Why do heating curves have plateaus? [duplicate]

The heating curve of a solid looks something like this:

Why do the plateaus occur, various explanations online say that it is because the energy is being used to break the bonds(why is not bond breaking a continous process?), but that reasoning is not very satisfactory(for various reasons, one of them being my bolded text) is there a better explanation?

• Commented Feb 26, 2023 at 14:51
• Bond-breaking is a continuous process, but the temperature is not increasing, since this is change in potential rather than kinetic energy. Commented Feb 26, 2023 at 14:53
• @RogerVadim Got that, but why can not the temp increase side by side? Why must all the heat be used after a certain point to overcome the potential energy? Commented Feb 26, 2023 at 14:55
• I think this is a good question - the problem here is that phase transitions are mostly discussed as equilibrium thermodynamics, whereas what you ask is manifestly a process developing in time Commented Feb 26, 2023 at 14:58
• @Shinrin-Yoku if intuition were enough to explain complex phenomena, there would not be point studying in the university etc. Commented Feb 26, 2023 at 15:00

What is tricky here from the point of view of the usual discussion of phase transitions is that we are dealing with a process, rather than a phase diagram (which presupposes thermal equilibrium at every point.)

If, e.g., we consider boiling water, it is heated inhomogeneously and it is very likely that some parts of it are attaining temperature higher than the boiling temperature - like the vapor in the rising bubbles. However, these parts of liquid quickly leave it, taking away the energy, and thus cooling the liquid. Thus the temperature (of the liquid) remains always around the boiling point.

I admit that this is the first thing that came to my mind, so the criticism and better ideas are welcome

• This is nice, so the ice melting essentially cools the water and so there is no temperature increase until all of it has melted? Commented Feb 26, 2023 at 15:25
• So the plateau only occurs because we're only measuring the temperature of the liquid? If we measured the average temperature of liquid+gas system, there would be no plateau? Commented Feb 27, 2023 at 1:10
• @BlueRaja-DannyPflughoeft no. You can still assume (at least rough) thermal equilibrium, liquid and gas both at the boiling temp, and the thing that changes as you add energy is the ratio of gas to liquid. Commented Feb 27, 2023 at 2:58

First: bond breaking IS a continuous process! every minute the same amount of water (or vapor) is produced by the same amount of energy. So all the energy goes in bond breaking, the water and ice stays 0°C or vapor and water at 100° C. Close to ice of 0°C you can not have water of higher temperature. ( just pour some Water of more than 0° on ice, the water will cool down the ice melt.

The picture of Roger Vadim is correct, but the explanation is not quite right.

The heating procedure you've outlined is implicitly presumed to be very slow so that the system can reach approximate thermodynamic equilibrium at every point in time. This just means that you assume the system "mixes" the input heat around sufficiently well so that you can ignore unstable states (see superheating/supercooling). It is always important to remember that in equilibrium thermodynamics, we wait a "long time" before measuring anything.

Here is a rigorous argument. We know that liquid water is unstable above its boiling point, and water vapor is unstable below its boiling point. Thus, if I have a coexisting mixture of gas and liquid in equilibrium, they must both be sitting at the same temperature (from the definition of equilibrium), which can only be the boiling point. Therefore, heating such a mixture cannot change the temperature until one is no longer in a coexisting phase (i.e. boiling off all the water). As Roger Vadim pointed out, you can think of this as the liquid water cooling the gas and preventing the system from changing temperature as you heat the system.

As a technical aside, the reason for the first-order discontinuity in the free energy is that the original state variables (e.g. $$P,T$$) are no longer sufficient to completely characterize the state. A degeneracy arises due to the mixture of two coexisting phases; accounting for the "phase-mixture" degree of freedom (e.g. through total energy) removes this degeneracy. Moreover, there is no kinetics involved and it is not an issue of process or physics outside of equilibrium thermodynamics. One can always stop heating in the middle of the phase transition and let the system reach equilibrium, where you'll find phase coexistence as its equilibrium fate.

I also remark that coexistence plateaus do not always happen; this is a peculiarity of first-order phase transitions which have coexistence regions. For instance, at a specific value of pressure and temperature, you can vaporize liquid water with zero latent heat and have no coexistence region.

• For instance, at a specific value of pressure and temperature, you can vaporize liquid water with zero latent heat and have no coexistence region. – I would rather say that the distinction between water and vapour vanishes and thus the term vaporise loses its meaning. Commented Feb 26, 2023 at 21:48
• +1 Thank you for adding much clarity. Commented Feb 27, 2023 at 14:43

The heating curve you've shown is assuming that heat energy is being slowly added to the substance over time1. So that time axis is really an energy axis.

In the diagonal parts of that chart, as we add more heat energy to the substance (and wait for it to distribute evenly throughout) its average temperature gradually increases, directly in proportion to the amount of heat we're adding. But things change once we reach a phase change temperature.

Breaking molecular bonds to transition from a solid to a liquid or a liquid to a gas takes energy. So every bit of solid that melts to a liquid "steals" a bit of the heat energy that is being added to the system. It doesn't really matter whether you think of the energy as breaking bonds "instead of" raising temperature, or think of it as raising the temperature of a bit of the solid above its melting point and then its bonds break and cool it back down. Either way, the average temperature of the substance stops going up at the same rate as it was before - that alone is enough to see that the line on the chart can't continue at exactly the same rate once it hits a phase change temperature.

The total amount of energy needed to melt/vaporise all of the substance is a non-negligible finite quantity. We're adding heat energy gradually and slowly, so to build up that total quantity of energy is going to take some time. That means we're not going to transition the entire bulk of the substance from solid to liquid (or liquid to gas) all at once. It's a gradual process that takes some time. During that time we're adding more energy but the substance is also "spending" energy to change phase, so there are two processes changing its average temperature; one positive and one negative. And if we add heat faster it simply means there is more excess energy available to break bonds faster, so the rate of conversion increases to match. This ends up meaning the average temperature stays pretty much constant - instead of the temperature changing in proportion to the energy being added the rate of phase change is in proportion to the energy being added. (But the rate of phase change isn't something directly represented in this chart, so it looks like "nothing is happening").

Finally, once the substance has fully transitioned, adding more heat raises the temperature again as the substance can't "steal" this energy to break bonds instead; all the bonds thermal energy is expected to break at this temperature are already broken.

Also note that the phase change is a continuous process, and that's important to explain the flat areas of the chart. On the scale of individual molecules, bond breaking is not continuous; either there are bonds or there aren't, there are no partially broken bonds. But the sort of model behind the chart is not used for individual molecules, it's used for relatively large bulk masses. And a bulk mass does change phase gradually as it gains/loses heat energy; not in the sense that it can be in a state that is "in between one phase and another", but in the sense that the fraction of it that has transitioned gradually increases from 0% to 100%.

You've probably seen an ice cube melt; the ice cube gradually gets smaller and smaller as the bulk gradually transitions to water. It doesn't suddenly liquefy all at once! And this is why there's a flat area on the chart; the energy needed to melt the ice is "spent" continuously over a period of time, and over that time the average temperature isn't increasing so the chart of temperature vs time/energy is flat. If the phase change were not a continuous process the substance would instead "save up" the energy we're slowly adding in the form of temperature until there was enough excess energy to transition the whole bulk. If that were the case you wouldn't see these plateaus in the chart, you'd instead see the temperature continuing to increase and then a vertical jump down as it lost temperature by putting some of its internal heat energy into breaking bonds. So this chart actually is evidence that "breaking bonds" is a continuous process (on a macroscopic scale).

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1 It's important that this is a slow process, meaning the rate we're adding heat is low relative to the rate at which heat can distribute throughout the substance. That lets us model the substance reasonably accurately as a single "thing" with one average temperature.

Things get more complicated if the heat is not evenly distributed throughout the substance (we have to model the various parts of it at different temperatures interacting), or if we're talking about very small scale (we need to model individual molecules and transitions).