Specific latent heat
When a substance changes state — melting, boiling, condensing, freezing — energy is transferred without any change in temperature. The energy needed to change the state of 1 kg of substance is the specific latent heat $L$:
$E = mL$
- $E$ in J.
- $m$ in kg.
- $L$ in J/kg.
There are two values:
- Specific latent heat of fusion $L_f$ — for melting (or freezing). For water: 334 000 J/kg.
- Specific latent heat of vaporisation $L_v$ — for boiling/condensing. For water: 2 260 000 J/kg.
$L_v$ is much larger than $L_f$ because all bonds between particles must be fully broken to escape into the gas phase.
What's happening to particles?
- Melting: lattice bonds break partly; particles become free to slide. PE rises; KE constant.
- Boiling: forces between particles overcome completely; they fly off as gas. Big PE jump.
Throughout both, temperature stays constant because energy goes into PE not KE.
Heating curve recap
Plotting temperature vs energy supplied for steady heating gives:
- Sloped section (solid heating).
- Flat plateau at melting point (length proportional to $mL_f$).
- Sloped (liquid heating).
- Flat plateau at boiling point (length proportional to $mL_v$).
- Sloped (gas heating).
The horizontal length of each plateau on the energy axis equals $mL$.
✦Worked example— Worked example 1
How much energy melts 0.30 kg of ice at 0 °C?
- $E = mL_f = 0.30 \times 334,000 = 100,200$ J ≈ 100 kJ.
✦Worked example— Worked example 2
How much energy is needed to turn 200 g of water at 100 °C completely into steam?
- $m = 0.20$ kg.
- $E = mL_v = 0.20 \times 2,260,000 = 452,000$ J = 452 kJ.
✦Worked example— Worked example 3 — combined heating + state change
How much energy heats 0.50 kg of ice at −10 °C to liquid water at 20 °C?
- Step 1, ice 0 °C: $E_1 = mc_{\text{ice}}\Delta\theta = 0.50 \times 2100 \times 10 = 10,500$ J.
- Step 2, melt: $E_2 = mL_f = 0.50 \times 334,000 = 167,000$ J.
- Step 3, water to 20 °C: $E_3 = 0.50 \times 4200 \times 20 = 42,000$ J.
- Total = 219 500 J ≈ 220 kJ.
⚠Common mistakes
- Using $c$ when you should use $L$ (or vice versa).
- Skipping a step in a multi-stage heating problem.
- Forgetting that the plateau is at the melting/boiling point, not above or below.
- Using mass in grams.
➜Try this— Quick check
How much energy condenses 0.10 kg of steam at 100 °C to liquid water at 100 °C?
- $E = mL_v = 0.10 \times 2,260,000 = 226,000$ J = 226 kJ. The energy is released by the condensing steam.
AI-generated · claude-opus-4-7 · v3-deep-physics