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Physics (IAL)

Thermal Energy Transfer

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Edexcel IAL Physics · Unit 3

Thermal Energy Transfer

The big idea: Heating something either makes its particles jiggle faster (temperature goes up) or breaks the bonds holding them together (state changes) — and it can never do both at the same time.

Summary — What This Chapter Covers

Specific heat capacity — how much energy it takes to warm 1 kg of a substance by 1 K, and why water and metals behave so differently when heated.
Specific latent heat — the "hidden" energy needed to change state (melt or boil something) without changing its temperature at all.
Core Practical 12 — calibrating a thermistor, i.e. turning a resistance reading into a trustworthy temperature reading.
Core Practical 13 — experimentally measuring the specific latent heat of fusion of ice, including how to correct for heat lost to the surroundings using a control funnel.

1. Specific Heat Capacity

What's actually going on?

Every substance is made of particles that are constantly jiggling around — vibrating in a solid, sliding past each other in a liquid, or zooming freely in a gas. Heat energy is just the total kinetic and potential energy tied up in all that jiggling.

When you pump heat energy into a substance, the particles move faster on average. We measure "how fast the particles are moving on average" using temperature. So more heat in → faster particles → higher temperature. Simple so far.

But here's the catch that trips people up: not every substance heats up at the same rate for the same energy input. Give copper and water the exact same amount of energy, and copper's temperature will shoot up while water barely warms at all. Why? Because different materials have different molecular structures — different numbers of ways their particles can store energy (vibrating, rotating, bond-stretching etc.), and different strengths of intermolecular bonds. Water is unusually good at "soaking up" energy without its temperature rising much — this is exactly what specific heat capacity measures.

Real-world analogy

Think of specific heat capacity like the size of a bucket. A big bucket (water, c = 4200) needs a lot of water poured in before the level (temperature) rises noticeably. A small bucket (copper, c = 390) fills up — and its level rises — much faster for the same pour. That's why a metal spoon in hot tea gets hot almost instantly, but the tea itself stays hot for ages.

Definition to know word-for-word: Specific heat capacity is the energy required to raise the temperature of one kilogram of a substance by one kelvin.

The temperature rise of an object heated by a certain amount of energy depends on three things:

  • The amount of heat energy transferred to it
  • The mass of the object (more mass = more particles to speed up = smaller temperature rise for the same energy)
  • The specific heat capacity of the material it's made from

The formula

ΔE = mcΔθ
ΔE = change in heat energy (J)  |  m = mass (kg)  |  c = specific heat capacity (J kg⁻¹ K⁻¹)  |  Δθ = change in temperature (K or °C)
Watch out

Δθ is a change in temperature, so it's the same numerical value whether you work in kelvin or Celsius (a 5 K rise = a 5 °C rise). Don't waste time converting Celsius to Kelvin for Δθ — you only need to convert if you're given an absolute temperature, not a change.

Why specific heat capacity values differ so much

SubstanceSpecific Heat Capacity (J kg⁻¹ K⁻¹)
Water4200
Ice2200
Aluminium900
Copper390
Gold130

Good electrical conductors like copper are also excellent conductors of heat because their free electrons transfer energy quickly, and they tend to have low specific heat capacities — they warm up and cool down fast. Water's very high specific heat capacity is exactly why it's used in radiators: it stores huge amounts of energy and releases it slowly, keeping a room warm for a long time.

Worked example

Water of mass 0.48 kg is increased in temperature by 0.7 K. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹. Calculate the energy transferred.

Step 1 — List knowns: m = 0.48 kg, Δθ = 0.7 K, c = 4200 J kg⁻¹K⁻¹

Step 2 — Write the equation: ΔE = mcΔθ

Step 3 — Substitute: ΔE = 0.48 × 4200 × 0.7 = 1411.2

Step 4 — Round sensibly: ΔE ≈ 1400 J

Practice Question 1

A 2.0 kg aluminium block (c = 900 J kg⁻¹ K⁻¹) is heated from 20 °C to 65 °C. Calculate the energy transferred to the block.

Practice Question 2

A metal block of mass 0.5 kg absorbs 4500 J of energy and its temperature rises by 25 °C. Calculate its specific heat capacity, and suggest which metal (from the table above) it might be.

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Also in the full note
  • 2. Specific Latent Heat
  • 3. Core Practical 12: Calibrating a Thermistor
  • 4. Core Practical 13: Investigating Specific Latent Heat
  • What to Memorise
  • Concepts Checklist
  • Exam Tips
  • Why does temperature "pause" during melting/boiling?
  • Changes of state — the six names
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