defined as the amount of heat required to raise the temperature of 1 kg of the substance by 1 K (or 1°C), with units $J\cdot k{g}^{-1}\cdot K^{-1}$. Specific heat capacity is an intrinsic property, so it does not depend on the mass of the sample.

The relationship between heat transferred ($Q$), mass, specific heat capacity, and temperature change is: $$Q=mc\Delta T$$ Where $Q$ is positive if heat is added to the system, and negative if heat is removed from the system. Values of $c$ for common substances are provided in the IB data booklet, so you do not need to memorize them.

Worked example: How much heat must be removed to cool 3 kg of olive oil (c = 1970 $J\cdot k{g}^{-1}\cdot K^{-1}$) from 180°C to 20°C for storage?

1. Calculate temperature change: $\Delta T=20-180=-160$ K
2. Plug into formula: $Q=3\ast 1970\ast (-160)=-945600$ J = -946 kJ The negative sign indicates 946 kJ of heat is removed from the oil.

5. Phase changes and latent heat

Phase changes (solid ↔ liquid ↔ gas) occur at a fixed temperature for a pure substance at constant pressure. During a phase change, all heat transferred to or from the system goes to changing intermolecular potential energy, so average kinetic energy (and temperature) stays constant.

Specific latent heat ($L$) is the energy per unit mass required to change the phase of a substance without changing its temperature, with units $J\cdot k{g}^{-1}$. The two key values you will encounter are:

• Specific latent heat of fusion ($L_f$): Energy to melt 1 kg of solid to liquid, or released when 1 kg of liquid freezes to solid
• Specific latent heat of vaporization ($L_v$): Energy to vaporize 1 kg of liquid to gas, or released when 1 kg of gas condenses to liquid

The formula for heat transferred during a phase change is: $$Q=mL$$ Where $Q$ is positive for melting/vaporization (heat added), and negative for freezing/condensation (heat removed). $L_v$ is always much larger than $L_f$ for all substances, because vaporization requires breaking all intermolecular bonds, while fusion only loosens bonds between particles.

Worked example: How much energy is required to convert 200 g of liquid water at 100°C to steam at 100°C, if $L_v$ for water is $2.26\times 10^6$ $J\cdot k{g}^{-1}$?

1. Convert mass to kg: 200 g = 0.2 kg
2. Plug into formula: $Q=0.2\ast 2.26\times 10^6=4.52\times 10^5$ J = 452 kJ

6. Ideal gas equation

An ideal gas is a theoretical model that simplifies calculations for real gases at low pressure and high temperature (far from their condensation point). The key assumptions of the ideal gas model are:

1. No intermolecular forces between gas particles
2. The volume of individual gas particles is negligible compared to the total volume of the gas
3. Collisions between particles and the container walls are perfectly elastic (no kinetic energy is lost)
4. Particles move in constant, random straight-line motion

The ideal gas equation relates the macroscopic properties of an ideal gas: $$pV=nRT$$ Where:

• $p$ = pressure in Pascals (Pa)
• $V$ = volume in cubic meters ($m^3$)
• $n$ = number of moles of gas
• $R$ = universal gas constant = 8.31 $J\cdot mo{l}^{-1}\cdot K^{-1}$ (given in data booklet)
• $T$ = absolute temperature in Kelvin (K)

For a fixed amount of gas (n constant), you can use the combined gas law to relate conditions before and after a change: $$\frac{p_1{V}_1}{T_1}=\frac{p_2{V}_2}{T_2}$$

Worked example: 0.5 moles of nitrogen gas are stored in a 0.002 $m^3$ container at 27°C. What is the pressure of the gas?

1. Convert temperature to Kelvin: $27+273=300$ K
2. Rearrange formula to solve for p: $p=\frac{nRT}{V}$
3. Substitute values: $p=\frac{0.5\ast 8.31\ast 300}{0.002}=623250$ Pa ≈ $6.2\times 10^5$ Pa

7. Kinetic theory of gases

The kinetic theory of gases links the microscopic behavior of ideal gas particles to their macroscopic measurable properties. The core derived result of the model is that the average translational kinetic energy of gas particles is directly proportional to the absolute temperature of the gas: $$⟨E_k⟩\propto T$$ This means all ideal gases at the same temperature have identical average kinetic energy, regardless of their molar mass. Lighter gas molecules (e.g., hydrogen, molar mass 2 g/mol) will have higher average speeds than heavier molecules (e.g., oxygen, molar mass 32 g/mol) at the same temperature, to keep their average kinetic energy equal.

The full formula for average kinetic energy is $⟨E_k⟩=\frac{3}{2}{k}_{B}T$, where $k_B$ is the Boltzmann constant (given in the data booklet). You do not need to derive this formula for SL, but you must understand the proportionality between temperature and average kinetic energy.

Exam tip: Examiners often ask you to explain why the pressure of a fixed volume of gas increases when its temperature rises, using kinetic theory: higher temperature = higher average particle speed = more frequent and more forceful collisions with the container walls = higher pressure.

8. First law of thermodynamics

The first law of thermodynamics is the law of conservation of energy applied to thermal systems. It states that the net heat added to a system is equal to the sum of the change in internal energy of the system and the net work done by the system on its surroundings: $$Q=\Delta U+W$$

The IB sign conventions (critical for scoring full marks) are:

• $Q$ = positive if heat is added to the system, negative if heat is removed
• $W$ = positive if work is done by the system (e.g., gas expands pushing a piston outwards), negative if work is done on the system (e.g., piston compresses the gas)
• $\Delta U$ = positive if internal energy increases (temperature rises or phase change to higher energy state), negative if internal energy decreases

Worked example: 150 J of work is done on a fixed mass of gas, and 80 J of heat is lost to the surroundings. What is the change in internal energy of the gas?

1. Identify values: $Q=-80$ J (heat lost), $W=-150$ J (work done on the system, so W is negative)
2. Rearrange formula: $\Delta U=Q-W$
3. Calculate: $\Delta U=-80-(-150)=70$ J The internal energy of the gas increases by 70 J, so its temperature will rise.

9. Heat transfer mechanisms

Heat is energy transferred between systems at different temperatures, and moves spontaneously from higher to lower temperature. There are three distinct transfer mechanisms, which examiners often ask you to identify in context:

1. Conduction: Heat transfer through a solid or stationary fluid without bulk movement of the substance, occurring via collisions between adjacent high-energy and low-energy particles. Metals are excellent conductors due to free electrons, while gases and insulators (e.g., wood, plastic) are poor conductors.
2. Convection: Heat transfer via bulk movement of liquids or gases (fluids). Warmer fluid is less dense, so it rises, while cooler, denser fluid sinks, creating convection currents that distribute heat. Examples include heating water in a pot, or global atmospheric wind patterns.
3. Radiation: Heat transfer via infrared electromagnetic waves, which does not require a medium and can travel through a vacuum. All objects emit and absorb thermal radiation: dark, matte surfaces are perfect emitters and absorbers, while shiny, light surfaces are poor emitters/absorbers and good reflectors. The heat from the Sun reaching Earth is an example of radiation.

10. Common Pitfalls (and how to avoid them)

• Wrong move: Using Celsius temperature in the ideal gas equation or average kinetic energy calculations. Why students do it: They forget these formulas rely on absolute temperature relative to absolute zero. Correct move: Always convert temperature to Kelvin by adding 273 before using these formulas, and show the conversion step to earn method marks.
• Wrong move: Using $Q=mc\Delta T$ for phase change calculations. Why students do it: They assume adding heat always increases temperature. Correct move: Split multi-step heating/cooling problems into separate steps: heat to reach phase change temperature, latent heat for phase change, then heat to change temperature of the new phase.
• Wrong move: Mixing up sign conventions for the first law of thermodynamics, especially work done on vs by the system. Why students do it: Different resources use different conventions. Correct move: Write the IB sign convention at the top of your working for every first law question: Q = + in, W = + out, ΔU = + increase.
• Wrong move: Using grams instead of kilograms for mass in specific heat or latent heat calculations. Why students do it: Questions often give mass in grams, and units for c/L are per kg. Correct move: Always check units, and convert mass to kg by dividing by 1000 before plugging into formulas.
• Wrong move: Confusing internal energy with the total energy of a system. Why students do it: They include macroscopic kinetic/potential energy of the whole object. Correct move: Remember internal energy only counts the microscopic kinetic and potential energy of particles inside the system, not the motion or position of the system as a whole.

11. Practice Questions (IB Physics SL Theme B Style)

Question 1

A 2 kg block of copper (c = 385 $J\cdot k{g}^{-1}\cdot K^{-1}$) is heated from 25°C to its melting point of 1085°C. It then absorbs 4.1 × 10⁵ J of energy to melt completely to liquid copper at 1085°C. a) Calculate the total energy required to heat the block from 25°C to fully melted liquid. b) Calculate the specific latent heat of fusion for copper.

Solution 1

a) First calculate the energy to raise the temperature of the solid copper: $\Delta T=1085-25=1060$ K $Q_1=mc\Delta T=2\ast 385\ast 1060=816200$ J = 8.16 × 10⁵ J Total energy = $Q_1+Q_{\text{fusion}}=8.16\times 10^5+4.1\times 10^5=1.23\times 10^6$ J (3 sig figs)

b) Rearrange $Q=mL$ to solve for $L_f$: $L_f=\frac{Q}{m}=\frac{4.1\times 10^5}{2}=2.05\times 10^5$ $J\cdot k{g}^{-1}$

Question 2

A 0.01 $m^3$ scuba tank contains 40 moles of compressed air at 27°C. Assume air behaves as an ideal gas. a) Calculate the pressure inside the tank. b) If the tank is left in the sun and its temperature rises to 47°C, what is the new pressure inside the tank, assuming no gas escapes?

Solution 2

a) Convert temperature to Kelvin: $T_1=27+273=300$ K Use $pV=nRT$: $p_1=\frac{nR{T}_1}{V}=\frac{40\ast 8.31\ast 300}{0.01}=9972000$ Pa ≈ 1.0 × 10⁷ Pa (2 sig figs)

b) New temperature: $T_2=47+273=320$ K, n and V are constant, so $\frac{p_1}{T_1}=\frac{p_2}{T_2}$ $p_2=p_1\ast \frac{T_2}{T_1}=9.972\times 10^6\ast \frac{320}{300}\approx 1.06\times 10^7$ Pa ≈ 1.1 × 10⁷ Pa

Question 3

A gas in a piston expands, doing 200 J of work on the surroundings, while absorbing 350 J of heat from the environment. a) State the first law of thermodynamics. b) Calculate the change in internal energy of the gas. c) State whether the temperature of the gas increases, decreases, or stays the same, justifying your answer.

Solution 3

a) The first law of thermodynamics states that the heat added to a system is equal to the sum of the change in internal energy of the system and the work done by the system, expressed as $Q=\Delta U+W$.

b) $Q=+350$ J (heat added), $W=+200$ J (work done by the gas) $\Delta U=Q-W=350-200=150$ J Internal energy increases by 150 J.

c) The temperature of the gas increases. Internal energy of an ideal gas is directly proportional to its absolute temperature, so an increase in internal energy corresponds to an increase in temperature.

12. Quick Reference Cheatsheet

13. What's Next

Thermal physics is a foundational topic that connects directly to multiple later IB Physics SL units, including Theme D (Energy Production), where you will apply the first law of thermodynamics and heat transfer principles to calculate the efficiency of fossil fuel, nuclear, and renewable power plants. If you choose the Astrophysics optional theme, you will also use thermal radiation and temperature relationships to calculate stellar luminosity, surface temperature, and lifetime. Mastering the unit conversion and sign convention rules covered here will also eliminate common mistakes in these later topics, and give you a strong base if you pursue physics, engineering, or environmental science at university.

To reinforce your understanding, work through official IBO past paper questions for Theme B, paying close attention to mark scheme wording for explanation questions (e.g., the difference between heat and temperature, or why temperature stays constant during phase changes). If you get stuck on any problem, or want to generate custom practice quizzes tailored to your weak spots, ask Ollie, our AI tutor, anytime on the homepage.