Changing State and Latent Heat - Physics for Pilot Assessments

Heating something does not always make it hotter. While ice is melting or water is boiling the temperature sits perfectly still, however hard you heat it, and this lesson explains where that energy is going instead. By the end of it you should be able to read a heating curve and say what each part of it means, calculate the energy needed to melt or boil a known mass of a substance, run that calculation backwards to find the mass, and explain why evaporation leaves what remains behind colder than it was.

Warming something up stops working at the melting point

Take a block of ice at −20 °C, put a thermometer in it, heat it steadily and watch. At first the thermometer behaves exactly as the previous lesson predicts: the temperature climbs at a steady rate, and the rate is set by the mass of ice and by ice's specific heat capacity of 2100 joules per kilogram per degree. Then it reaches 0 °C, and stops. The heater is still delivering energy at exactly the same rate, and for a long time nothing on the thermometer moves — while the ice steadily turns into water at 0 °C. Only when the last of the ice has gone does the temperature start climbing again, now more slowly than before because liquid water's specific heat capacity is 4200, twice ice's. At 100 °C it stops for a second time, for very much longer than it stopped at 0 °C, while the water turns into steam. After that the steam's temperature climbs once more.

Energy goes in at a constant rate throughout, so the horizontal axis is really time as well. On the two flat sections the energy is rearranging particles instead of speeding them up, so the thermometer shows nothing. The boiling plateau is much longer than the melting one — that is a real fact about water, not an artefact — though both plateaus are drawn compressed relative to the sloping sections, because at a true single scale boiling alone would swallow most of the width and squeeze the three ramps to a few pixels.

Three details of that curve are examined. The sloping sections have different steepnesses, because the three states of the same substance have different specific heat capacities. The flat sections are called plateaus, and while a plateau lasts the substance is a mixture of two states at one single temperature — ice and water together at 0 °C, or water and steam together at 100 °C. And the plateau temperatures have names: the melting point is the temperature at which a solid becomes a liquid, which is exactly the same temperature as the freezing point at which the liquid becomes solid again, and the boiling point is the corresponding temperature for liquid and gas.

Where the energy goes instead

The particle picture answers this cleanly. In a solid the particles sit in fixed positions, held there by forces of attraction to their neighbours, and vibrate about those positions; supplying heat makes them vibrate harder, which is a rise in their average kinetic energy and therefore a rise in temperature. That works right up to the melting point. At the melting point the vibration is already violent enough to be on the verge of tearing the fixed pattern apart, and any further energy supplied goes into breaking particles free of their positions — into working against the forces holding the structure together — rather than into making them move faster. Their average kinetic energy is therefore unchanged, and since temperature measures exactly that average, the thermometer does not move. Nothing is being wasted: the energy is stored in the new, looser arrangement.

Boiling is the same argument taken further. In a liquid the particles are still touching one another and still attracted to their neighbours, just no longer in a fixed pattern; to become a gas each particle has to be pulled completely clear of all of them, which means overcoming that attraction entirely rather than merely loosening it. On top of that, the escaping particles have to push the surrounding atmosphere out of the way to make room for a volume of gas around 1600 times the volume of the liquid it came from. Both of those cost energy, and neither shows on a thermometer.

Energy transferred in this way, changing a substance's state at constant temperature, is called latent heat. The word latent means hidden, and it is well chosen: this is the one kind of heating that no thermometer can detect. The amount required per kilogram is a property of the substance, exactly as specific heat capacity is, and it is called the specific latent heat — again with specific meaning per kilogram. There are two of them for any substance, because there are two state changes: the specific latent heat of fusion covers melting and freezing, fusion being the old word for melting, and the specific latent heat of vaporisation covers boiling and condensing.

Heat needed for a change of state

Q = mL

$Q$: the heat energy supplied or released, in joules (J)
$m$: the mass changing state, in kilograms (kg)
$L$: the specific latent heat of the substance for that change, in joules per kilogram (J/kg)

Notice what is missing from that expression compared with the one in the previous lesson: there is no temperature change in it, because there is no temperature change happening. A latent-heat calculation and a specific-heat calculation are two different sums, and a question that spans a plateau needs both of them added together.

Values, and why one is so much larger

For water the two figures are 340 000 joules per kilogram to melt ice, and 2 260 000 joules per kilogram to boil water — the second is about six and a half times the first, which is why the upper plateau on the heating curve is so much longer than the lower one. Both are large numbers next to a specific heat capacity of 4200: melting a kilogram of ice takes the same energy as heating a kilogram of liquid water by about 81 degrees, and boiling a kilogram of water takes as much as heating that kilogram from freezing to boiling more than five times over. The reason the vaporisation figure dwarfs the fusion figure is the one given above — melting only loosens the arrangement while the particles stay in contact, whereas boiling separates them completely and has to shove the atmosphere aside as well.

The same energy comes back out

Latent heat is not consumed, it is stored, and running the change backwards releases every joule of it. Freezing a kilogram of water gives out 340 000 joules to its surroundings, and condensing a kilogram of steam gives out 2 260 000. Two familiar facts follow directly. A steam burn is far worse than a scald from boiling water at the same 100 °C, because the steam condenses on your skin first and dumps that vast latent heat into it before the resulting water has even begun to cool. And ice cools a drink far more effectively than the same mass of water at 0 °C would, because the ice has to absorb 340 000 joules per kilogram from the drink simply to become water, before it starts taking energy to warm up.

Evaporation cools what is left behind

Water does not have to reach 100 °C to become a gas. At any temperature the particles in a liquid have a spread of speeds around the average, and a particle near the surface that happens to be moving unusually fast, in the right direction, can break free of its neighbours' attraction and leave. That is evaporation, and it happens from any exposed liquid surface at any temperature — a puddle dries without ever being heated. Boiling is different in kind, not just in degree: boiling happens throughout the liquid, forms bubbles, and only occurs at one particular temperature for a given pressure.

Now the consequence that gets examined. The particles that escape are the fastest ones, by definition — the slow ones cannot break free. Removing the fastest members of a population lowers the average of what remains, and the average kinetic energy of the particles left in the liquid is its temperature. So evaporation cools the remaining liquid, and it cools whatever the liquid is in contact with. This is not the same as saying the liquid loses energy because it is warmer than the air; it happens even when the air is cooler than nothing at all would explain, and it is the mechanism behind sweating, behind a wet cloth on the back of the neck, behind water kept cool in a porous clay jug, and behind the surprising chill of spilled fuel or alcohol on your hand. Anything that speeds evaporation up — a larger surface area, a higher temperature, dry air, or a draught to carry the escaped particles away before they can return — increases the cooling.

Example 1 — melting a known mass

How much energy is needed to melt 0.5 kilograms of ice that is already at 0 °C? The specific latent heat of fusion of water is 340 000 joules per kilogram. What is the temperature of the water produced?

Use heat = mass × specific latent heat, so the sum is 0.5 × 340 000.
Multiplying by 0.5 is halving: $Q = \frac{340\,000}{2} = 170\,000\ \text{J}$
So 170 000 joules, which is 170 kilojoules.
The water produced is at 0 °C — exactly the temperature the ice started at. Every one of those 170 000 joules went into rearranging particles, and none of it into raising the temperature.
For scale, note that 170 000 joules would instead have raised the temperature of that same half-kilogram of liquid water by about 81 °C, using the specific heat capacity of 4200 from the previous lesson.

Example 3 — a sum that spans a plateau

0.2 kilograms of water at 80 °C is heated until it has all boiled away. How much energy does that take? Use 4200 joules per kilogram per degree for water and 2 260 000 joules per kilogram for its specific latent heat of vaporisation.

There are two separate stages, so there are two separate sums. First the water has to be raised from 80 °C to its boiling point of 100 °C, a change of 20 °C.
For that stage use heat = mass × specific heat capacity × temperature change: $Q = 0.2 \times 4200 \times 20$ Take it in order: $0.2 \times 4200 = 840 \quad\text{and}\quad 840 \times 20 = 16\,800\ \text{J}$
Then the water at 100 °C has to be turned into steam at 100 °C, which is heat = mass × specific latent heat: $Q = 0.2 \times 2\,260\,000 = 452\,000\ \text{J}$
Add the two stages: $16\,800 + 452\,000 = 468\,800\ \text{J}$
Look at the proportions. Over 96 per cent of the total went into the plateau, where the thermometer showed nothing at all happening. That is the whole reason a pan boils dry so much more slowly than it comes to the boil.

Why it matters in aviation

This lesson is the physics of airframe icing. Water droplets in cloud can stay liquid well below 0 °C, because freezing has to start somewhere and a small clean droplet suspended in air often has nowhere to start; the droplet is then supercooled, and the arrival of a solid surface is the trigger it was waiting for. It freezes on contact, and as it freezes it releases its latent heat of fusion into that surface and into the water still around it.

That release is why ice does not simply appear wherever the air is cold enough. How much of an impacting droplet turns to ice immediately depends on how fast the latent heat can be carried away, and that in turn decides whether the deposit builds up as an opaque mass at the point of impact or as a clearer layer that has spread back along the surface as liquid first. The same latent heat, given out as water vapour condenses instead of freezing, is what keeps a growing cloud growing. And the process run backwards matters as well: evaporation takes latent heat out of whatever is left behind, which is why fuel and induction systems can sit below the temperature of the air around them.

What the test asks

The dominant distractor is the one the heating curve exists to kill: an item describes ice being heated at a steady rate and offers an answer in which the temperature climbs continuously through 0 °C, or asks what the temperature of the water is immediately after the last of the ice has melted and offers something above zero. It is 0 °C, and it stayed at 0 °C throughout. Expect also the mirror-image error on a cooling curve, where a substance is said to keep cooling while it freezes. Two further traps are worth naming. One substitutes the wrong constant — using the specific heat capacity where the specific latent heat is needed, or using water's 4200 for ice instead of 2100. The other concerns evaporation: an option will explain the cooling by saying the liquid loses heat to the cooler air, which is a different mechanism and does not account for cooling below the air's own temperature. The escaping particles are the fast ones, and it is the average that is left behind that has fallen.