IGCSE Physics revision · Thermal physics
Thermal properties of matter
Home of the two most misused ideas on the paper: specific heat capacity and the difference between boiling and evaporation. The calculation is routine; the marks die in the explanations.
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What the syllabus demands
- —Describe thermal expansion of solids, liquids and gases and its applications
- —Define specific heat capacity and use E = mcΔθ
- —Describe melting and boiling in terms of energy input without temperature change
- —Distinguish boiling from evaporation
- —Explain the factors that affect the rate of evaporation and its cooling effect
Definitions that earn marks
Clear definitions to practise — check your course mark scheme
- Specific heat capacity
- The energy required to raise the temperature of 1 kg of a substance by 1 °C.
- Melting point
- The temperature at which a solid changes to a liquid, with no temperature change while the change of state occurs.
- Evaporation
- The escape of the most energetic particles from the surface of a liquid, at any temperature below boiling — leaving the remaining liquid cooler.
The equations
More equations to practise: the IGCSE formula sheet.
Where the marks die
Common mistakes to check
- 01
Using the full temperature instead of the temperature change in E = mcΔθ. It is Δθ — final minus initial.
- 02
Saying temperature rises during melting or boiling. It does not: the energy input goes into breaking bonds between particles (increasing potential energy), not into raising temperature. Flat sections on a heating curve are worth marks only if you say why.
- 03
Confusing evaporation with boiling. Boiling happens at one temperature, throughout the liquid, with bubbles. Evaporation happens at any temperature, only at the surface.
- 04
Explaining evaporation's cooling effect without 'most energetic': the fastest particles escape, so the average kinetic energy of those remaining falls — and average kinetic energy is temperature.
One worked example, done properly
Question
How much energy is needed to heat 2.0 kg of water from 20 °C to 70 °C? (c of water = 4200 J/(kg °C))
Method
- 1.Δθ = 70 − 20 = 50 °C.
- 2.E = mcΔθ = 2.0 × 4200 × 50.
E = 420,000 J = 420 kJ