Thermal Physics

Thermal Physics is the second topic in the Cambridge IGCSE Physics (0625) syllabus. It typically supplies 6 to 8 marks on Paper 1 or 2 and at least one full structured question on Paper 3 or 4. Students who master two equations and one model can secure most of those marks.

What does Thermal Physics cover in IGCSE 0625?

Four strands: the kinetic particle model of solids, liquids and gases; thermal properties (specific heat capacity, latent heat, expansion); gas behaviour with the Kelvin scale; and thermal energy transfer by conduction, convection and radiation. Extended candidates add the gas pressure, specific heat capacity and specific latent heat calculations. Core candidates stay descriptive on those areas.

The particle model is the spine of the whole topic. Every explanation question (why gases exert pressure, why evaporation cools, why solids expand) earns marks only when you answer in terms of particles. Examiners award nothing for “heat rises” or “cold gets in”.

Why do students lose marks here?

Three repeat offenders. First, mixing up heat and temperature: thermal energy is measured in joules, temperature in °C or kelvin. Second, dropping the $\Delta$ in $\Delta E = mc\Delta\theta$ and substituting a single temperature instead of the change. Third, writing vague explanation answers. “Particles vibrate more” scores; “it gets hotter” does not. On six-mark conduction, convection and radiation questions, mark schemes demand the mechanism, not the name.

There is also a sigh of relief for Core students: $pV = \text{constant}$, $c = \dfrac{\Delta E}{m\Delta\theta}$ and $L = \dfrac{E}{m}$ sit on the Extended (Supplement) tier only. Check your tier before revising. Our subtopic pages flag this on every page.

How should you revise it?

Learn the three equations cold: $\Delta E = mc\Delta\theta$, $E = mL$, and $pV = \text{constant}$ (all three Extended). Then drill explanation answers. Write a four-line particle-model explanation for melting, boiling, evaporation, conduction, convection and radiation. Compare each against a real mark scheme. Finally, practise the standard experiments: measuring specific heat capacity of a metal block appears regularly on Paper 6.

Worth knowing for Malaysian candidates: thermal questions love local contexts. Sea breezes, car radiators and why a tiled floor feels cold are all convection and conduction in disguise.

Our tutors give Thermal Physics roughly one week of an eight-week revision cycle, because its mark-per-hour payoff is high. That is 1-to-1 at RM80/hr, one tutor on one student. Your first hour is a free taught trial, and it can target this exact topic. Message us on WhatsApp.

How Thermal Physics Is Assessed Across the Papers

Thermal Physics is compact but reliable, so it appears every session. Papers 1 (Core) and 2 (Extended) usually carry two or three multiple-choice items: a particle-model picture, a conduction or convection scenario, or a one-line specific heat calculation. Papers 3 (Core) and 4 (Extended) give at least one structured question, often worth 6 to 10 marks, mixing a particle-model explanation with a numerical part. The Core and Extended split is sharp here. Core candidates describe the ideas qualitatively, while the Extended (Supplement) tier owns the real calculations: specific heat capacity $\left(c = \dfrac{\Delta E}{m\Delta\theta}\right)$, specific latent heat $\left(L = \dfrac{E}{m}\right)$, and the gas law $pV = \text{constant}$ at fixed temperature. Check your tier before you revise the maths. Papers 5 (Practical) and 6 (Alternative to Practical) test the cooling-curve and heat-capacity experiments: reading a thermometer, plotting temperature against time, and spotting the plateau during a change of state.

A Worked Example That Shows the Method

A heater supplies energy to $0.50\ \text{kg}$ of water, raising its temperature from $20\ \text{°C}$ to $80\ \text{°C}$. The specific heat capacity of water is $4200\ \text{J/(kg °C)}$. Calculate the energy transferred. [3]

Worked solution:

$\Delta\theta = 80 - 20 = 60\ \text{°C}$

$\Delta E = mc\Delta\theta = 0.50 \times 4200 \times 60 = 126\,000\ \text{J}$

So the energy transferred is $126\,000\ \text{J}$, which is $126\ \text{kJ}$ (3 significant figures). The single most common error here is substituting a temperature instead of a temperature change, so calculate $\Delta\theta$ as a separate first step and write it down. Note that a change of $60\ \text{°C}$ equals a change of $60\ \text{K}$, because the two scales share the same size of degree, so the substitution works whichever scale the question prefers.

Mark scheme:

• M1: $\Delta E = mc\Delta\theta$ with correct substitution (using $\Delta\theta = 60$, not $80$)
• A1: $126\,000\ \text{J}$ with the unit
• B1: correct conversion to $126\ \text{kJ}$, or answer to a sensible number of significant figures

Start with the full equations list for this topic, then work through our guide to answering physics calculation questions step by step.