My counterargument:

Intermolecular forces between molecules are either intact or broken. There is no in-between. Therefore, the change from intact to broken is instantaneous.

Applied heat energy increases the movement of particles in a substance, which in turn increases its temperature. When movement reaches a critical point, intermolecular force is overpowered, and a change of state has occurred.

The explanation that “energy goes to break the bonds, and not to increase temperature” makes no sense to me. In order to break the bond, the movement of the molecules needs to be intensified, which would increase the temperature of the water. That’s why I think that there should be no period of constant temperature.

My question: Where does that energy go? It’s not increasing temperature; so what is it doing?

  • $\begingroup$ Are you not over-engineering something much more simple than you thought it? Doesn't temperature remain constant during a state change only because other factors like volume or pressure alter? $\endgroup$ Commented Feb 21, 2021 at 0:40
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    $\begingroup$ I seriously doubt that all the bonds break at once. Yes, breaking the bond is an instantaneous process, but not for the entire solid at once. $\endgroup$
    – cst1992
    Commented Feb 21, 2021 at 8:05
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    $\begingroup$ Temperature's for equilibrium. If you want to talk dynamics, then there's super-cooling/super-heating/etc.. If you want to talk about transient behavior in short time-frames, then you may need to pick a non-equilibrium framework since the pseudo-equilibrium approximation can't be assumed to hold well. $\endgroup$
    – Nat
    Commented Feb 21, 2021 at 13:39
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    $\begingroup$ @Raen "Where does that energy go? It’s not increasing temperature; so what is it doing?" Simply put, that energy goes towards increasing the entropy of the system. Hence why there's no increase in temperature i.e. the system's ability to transfer more heat towards some external temperature-measuring device we call a "thermometer". Or, mathematically, why $\Delta G = \Delta H - T \Delta S = 0$ or as I say $T = \Delta H / \Delta S = \text{constant}$ for phase changes at equilibrium! $\endgroup$
    – ManRow
    Commented Feb 22, 2021 at 20:16

8 Answers 8


Why does the temperature remain constant during a change in the state of matter?

It doesn't exactly, as we see below, but it almost does under typical conditions where something called heterogeneous nucleation (that is, easy formation of molecular-scale clusters of the new phase) dominates. This is the case for water boiling right around 100°C at sea level in a standard kettle, for instance.

Two effects arise simultaneously.

First, thermodynamically, one phase becomes the more stable phase at equilibrium.

Second, kinetically, creation of the new phase proceeds in an enormously fast process that moves quickly toward that equilibrium.

Consequently, essentially any overheating is immediately absorbed by bond breaking; conversely, any undercooling is immediately offset by energy released through bond formation. The process is fast—like exponential-function-applied-to-a-squared-term fast, as derived below.

Both factors (thermodynamic and kinetic) are essential. Without the first, the driving is absent to form regions of the new phase. Without the second, the formation process is slow or unpredictable (as in the case of room-temperature diamond decomposing to form more stable graphite, say, or a cup of microwaved water in a smooth, clean container that suddenly erupts partially into steam, or a bottle of ice from the freezer that quickly freezes when tapped).

Everything above repeats the standard explanation; the point of this answer is to present quantitative models, which are less often surveyed.

To address a point in the body of your question:

My counterargument: Intermolecular forces between molecules are either intact or broken. There is no in-between. Therefore, the change from intact to broken is instantaneous.

I don't think this is a great argument, as bonds are forming and breaking all the time in real materials at finite temperatures. Liquids have some fraction of broken bonds compared to the solid state, and the bonds themselves are fluctuating constantly. Vacancies—atomic-scale defects—are themodynamically required at any temperature. This certainly seems like an "in-between" to me.

In any case, the conceptual view can be shored up by replacing forces and bonds with the macroscale Gibbs free energy $G = H - TS$, where $H$ is the enthalpy (a measure of bonding, which Nature prefers), $T$ is the temperature, and $S$ is the entropy (a measure of broad possibilities, which Nature also prefers). Note that close bonding and broad dispersion act in opposition here. Note also that the entropy-containing term includes the temperature, predicting a temperature dependence that ends up being pivotal.

The phase with the lowest Gibbs free energy is always the equilibrium state, as mediated by the tradeoff between enhanced bonding and enhanced entropy. In addition, the higher-entropy phase is always the higher-temperature stable phase. For the typical single-component system discussed here, two phases can have the same $G$ at only one temperature: the phase-change temperature. This is part of why we would see a sudden transition at this temperature.

We can visualize this behavior in the following image. The lowest curve is the most stable. On the left side, the curves are more shallow because entropy $S$ (the negative slope) decreases toward zero with decreasing $T$. But the larger-entropy phases always win out with increasing temperature because of the $-TS$ term. (If the liquid curve were a little higher, we'd have a material such as carbon dioxide that doesn't exhibit an equilibrium liquid phase—at least not at atmospheric pressure—and sublimates directly from the solid to the gas state upon heating.)

Now let's get into why the temperature seems to stay constant during a phase change. It doesn't, in fact, but it is true that a only a very slight amount of undercooling or overheating can be sufficient to trigger a complete rapid transformation from the less stable phase to the more stable phase. The following derivation follows Porter & Easterling, Phase Transformations in Metals and Alloys, but does not assume any particular material class.

Consider slight overheating past one of the transition points in the diagram above. The volumetric driving force for the phase transformation is the reduction in Gibbs free energy, $\Delta G=\Delta H - T\Delta S$, where $\Delta$ compares the state values between the two phases. Close to the intersection, $\Delta H$ is just the latent heat $L$ of the transformation. In addition, we know that $\Delta G =0$ at the transformation temperature $T_t$, so by linear approximation $$\Delta G = L\left(1-\frac{T}{T_t}\right)=\frac{L\Delta T}{T_t}.$$ Thus, the driving force $|\Delta G|$ for transformation is modeled as increasing linearly with increasing overheating or undercooling $\Delta T$.

Now consider the benefits and penalties of nucleating a new phase in the form of a spherical molecular cluster with radius $r$. The benefit is the reduction in Gibbs free energy $\frac{4}{3}\pi r^3\Delta G$. The penalty is the new interface area $4\pi r^2\gamma$, where $\gamma$ is the surface tension of the new phase with respect to the old phase.

Clusters form and dissipate constantly from random particle motion; for just a few molecules, $r^3\ll r^2$ (a sloppy comparison, but apply the relevant coefficients for dimensional consistency). If the overheating/underheating is low, there's no energetic driving force for further growth, and the cluster is likely to just come apart. But a larger temperature excursion alters this balance substantially.

A cluster becomes stable (at radius $r^\star$) when a size increase starts to produce a net energy decrease, i.e., when a maximum exists and the derivative is zero: $$\frac{d}{dr}\left(4\pi r^2\gamma-\frac{4}{3}\pi r^3\Delta G\right)=0.$$ Solving this equation, we find that at the stable radius, the activation energy or energy peak that needs to be overcome is $$\Delta G^\star=\frac{16\pi\gamma^3}{3(\Delta G)^2}.$$ The corresponding expected volumetric nucleation rate $N$ from Arrhenius theory is then $$N=f\exp\left(-\frac{\Delta G^\star}{kT}\right)=f\exp\left(-\frac{16\pi\gamma^3}{3(\Delta G)^2kT}\right)=f\exp\left(-\frac{A\gamma^3}{(\Delta T)^2}\right),$$ where $k$ is Boltzmann's constant and where $f$ is a characteristic (very high) vibration frequency of atomic activity in the material; Porter & Easterling use $f\approx 10^{11}\,\mathrm{Hz}$, for example. To simplify the discussion, take $A$ to be a constant that incorporates everything except two parameters of interest: the surface tension difference $\gamma$ and the overheating/undercooling $\Delta T$.

The surface tension $\gamma$ would be important if every cluster interface routinely separated only the old phase from the new phase. However, $\gamma$ can be much lower if the cluster initiates on a particle of dirt or a crack in a container wall that has a high energy anyway because of unsatisfied bonds. The former case is called homogenous nucleation, and the latter—which we often see in our dirty, defect-filled world—is called heterogenous nucleation.

For small $A\gamma^3$, consider the nucleation rate $N\sim \exp\left(-\frac{A\gamma^3}{(\Delta T)^2}\right)$. We see a tremendously rapid increase in nucleation for slight overheating/undercooling because we're applying the exponential function to a squared temperature difference:

The kinetics of the process are so strong that essentially any overheating is immediately shunted into breaking bonds to form the now-favored higher-temperature phase. (Conversely, any undercooling past the phase transition temperature is immediately inundated with energy released from bond formation to provide the now-favored lower-temperature phase.)

This is ultimately why phase transitions seem to occur entirely at a single well-defined temperature. That's not quite the case, as shown above, but it can be a reasonably good approximation.

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    $\begingroup$ This is the correct answer and should be at the top! $\endgroup$
    – R. Rankin
    Commented Feb 20, 2021 at 6:31
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    $\begingroup$ It would be good to add a little further explanation of the "sudden transition". e.g., explicitly noting that at the transition temperature, both phases coexist in equilibrium, and adding or removing heat causes one phase to grow and the other to shrink. $\endgroup$
    – nanoman
    Commented Feb 21, 2021 at 6:42
  • $\begingroup$ @nanoman Good suggestion; done. $\endgroup$ Commented Feb 21, 2021 at 23:14
  • $\begingroup$ Another, essentially equivalent way to view this is looking at the entropy as the natural log of the number of microstates. Then temperature is related to the change in internal energy with the change in microstates. During phase transitions, the slope of internal energy appears constant on average. $\endgroup$
    – R. Rankin
    Commented Mar 3, 2021 at 19:10
  • $\begingroup$ @Chemomechanics hey, U seem like a knowledgeable person who know very much about phases and thermodynamics, I am a high school student and unable to understand this question:physics.stackexchange.com/questions/718145/… ......can u please explain this to me in simple language considering that I am in high school .If u are not ready to post the an answer ,,we can discuss this in comments also .Please help me with this. $\endgroup$ Commented Jul 15, 2022 at 8:55

In order to undergo a phase change from a solid to liquid, or liquid to gas, the heat added needs to cause molecules to move farther apart from one another, in opposition to the intermolecular attraction forces, which in turn means it increases the internal potential energy of the substance. Thus the heat added increases internal potential energy and not internal kinetic energy (an thus not the temperature)

The gravitational potential energy (GPE) analogy is increasing the vertical separation between an object and the surface of the earth by raising it to rest at a certain height $h$ above the surface. Work (energy) input by the agent lifting the object increases the GPE of the object without increasing the kinetic energy of the object.

Hope this helps.

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    $\begingroup$ yes yes, the idea of separating the internal potential energy and the kinetic energy makes this perfectly clear, thanks for your answer! I'm glad I didn't try to answer this one... $\endgroup$ Commented Feb 19, 2021 at 22:57
  • $\begingroup$ Hm, I hadn’t thought about potential energy. But, in this case, isn’t kinetic energy needed in order to move the molecules in order to increase potential energy? And why wouldn’t your explained effect apply to the substance before it reaches that state-changing point? How I see it, applied heat increases both potential and kinetic energy regardless of the temperature of the substance. $\endgroup$
    – Raen
    Commented Feb 19, 2021 at 23:43
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    $\begingroup$ @Raen Not really, no. Consider the gravitational case. Raising an object from a low circular orbit to a higher one requires an energy input that increases the GPE, but the KE decreases! Similarly for a phase change: the bonding forces seal the excess KE of an escaping molecule away as PE. (If there isn't enough KE to escape, then that energy oscillates between PE and KE; the molecule wiggles.) The molecule cannot come back to the condensed phase because that PE will turn into KE and either make it separate again or knock another molecule out in its place. $\endgroup$
    – HTNW
    Commented Feb 20, 2021 at 22:54
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    $\begingroup$ This answer doesn't clearly resolve the question IMO, because the ideas of equipartition and virial relations indicate that potential and kinetic energy are linked. So from this viewpoint it still does seem odd (as OP notes) to increase one with no effect on the other. $\endgroup$
    – nanoman
    Commented Feb 21, 2021 at 5:27

In order to break the bond, the movement of the molecules needs to be intensified, which would increase the temperature of the water. That’s why I think that there should be no period of constant temperature.

Fundamentally, temperature is not really about the "intensity of movement" of the particles in a system. This is the reason behind your confusion.

Think about how we measure temperature: we use a thermometer. The thermometer cannot directly detect the movement of particles in the system that it is in contact with. What the thermometer does do is to either yield heat to the system, or absorb heat from the system. When at last equilibrium is reached, where the rates of heat flow from the thermometer to the system and from the system to the thermometer are equal, then the amount of thermal energy remaining in the thermometer allows us to infer the temperature.

When two systems (or two phases within a system) are in thermal equilibrium with each other, we define them to have the same temperature (regardless of what is actually going on at the microscopic level within each system). When one system has a tendency to lose heat to the other, the former, by definition, has a higher temperature than the latter. It is only because temperature is defined in this way, that a thermometer has the ability to reliably measure temperature.

Water that is at its boiling point is in equilibrium with water that has just boiled. If the water continues boiling, the heat that makes that happen is not coming from the steam, but from the heating element. If you remove a kettle from the stove, you have a situation where the liquid water will not lose heat to the steam nor will the steam lose heat to the liquid water. The water will not continue to boil, since there is no longer energy input for breaking bonds. The steam will also not spontaneously condense into water, because doing so would release energy and there is nowhere for that energy to go. So the proportion of the water in the system that is liquid, and the proportion that is gaseous, will remain constant unless the system is interacting with its environment somehow. Internally, the system is at equilibrium. This proves that the liquid water and the steam have the same temperature, even though the steam has more energy per molecule than the liquid water.

It similarly follows that any first-order phase transition (that is, a discontinuous phase transition that involves a latent heat) is a constant-temperature phase transition.


The change from intact to broken is instantaneous, but for each bond. Phase changes for a bulk of material takes time. While it is happening, it is the potential energy not the kinetic energy that is changing.

For any region where the temperature starts to rise $\delta T$, that energy is transfered to not yet transformed neighborhood. At least for phase change near to equilibrium.


In order to break the bond, the movement of the molecules needs to be intensified...

This is incorrect in the sense you've stated.

With every bound system is associated a binding energy, which, when supplied in appropriate form to the system, unbinds the system. For example, a satellite orbitting the earth is bound by gravitational energy. In order to unbind the satellite significant energy would need to be added to remove it from orbit. However, this does not imply that the satellite must be moving faster (implying higher temperature by analogy) once the bond is broken.

  • $\begingroup$ He's talking about quantum-mechanical bonds. These can't break in a continuous fashion like a satellite breaking away from earth, they break spontaneously (i.e. without an observable time interval where the bond is in some intermediate state). $\endgroup$
    – user132372
    Commented Feb 21, 2021 at 14:52

I also don't like explanations on a broad energy level. They explain how the bookkeeping works, but not really why that's the right way to do the bookkeeping.

So consider things on a molecular level. Have you ever tried to throw a loose wad of tape into the trash, but it stuck to your hand a little and only got halfway to the trash can? The breaking of bonds during a phase change is like that. The molecules are oscillating rapidly in a deep potential well. Every so often, one gets jostled hard enough to escape the potential well, but it loses speed on the way out, so that it is moving slower when it hits another molecule of the surrounding liquid. Meanwhile, the remaining molecules, instead of bouncing back off their fast-moving neighbor when they approach it, are given a tug in the direction they were moving that has the effect of softening their rebound caused by the attraction of their neighbors. So molecules on both sides are now moving more slowly. This process can happen in reverse too, and at the melting point both are basically equally likely. So adding heat just makes more molecules detach, compared to the number reattaching. Only after melting is complete and there's basically no more reattaching does the average speed go up.

This explanation, I should say, is inspired by Feynman's description of the atomic hypothesis:

that all things are made of atoms — little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repelling upon being squeezed into one another.


Let's look at this from the energy side:

  • From 0°C to 100°C, energy increase translates to a (mostly) proportional amount of temperature increase.
  • At 100°C, energy increase translates not to temperature increase but to an amount of water vaporized. The water vapor will be 100°C, too, i.e. vapor and liquid are still in thermal equilibrium.
  • Above 100°C, energy increase will again translate to temperature increase.

(It's not as linear as I'm describing things here, but the details don't change the picture significantly.)

Note that the entire thing is completely independent of the mechanics of the phase change; all that matters is that one phase has a different energy content than the other, and that the material can partly undergo the phase change.
(This means that your question does make sense for systems that cannot change phase in a piecemeal fashion but must do it all at once. I am not aware of any example, but I could imagine that some large quantum-entangled systems could expose such behaviour.)

This mechanism applies to all phase changes: superheating, subcooling, crystallization, crystal restructuring. It is also independent of whether the phase change is induced by heating, cooling, (de)pressurizing, external magnetic field, or whatever: At the moment of the phase change, whatever is causing it will have its effect absorbed into the phase change until everything has changed, then the normal effect will resume.


The bonds themselves represent a decrease is chemical potential energy. The energy needs to go into this to break it. As a metaphor consider you have a ball in a grass field. If there is a hole in the field and the ball goes in it,it will come to rest at the bottom of the hole and won't leave. For you to put in back in the field where it can roll around takes adding gravitational potential energy. If you kick the ball in the hole it will come out with the kinetic energy of the kick minus the amount needed to get the ball out of the hole.

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    $\begingroup$ While a chemical reaction involves the breaking and forming of bonds within molecules, phase transitions involve the breaking or forming of intermolecular forces. $\endgroup$
    – Bob D
    Commented Feb 19, 2021 at 22:04
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    $\begingroup$ You are absolutely correct. But the base intuition is identical $\endgroup$ Commented Feb 19, 2021 at 22:12
  • $\begingroup$ The base intuition, yes, phase transitions are similar to chemical reactions as they have an associate enthalpy change. But. as I think you'll agree, the nature of those enthalpy changes are not necessarily the same. $\endgroup$
    – Bob D
    Commented Feb 19, 2021 at 22:28

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