Key Idea: Topic 2.1 is about how thermal energy is stored in matter and how it moves from hot to cold. It ties together four ideas: what internal energy is (and the particle model), how much energy a temperature change needs (Q = mcΔT), the hidden energy of a state change (Q = mL), and the three ways heat travels — conduction, convection and radiation. It is examined on Paper 1A (quick MCQs — define internal energy, compare densities, spot which formula a heating-curve part needs) and on Paper 2 (rearrange Q = mcΔT, energy-balance/calorimetry with latent heat, and the conduction rate ΔQ/Δt = kA·ΔT/Δx with its unit, the watt).
📐 Key formulas (all four are given)
Every equation in this topic is given in the data booklet — so you do not memorise them, but you must know which one to reach for and how to rearrange it.
- density (kg m⁻³)
- mass (kg)
- volume (m³)
- thermal energy added or removed (J)
- mass (kg)
- specific heat capacity (J kg⁻¹ K⁻¹)
- temperature change (K, or °C — same size)
- thermal energy transferred (J)
- mass changing state (kg)
- specific latent heat (J kg⁻¹)
- rate of heat flow — energy each second (W, i.e. J s⁻¹)
- thermal conductivity of the material (W m⁻¹ K⁻¹)
- cross-sectional area the heat flows through (m²)
- temperature difference across the slab (K or °C)
- thickness of the slab (m)
🧭 Which equation, and when?
The single most-tested decision in this topic: is the temperature changing (a slope) or is the state changing (a flat plateau at constant temperature)?
🔥 The three ways heat travels
🧊 Latent heats — fusion vs vaporisation
✍️ IB-style worked examples
IB-style question — energy to warm water (Q = mcΔT)
A heater warms 0.40 kg of water from 15 °C to 65 °C. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹. Calculate the thermal energy supplied.
Solution:
There is no change of state, so use the given specific-heat equation:
Find the temperature change first (final − start = 65 − 15):
Substitute m = 0.40, c = 4200, ΔT = 50:
Work it out — keep the unit:
Q = 8.4 × 10⁴ J (84 kJ). ΔT is the CHANGE (50 K), never the actual temperature.
IB-style question — warm then melt (Q = mcΔT then Q = mL)
0.20 kg of ice at −10 °C is heated until it is water at 0 °C. Take c(ice) = 2.1 × 10³ J kg⁻¹ K⁻¹ and L(fusion) = 3.3 × 10⁵ J kg⁻¹. Find the total energy needed.
Solution:
Step 1 — warm the ice from −10 °C to 0 °C (a sloping part), use the given formula:
Step 2 — melt the ice at 0 °C (a flat part, constant temperature), use the given latent-heat formula:
Add one Q-term per step:
Work it out — keep the unit:
Total ≈ 7.0 × 10⁴ J. Melting (Q₂) dominates — the state change costs far more than the 10 °C warming step.
IB-style question — rate of conduction (ΔQ/Δt = kA·ΔT/Δx)
Heat conducts through a wall of area 5.0 m² and thickness 0.25 m, with k = 0.60 W m⁻¹ K⁻¹. Inside is 21 °C and outside is 6 °C. Calculate the rate of heat loss, give its unit, and state what happens to it if the wall is made twice as thick.
Solution:
Use the given conduction equation:
Find the temperature difference (21 − 6 = 15):
Substitute k = 0.60, A = 5.0, Δx = 0.25:
Work it out — a rate of energy is in watts (W):
Δx is on the bottom, so doubling the thickness halves the rate:
≈ 1.8 × 10² W (180 W). Doubling the thickness halves the rate to ≈ 90 W — thicker walls insulate better.
IB-style question — calorimetry: mass of ice melted
0.15 kg of water at 40 °C is poured onto ice already at 0 °C; the water cools to 0 °C and some ice melts. Take c(water) = 4.2 × 10³ J kg⁻¹ K⁻¹ and L(fusion) = 3.3 × 10⁵ J kg⁻¹. Assuming no energy is lost, find the mass of ice melted.
Solution:
Energy released by the cooling water (temperature change), use the given formula:
No losses ⇒ energy lost by the water = energy gained by the ice, which melts at constant temperature:
Rearrange for the mass of ice melted:
Work it out — keep the unit:
About 0.076 kg (76 g) of ice melts. The water's lost heat (Q = mcΔT) becomes the ice's gained latent heat (Q = mL).
✅ Quick self-check
Tap each card to reveal the answer.
🎯 Highest-yield exam reminders
Exam Tips
- Internal energy = random KE + intermolecular PE — always name BOTH parts; never forget the PE. Temperature tracks only the KE part.
- ΔT in Q = mcΔT is a temperature CHANGE (final − start), and a change in K equals a change in °C — never convert ΔT to kelvin.
- Decide slope vs flat: temperature changing ⇒ Q = mcΔT; state changing at constant temperature ⇒ Q = mL (no ΔT). Multi-step problems need one Q-term per step.
- Lv ≫ Lf for the same substance, so boiling needs much more energy than melting — that is the longer plateau and the reason steam burns are worse than hot-water burns.
- Calorimetry with no losses: energy lost by the hot part = energy gained by the cold part. A measured value is usually 'off' because heat escapes to the surroundings or the container.
- The conduction rate ΔQ/Δt = kA·ΔT/Δx is a RATE — its unit is the watt (W). Work out ΔT first; thickness Δx is on the bottom, so a thicker layer conducts more slowly (rate ∝ 1 ÷ thickness).
- A cooling curve flattens because the temperature difference driving the heat loss keeps shrinking — smaller difference, smaller gradient. Only radiation crosses a vacuum.