VCE Physics (Year 11) · Unit 1 — Thermodynamics · 12m read

Kinetic Theory and Specific Heat

Ever wonder why coastal towns stay warm in winter while inland areas freeze? It's all about specific heat—basically how much energy different materials need to change temperature. You'll learn to use the formula Q = mcΔT to calculate exactly how much heat energy is required to warm something up, which explains why water (with its huge specific heat) acts like Earth's natural temperature buffer. Mastering this unlocks real problems from climate science to cooking to designing safer buildings.

Learning objective: Use Q = mcΔT to solve thermal energy problems.

By the end of this lesson you can:

• I apply Q = mcΔT correctly.
• I use latent heat for phase changes.
• I solve energy balance problems.

Temperature is how fast particles dance. Heat a substance and its particles dance faster. Specific heat tells you how much energy it takes to speed up the dance.

Specific heat and phase change

KINETIC THEORY (recap):
• All matter made of particles in motion.
• T = measure of average kinetic energy per particle.

SPECIFIC HEAT CAPACITY (c):
• c = energy per kg per °C to raise T.
• Water: c = 4180 J/kg°C (huge!).
• Metals: c ~ 400-900 J/kg°C.

Q = m c ΔT
• Q = heat (J).
• m = mass (kg).
• c = specific heat (J/kg°C).
• ΔT = temperature change (°C).

LATENT HEAT:
• Energy absorbed during PHASE CHANGE without T change.
• LATENT HEAT OF FUSION (L_f) — solid ↔ liquid.
• LATENT HEAT OF VAPORISATION (L_v) — liquid ↔ gas.
• Q = m L.

WATER SPECIFIC VALUES:
• c_water = 4180 J/kg°C.
• c_ice = 2100 J/kg°C.
• c_steam = 2000 J/kg°C.
• L_fusion (water) = 334,000 J/kg.
• L_vaporisation (water) = 2,260,000 J/kg.

PHASE-CHANGE GRAPH:
• Heating ice at −10 °C → water → steam at 110 °C.
• Flat plateaus at 0 °C (melt) and 100 °C (boil) = latent heat.
• Sloped sections = Q = mcΔT.

ENERGY BALANCE:
• Q_lost = Q_gained (for isolated systems).
• E.g., hot metal in cold water → m₁c₁(T_hot − T_f) = m₂c_water(T_f − T_cold).

WHY WATER HAS HIGH c:
• Hydrogen bonding requires extra energy to break.
• Huge thermal inertia — oceans moderate climate.

EFFICIENCY:
• Real systems lose energy to surroundings.
• η = useful energy out / total energy in.

APPLICATIONS:
• Cooling systems.
• Radiators (copper / aluminium — moderate c, high thermal conductivity).
• Sweating (evaporation cooling — huge L_v).

Energy equations

• Temp change: Q = mcΔT.
• Phase change: Q = mL.
• Heat balance: Q_lost = Q_gained.

Kinetic theory

• Average KE: \bar{KE} = \tfrac{3}{2} k_B T
• Specific heat: Q = mc\Delta T
• Phase change: Q = mL

Quick reference

| Formula | Meaning |
|---|---|
| Q = mcΔT | Heat to change temperature |
| Q = mL | Latent heat (change of state) |
| c (water) | 4200 J/kg°C (approx.) |
| Kelvin | K = °C + 273 |
| Kinetic theory | Temperature ∝ average KE |

Worked example (Medium)

How much energy to heat 2 kg of water from 20 °C to 100 °C and then boil it all? c = 4180, L_v = 2.26 × 10⁶ J/kg.

1. Heat to 100 °C. Q₁ = mcΔT = 2 × 4180 × 80 = 668,800 J.
2. Boil. Q₂ = mL = 2 × 2,260,000 = 4,520,000 J.
3. Total. Q = 668,800 + 4,520,000 = 5,188,800 J ≈ 5.19 MJ.

Answer: Approximately 5.19 MJ — note phase change dominates.

Worked example (Hard)

How much energy is needed to convert 500 g of ice at −10 °C to steam at 110 °C? Use $c_{ice} = 2100$, $c_{water} = 4180$, $c_{steam} = 2000$ J/kg°C, $L_f = 334000$ J/kg, $L_v = 2260000$ J/kg.

1. Heat ice from −10 to 0 °C. $Q_1 = 0.5 \times 2100 \times 10 = 10500$ J.
2. Melt ice at 0 °C. $Q_2 = 0.5 \times 334000 = 167000$ J.
3. Heat water from 0 to 100 °C. $Q_3 = 0.5 \times 4180 \times 100 = 209000$ J.
4. Boil water + heat steam. $Q_4 = 0.5 \times 2260000 = 1130000$ J. $Q_5 = 0.5 \times 2000 \times 10 = 10000$ J. Total = $10500 + 167000 + 209000 + 1130000 + 10000 = 1526500$ J $\approx 1.53$ MJ.

Answer: Approximately 1.53 MJ. The vaporisation stage alone accounts for ~74% of the total energy.

Common mistakes

• Using $Q = mc\Delta T$ during a phase change Temperature doesn't change during melting or boiling, so $\Delta T = 0$ gives $Q = 0$ incorrectly Fix: During a phase change, use $Q = mL$. Only use $Q = mc\Delta T$ when the substance is changing temperature, not state.
• Forgetting to convert grams to kilograms Specific heat capacity is in J/kg°C, but mass is often given in grams Fix: Always convert mass to kg before substituting: 500 g = 0.5 kg.
• Using the wrong specific heat for the wrong phase Using $c_{water}$ when heating ice or steam Fix: Ice, water, and steam each have different specific heats. Check which phase the substance is in at each stage.

Phase changes and the molecular perspective

When you heat a substance, the added energy does one of two things: increases the kinetic energy of particles (temperature rises) or breaks intermolecular bonds (phase change, temperature stays constant).

During melting, energy breaks the rigid bonds of the solid lattice, allowing particles to slide over each other. The temperature stays at the melting point until all bonds are broken. Similarly, during boiling, energy separates liquid particles into widely spaced gas particles, overcoming the remaining intermolecular attractions.

The latent heat of vaporisation is always much larger than the latent heat of fusion because gas particles must be completely separated from each other, while liquid particles remain close but mobile. For water: $L_v = 2260$ kJ/kg vs $L_f = 334$ kJ/kg — a factor of nearly 7.

This explains everyday observations: it takes far more energy to boil a kettle dry than to melt the same mass of ice. It also explains why sweating is such an effective cooling mechanism — each gram of sweat that evaporates absorbs 2260 J from your skin.

The heating curve (temperature vs energy added) has characteristic flat sections (phase changes) and sloped sections (temperature changes). The slope of each section equals $1/(mc)$ for that phase. VCAA frequently presents this graph and asks students to identify phases, calculate energy for specific stages, or explain what happens at the plateaus.

Practice questions

1. The heat needed to raise 0.5 kg of copper (c = 385) by 40 °C is:

   1. 7700 J
   2. 15400 J
   3. 770 J
   4. 19250 J

   (1 marks)

2. A 300 g iron (c = 460 J/kg°C) at 90 °C is dropped into 500 g of water (c = 4180) at 20 °C. Find the final temperature (ignore heat loss).

   (4 marks)

Summary

• Apply Q = mcΔT correctly.
• Use latent heat for phase changes.
• Solve energy balance problems.