Heat ice into steam and watch the particles change behaviour

Solid: packed tight and shivering

In a solid the attractions are so strong that particles cannot leave their positions. They only jiggle about them. That is why a block of wood stays the shape you put it in.

In the simulation the solid particles, drawn in the theme colour, sit in rows. Heating makes the shaking stronger, but nobody leaves, until the melting point is reached and the bonds give way.

Liquid: close, but free to slide

In a liquid the particles are still almost touching, so the volume barely changes. But their movement energy is now enough to loosen the grip a little, and they slide past each other. That is why a liquid takes the shape of its container and pours.

The green particles pile up in a pool at the bottom with empty space above. Because they are nearly as close as in a solid, liquids are almost impossible to squeeze into a smaller volume.

Gas: far apart, fast and free

In a gas the particles are so far apart that the attractions hardly matter. They travel in straight lines until they hit a wall or another particle, so a gas fills whatever container it is in.

The amber particles race around the whole box. In a real gas the gaps are far larger still; the box is small, so the gaps are drawn smaller than they really are.

Stretch 1: ice from −-20 °C to 0 °C (slope)

Solid ice is warming. Its specific heat is 2.1 J g⁻¹ K⁻¹, so raising it by -20 °C takes 2.1 × -20 = 42 J. This short slope on the far left is hard to see, because it is under two percent of the width of the x axis.

The particles just vibrate harder. The numbers go up, but nobody leaves a lattice site.

Stretch 2: melting at 0 °C (flat)

The temperature now stalls at 0 °C. Melting each gram of ice takes 334 J, and none of it speeds the particles up. It is spent breaking down the regular arrangement, loosening the grip that held each particle in place.

In the simulation solid (theme colour) and liquid (green) particles coexist during this stretch. They convert one by one while the line stays flat, and the last one is liquid once 334 J is in. At the default burner power of 100 J/g/s this stretch takes 3.34 seconds.

Stretch 3: water from 0 °C to 100 °C (slope)

Now everything is liquid. Water's specific heat is 4.18 J g⁻¹ K⁻¹, so warming 1 g by 100 °C takes 4.18 × 100 = 418 J. That is a little more than melting needed (334 J), but far less than boiling.

The line is nearly straight because the specific heat of liquid water barely changes. The mean speed of the green particles grows steadily, which you can see as slightly livelier motion.

Stretch 4: boiling at 100 °C (flat, the longest)

The temperature stalls again, this time at 100 °C, and each gram needs 2,260 J. That is 6.8 times as much as melting, and 5.4 times as much as warming the same water from 0 °C to 100 °C.

Why so much? In melting, particles only get free of their lattice sites and stay close to neighbours. To become vapour each particle must escape its neighbours' attraction altogether and travel far away, which costs far more energy. At 100 J/g/s this stretch takes 22.6 seconds.

Stretch 5: steam from 100 °C to 120 °C (slope)

Everything is gas now. Steam's specific heat is about 2.0 J g⁻¹ K⁻¹, so going on to 120 °C needs another 40 J. The line slopes again and temperature and mean kinetic energy rise together.

The full total is 42 + 334 + 418 + 2,260 + 40 = 3,094 J, the number the "Heat added" reading ends on.

What each symbol means

Q is the heat absorbed or released, measured in joules (J). m is the mass, in grams or kilograms, as long as it matches the units of c and L. c is the specific heat capacity, the heat needed to raise 1 g by 1 K. L is the specific latent heat, the heat needed to change the state of 1 g.

• Water: c = 4.18 J g⁻¹ K⁻¹ (ice 2.1, steam 2.0)
• Latent heat of fusion of ice: L = 334 J/g = 3.34 × 10⁵ J/kg (about 80 cal/g)
• Latent heat of vaporization of water: L = 2,260 J/g = 2.26 × 10⁶ J/kg (about 540 cal/g)

When both formulas are needed

To go from one state to another you add the stretches one by one: Q = mcΔT for each slope, Q = mL for each flat part, then sum them all. Never try to solve the whole journey with a single formula.

Why evaporation cools

The particles that escape from the surface are the most energetic ones. The ones left behind have a lower mean kinetic energy, so the liquid cools. That is why sweating cools you in a breeze and why water in a clay pot stays cool.

The boiling point depends on pressure

A liquid boils when its vapour pressure equals the outside pressure. Lower the outside pressure and the boiling point drops (on a high mountain water boils below 100 °C, so food takes longer to cook). Raise it and the boiling point rises (in a pressure cooker water boils above 100 °C and cooking is faster). The 100 °C figure is for normal atmospheric pressure.

Sublimation: gas without the liquid

Some solids turn straight into a gas when heated, without becoming a liquid first. This is sublimation. Camphor, naphthalene, iodine and ammonium chloride do it. Dry ice, which is solid carbon dioxide, sublimes at about −78 °C at normal pressure.

The reverse, a gas turning directly to a solid, is called deposition; frost forming on a freezer wall is an example. The simulation shows only the three states of water, but the idea in sublimation is the same: the particles break all their bonds in a single step, so the heat needed is close to the sum of the heats for melting and vaporizing.

A fourth state: plasma (at a glance)

At very high temperatures the molecules of a gas break up into ions and electrons, a mixture called plasma. The Sun, lightning and neon lamps all contain it. For most school exams the three states are the core and plasma is general knowledge.

Experiment 1: time the two plateaus

Leave the burner power at 100 J/g/s, run the simulation and keep a stopwatch. The melting plateau should take about 3.34 seconds and the boiling one about 22.6. Their ratio is 6.8, exactly the ratio of the two latent heats.

The whole run takes about 30.9 seconds, so the boiling plateau eats most of the time. Boil a kettle at home and you see the same thing: after it starts to boil, it takes a long time to boil dry.

Experiment 2: watch the particles, not the heat

Pause, choose "Melting" under "Jump to a stage", then drag the "Heat added" slider slowly to the right. Particles switch one at a time from the theme colour to green. Read the numbers in the "Particles: solid · liquid · gas" row.

Through all of this the temperature and the mean kinetic energy do not change. Only the heat added rises, and the number of solid particles drops.

Experiment 3: run it backwards

Tick "Cool the sample instead of heating it". Jump to "Gas (steam)" and press play. The temperature drops, stalls at 100 °C, and amber particles turn green one by one and pile up at the bottom: condensation. It stalls again at 0 °C as green particles return to the lattice: freezing.

Is heat going in or coming out? Coming out. Condensing gives out 2,260 J per gram and freezing gives out 334 J.

Experiment 4: change the burner power

Set the power to 400 J/g/s and run it again. Every plateau is crossed more quickly, but the graph has the same shape, because its x axis is the heat added and not time. The burner power only sets how fast you travel along the curve.

This matters in the lab, where a temperature-time graph is usually drawn. With a steady heater it is the same graph with the x axis divided by the power.

Problem 1: the whole simulation run

How much heat takes 1 g of ice at −-20 °C to steam at 120 °C?

Stage 1: 2.1 × -20 = 42 J. Stage 2: 334 J. Stage 3: 4.18 × 100 = 418 J. Stage 4: 2,260 J. Stage 5: 2.0 × 20 = 40 J. The total is 3,094 J. Boiling alone takes 73% of it and melting 11%.

Problem 2: heating water for tea

How much heat raises 250 g of water from 25 °C to 100 °C? (c = 4.18 J g⁻¹ K⁻¹)

Q = mcΔT = 250 × 4.18 × 75 = 78,375 J ≈ 78.4 kJ. Once the water boils the temperature stops rising and any further heat makes steam.

Problem 3: melting ice at 0 °C

50 g of ice is exactly at 0 °C. How much heat melts it completely?

The temperature does not change, only the state, so Q = mL = 50 × 334 = 16,700 J ≈ 16.7 kJ. The water that results is still at 0 °C.

Problem 4: ice to water at 20 °C

10 g of ice at −10 °C is to become water at 20 °C. How much heat is needed?

Three stages. (a) Ice from −10 to 0 °C: 10 × 2.1 × 10 = 210 J. (b) Melting: 10 × 334 = 3,340 J. (c) Water from 0 to 20 °C: 10 × 4.18 × 20 = 836 J. Total 210 + 3,340 + 836 = 4,386 J. Notice that melting is the biggest part.

Problem 5: boiling water away

100 g of water is boiling at 100 °C. How much heat turns all of it to steam? Compare this with heating it from 0 °C to 100 °C.

Boiling off: Q = mL = 100 × 2,260 = 226,000 J = 226 kJ. Heating from 0 to 100 °C: 100 × 4.18 × 100 = 41,800 J ≈ 41.8 kJ. Boiling takes about 5.4 times as much.

Problem 6: a steam burn versus a scald

1 g of steam at 100 °C and 1 g of boiling water at 100 °C each land on skin and cool to 37 °C. How much heat does the skin receive from each?

Boiling water: 1 × 4.18 × 63 = 263.3 J. Steam: first condensing gives 2,260 J, then cooling gives 263.3 J, total 2,523.3 J. Steam delivers about 9.6 times as much, which is why steam burns are so severe.

Problem 7: ice in a glass of water

200 g of water at 30 °C has 20 g of ice at 0 °C dropped into it. Assuming no heat is lost to the surroundings, what is the final temperature?

Cooling the water to 0 °C would release 200 × 4.18 × 30 = 25,080 J. Melting the ice takes 20 × 334 = 6,680 J, so all the ice melts. The remaining 18,400 J warms the total 220 g of water: ΔT = 18,400 ÷ (220 × 4.18) = 20.0 °C. The final temperature is about 20.0 °C.

Problem 8: a 1000 W heater

One watt is one joule per second. A 1,000 W heater runs for one minute (60 s) on ice at 0 °C. How many grams melt? (Ignore losses.)

Heat supplied = 1,000 × 60 = 60,000 J. Mass melted = Q ÷ L = 60,000 ÷ 334 ≈ 180 g.

Problem 9: condensing steam

2 g of steam at 100 °C condenses into water at the same temperature. How much heat is released?

Q = mL = 2 × 2,260 = 4,520 J. That released heat is why the lid of a pot over a boiling kettle becomes so hot.

A sample structured question

Stimulus: 10 g of ice at −10 °C is heated steadily. (Ice c = 2.1 J g⁻¹ K⁻¹, water c = 4.18 J g⁻¹ K⁻¹, latent heat of fusion 334 J/g.)

(a) Define latent heat. (b) Explain, using the particle model, why the temperature stays at 0 °C while the ice melts. (c) Calculate the heat needed to turn the ice into water at 0 °C. (d) Calculate the heat needed to reach 20 °C and say which stage takes the most.

Answer to (c): 210 + 3,340 = 3,550 J. Answer to (d): 4,386 J in total; melting, at 3,340 J, is the largest part (see Problem 4).