D. Energy and Society · Thermal and Nuclear EnergyChanges of State and Latent Heat
D3.7 D3.3 D2.10 D2.1About 70 minutes
Learning goals- I can name the six changes of state and say whether each absorbs or releases energy.
- I can use the kinetic molecular theory to explain why temperature stays constant during a change of state.
- I can define specific latent heat of fusion and of vaporization and use Q = mL.
- I can explain everyday effects of latent heat, such as sweating, steam burns and frost protection.
Words to know
change of state (phase change)
A change between solid, liquid and gas, such as melting or boiling. Temperature stays constant while it happens.
latent heat
The energy absorbed or released during a change of state, with no change in temperature. "Latent" means hidden.
specific latent heat of fusion (L_f)
The energy needed to melt 1 kg of a solid at its melting point. Unit: J/kg.
specific latent heat of vaporization (L_v)
The energy needed to change 1 kg of a liquid to a gas at its boiling point. Unit: J/kg.
sublimation / deposition
Solid directly to gas / gas directly to solid, without passing through the liquid state.
evaporation
Vaporization from the surface of a liquid below its boiling point; the fastest particles escape, so the liquid left behind cools.
Why does a skating rink stay at 0 °C?
People who build backyard rinks in January know a strange fact: on a mild day the ice gets slushy, but a thermometer stuck in the slush reads 0 °C, not higher. The Sun keeps pouring energy in, yet the temperature does not rise until all the ice has melted.
Where does that energy go? It is not lost; it is doing a different job. In this lesson you will find out what that job is, and learn to calculate it.
Six changes of state
Matter changes state when energy is added or removed:
- Melting (solid → liquid) and vaporization (liquid → gas) absorb energy.
- Freezing (liquid → solid) and condensation (gas → liquid) release energy.
- Sublimation (solid → gas) absorbs energy; deposition (gas → solid) releases it. Snow slowly disappearing on a cold, sunny, dry day is partly sublimation, and frost forming on a car windshield is deposition.
Vaporization happens in two ways. Boiling happens throughout the liquid at the boiling point (bubbles of vapour form inside). Evaporation happens only at the surface, at any temperature.
What the KMT says is happening
Remember: temperature measures the average kinetic energy of the particles, but thermal energy also includes their potential energy, which depends on how far apart they are and how strongly they attract each other.
- When you heat ice at −10 °C, the energy makes the particles vibrate faster: their kinetic energy rises, so the temperature rises.
- At 0 °C, the particles are vibrating hard enough to start breaking free of their fixed positions. Now the added energy goes into pulling particles apart against their attractions. That increases their potential energy, not their kinetic energy. So the temperature stays at 0 °C until all the ice has melted.
- Once it is all liquid, added energy speeds the particles up again and the temperature rises.
- At 100 °C (at normal air pressure) the same thing happens for boiling: the energy separates the particles completely so they can escape as a gas.
The energy absorbed or released during a change of state, with no change in temperature, is called latent heat ("latent" means hidden: a thermometer cannot see it).
Calculating latent heat: Q = mL
The energy for a change of state depends on the mass and the substance:
Q = mL
- Q = heat (J)
- m = mass (kg)
- L = specific latent heat (J/kg). Use L_f (fusion) for melting or freezing and L_v (vaporization) for boiling or condensing.
Sign convention: Q = +mL when the substance absorbs energy (melting, vaporizing) and Q = −mL when it releases energy (freezing, condensing).
Notice there is no ΔT in this equation, because the temperature does not change during a change of state.
| Substance | Melting point (°C) | L_f (J/kg) | Boiling point (°C) | L_v (J/kg) |
|---|
| water | 0 | 3.34 × 10⁵ | 100 | 2.26 × 10⁶ |
| ethanol | −114 | 1.04 × 10⁵ | 78 | 8.5 × 10⁵ |
| lead | 327 | 2.5 × 10⁴ | about 1750 | 8.7 × 10⁵ |
| aluminium | 660 | 3.97 × 10⁵ | about 2500 | 1.1 × 10⁷ |
Why is L_v so much bigger than L_f? Melting only loosens the particles so they can slide past each other; they stay close together. Vaporizing must separate the particles completely, against all their attractions, and the gas must also push back the surrounding air as it expands. For water, vaporizing 1 kg takes almost 7 times the energy of melting it.
Worked exampleExample 1: Melting ice for a drink
How much energy is needed to melt 0.50 kg of ice that is already at 0 °C?
- The ice is at its melting point, so only a change of state happens: Q = mL_f.
- Q = (0.50 kg)(3.34 × 10⁵ J/kg) = 1.67 × 10⁵ J (to 2 significant digits, 1.7 × 10⁵ J).
- Compare: the same 167 000 J could warm 0.50 kg of liquid water by 167 000 ÷ (0.50 × 4180) ≈ 80 °C!
Q ≈ 1.7 × 10⁵ J. Melting ice takes a lot of energy, which is why ice is so good at keeping drinks cold.
Worked exampleExample 2: Why steam burns are serious
20 g of steam at 100 °C condenses on a person's skin and the water then cools to 37 °C. (a) How much heat is released by condensing? (b) How much more is released as the water cools to 37 °C? (c) Compare with 20 g of boiling water cooling from 100 °C to 37 °C.
- (a) Condensing releases energy: Q = −mL_v = −(0.020)(2.26 × 10⁶) = −4.52 × 10⁴ J.
- (b) Cooling the condensed water: Q = mcΔT = (0.020)(4180)(37 − 100) = (0.020)(4180)(−63) = −5.27 × 10³ J.
- Total from the steam: 45 200 + 5267 ≈ 5.0 × 10⁴ J released into the skin.
- (c) Boiling water releases only the part (b) amount, about 5.3 × 10³ J.
- The steam delivers about 9.6 times as much energy (50 467 ÷ 5267), almost all from latent heat. That is why steam from a kettle spout or a pot lid is so dangerous.
Steam releases about 5.0 × 10⁴ J, roughly ten times as much as the same mass of boiling water.
Worked exampleExample 3: Find a latent heat
It takes 1.5 × 10⁵ J to melt 0.60 kg of a solid at its melting point. What is its specific latent heat of fusion?
- Rearrange Q = mL: L = Q ÷ m.
- L_f = 1.5 × 10⁵ J ÷ 0.60 kg = 2.5 × 10⁵ J/kg.
Latent heat in daily life
- Sweating and evaporation: when sweat evaporates, the fastest-moving water particles escape. They carry energy away, so the remaining liquid and your skin cool down. Getting out of a lake on a breezy day feels cold for the same reason.
- Frost protection: on frosty spring nights some fruit growers spray water on blossoms. As the water freezes it releases latent heat, which helps keep the blossoms near 0 °C instead of dropping lower.
- Ice packs and coolers: melting ice absorbs a large amount of energy at a steady 0 °C.
- Refrigerators and heat pumps use a liquid that evaporates inside (absorbing latent heat) and condenses outside (releasing it). You will see this in Heat Pumps, Refrigerators and Thermal Technologies.
Just for funBoiling away 1 kg of water at 100 °C takes 2.26 MJ. Heating that same kilogram from 0 °C all the way to 100 °C takes only about 0.42 MJ. So if it feels like a pot takes forever to boil dry, physics agrees: the boiling part needs over 5 times more energy than the warming part.
Hands-on activityHidden energy in your kitchen
Part A: the 0 °C plateau. You need: a glass, lots of crushed ice or small ice cubes, a little cold water and a kitchen thermometer.
- Fill the glass with ice and add just enough cold water to fill the gaps. Stir for a minute and read the thermometer.
- Leave the glass on the counter. Stir and read the temperature every 5 minutes until all the ice has melted, then for 15 more minutes.
- Describe what the temperature does while ice remains and after it has all melted. Explain using the KMT and the words latent heat and potential energy.
Part B: evaporation cools. You need: two thermometers (or one used twice), a small piece of paper towel, a rubber band, room-temperature water and a fan or a piece of cardboard to fan with.
- Wrap the wet paper towel around the bulb of one thermometer with the rubber band. Leave the other bulb dry.
- Fan both for 3 minutes, then read both thermometers.
What to look for: in Part A the reading stays very close to 0 °C as long as there is ice, then climbs toward room temperature. In Part B the wet bulb reads several degrees lower, because evaporating water takes its latent heat of vaporization from the thermometer.
Your turn
1Name the change of state and say whether energy is absorbed or released: dew forming on grass overnight.
Condensation (gas → liquid): energy is released.
2Name the change of state: dry ice (solid carbon dioxide) turning directly into gas.
Sublimation: energy is absorbed.
3How much energy is needed to melt 2.0 kg of ice at 0 °C?
Q = mL_f = (2.0)(3.34 × 10⁵) = 6.7 × 10⁵ J.
4How much energy is needed to boil away 0.25 kg of water at 100 °C?
Q = mL_v = (0.25)(2.26 × 10⁶) = 5.7 × 10⁵ J (565 000 J).
5How much heat is released when 3.0 kg of water at 0 °C freezes?
Q = −mL_f = −(3.0)(3.34 × 10⁵) = −1.0 × 10⁶ J: about 1.0 MJ is released.
6Use the KMT to explain why the temperature of boiling water stays at 100 °C while the stove keeps adding energy.
The added energy goes into separating the particles completely (increasing their potential energy) so they escape as a gas. Their average kinetic energy does not increase, so the temperature stays at 100 °C.
7Why does sweating cool you down?
When sweat evaporates, the fastest particles escape and carry energy away. The latent heat of vaporization comes from your skin, so your skin cools.
More practice: warm-up
1What is the unit of specific latent heat?
2Which is larger for water, L_f or L_v?
L_v (2.26 × 10⁶ J/kg compared with 3.34 × 10⁵ J/kg).
3Does freezing absorb or release energy?
4What does the word "latent" mean?
Hidden (a thermometer does not show it as a temperature change).
5Give an example of deposition.
Frost forming on a cold window or windshield (water vapour → ice).
More practice: on your own
16.8 × 10⁴ J melts 0.20 kg of a solid at its melting point. Find L_f.
L_f = Q ÷ m = 68 000 ÷ 0.20 = 3.4 × 10⁵ J/kg (very close to ice).
2What mass of water at 100 °C can be vaporized by 1.0 × 10⁶ J?
m = Q ÷ L_v = 1.0 × 10⁶ ÷ 2.26 × 10⁶ ≈ 0.44 kg.
3How much energy is needed to melt a 4.0 kg lead block already at 327 °C? (L_f = 2.5 × 10⁴ J/kg)
Q = (4.0)(2.5 × 10⁴) = 1.0 × 10⁵ J.
4Explain why a pail of water left in a root cellar can help keep stored vegetables from freezing on a very cold night.
As the water in the pail cools to 0 °C and starts to freeze, it releases latent heat (334 000 J for every kilogram that freezes). That released energy slows the temperature drop in the cellar, holding it near 0 °C while the water freezes.
5Why do you feel cold stepping out of a pool on a windy day, even if the air is warm?
Water on your skin evaporates quickly in the wind. Evaporation absorbs latent heat from your skin, cooling it.
Challenge
1How much energy is released when 0.500 kg of steam at 100 °C condenses and the water cools to 20 °C?
Condense: −(0.500)(2.26 × 10⁶) = −1.13 × 10⁶ J. Cool: (0.500)(4180)(−80) = −1.672 × 10⁵ J. Total ≈ −1.30 × 10⁶ J (1.30 MJ released).
2A 1500 W kettle holds 1.0 kg of water at 100 °C. If the switch fails and the kettle keeps heating, how long until the water has all boiled away? (1 W = 1 J/s; ignore losses)
Q = mL_v = (1.0)(2.26 × 10⁶) = 2.26 × 10⁶ J. Time = 2.26 × 10⁶ ÷ 1500 ≈ 1500 s, about 25 minutes.
3Explain, using the KMT, why L_v for water is about 7 times L_f.
In melting, particles only need to loosen enough to slide past each other; they stay close and still attract strongly. In vaporizing, every particle must be pulled completely free of its neighbours, which takes far more energy, and the gas also does work pushing back the atmosphere as it expands.
Quiz
1. While ice at 0 °C melts, the temperature of the ice-water mixture:
2. Which change of state releases energy?
3. How much energy is needed to melt 0.20 kg of ice at 0 °C? (L_f = 3.34 × 10⁵ J/kg)
4. During boiling, the energy added to water mainly:
5. Why can a steam burn be much worse than a burn from the same mass of boiling water?
6. Which equation do you use for heat absorbed during melting?
Big idea: expectation D3.7 asks students to use the kinetic molecular theory to explain energy transfer during changes of state; D2.10 asks them to solve problems with Q = mL. The key insight: during a change of state, energy changes the particles' potential energy, not their kinetic energy, so the temperature is constant.
How to help: Do Part A of the activity together; seeing the thermometer sit at 0 °C is memorable. Ask why steam from the kettle is more dangerous than splashing water, and why a wet swimsuit feels cold in the wind.
Common difficulties:
- Using Q = mcΔT during a change of state (ΔT would be 0, giving Q = 0).
- Using L_f when L_v is needed, or the reverse. Fusion = melting/freezing; vaporization = boiling/condensing.
- Thinking that energy added during melting is "lost". It is stored as potential energy and released again on freezing.
- Forgetting that evaporation (surface, any temperature) and boiling (throughout, at the boiling point) are both vaporization.
Safety: never put your hand in steam to test anything. The activity uses only ice, room-temperature water and a fan.