# Latent heat vs temperature of phase transitions?

Is the latent heat associated with phase transitions correlated with the temperature at which they occur?

The latent heat is related to the difference in energy between the two phases, and the temperature of the phase transition occurs at the point where the difference in energy between the two phases is comparable to thermal fluctuations. What factors would lead to departures from a linear relationship between the two, e.g. in different materials or different phase transitions in the same material? Or is there an error in my assumptions?

Let me state first that one should not speak about energies but about free energies, as soon as temperature is not absolute zero. Discussion of energies has in this case no sense.

Second, a very simple answer to the original question is pointing out that according to its definition the heat, $Q = T \Delta S$, where $S$ is entropy, $\Delta S$ is its difference in the two states. You may find discussion of this point in the 5th volume of Landau and Lifshitz's book, Eq. (13.4) and around. This is valid also for the latent heat. Thus, the answer, is yes, it depends upon temperature. This alone does not give one too much, since one needs to know the entropies in the both phases.

Third, you are wrong to think that transition takes place "at the point where the difference in energy between the two phases is comparable to thermal fluctuations". That is simply an error. The condition of any phase transition is $F_1=F_2$, where $F_{1,2}$ are the phases free energies. One may use also other thermodynamic potentials instead depending upon details of the problem statement. In the case of the 2nd order transition one can use a more simple relation, but the equality $F_1=F_2$, nevertheless, stays behind that relation.

Your question "What factors would lead to departures from a linear relationship between the two, e.g. in different materials or different phase transitions in the same material?" is completely unclear to me, but in any case I would not strongly expect anything related to phase transitions to be linear.

• Welcome to Physics.SE, Alexei. We have MathJax---a LaTeX rendering engine---active on the site so that you can write neatly formatted mathematics. I've attempted to do this post just to give you the idea. As always you re free to revert or correct any edits that you don't like. Oct 4, 2012 at 17:10
• I know. Unfortunately that has been impossible the day I did it. No latex have been visible, and site was unstable. On the other hand there is no serious difference, to format equation like F1=F2 or not. It will be anyway understood. Oct 8, 2012 at 11:27

The link between temperature and latent heat associated to any phase transition can be found using classical thermodynamics.

Rigourosly talking, phase changes can be described using the Gibbs free energy potential, $G=U+PV-TS$. It follows from the Second Law of Thermodynamics that any closed system (which is one that cannot transfer any mass across its boundaries, but it may involve heat and/or work exchanges with its environment) that evolves from an initial state to an equilibrium state at constant pressure and temperature, will minimize $G$, which is nothing but an extensive property.

You can think of Gibbs free energy as a way to physically represent and follow an evolution for most natural systems and find equilibrium states. Equilibrium states are static: once you've reached them, the system will not abandon them unless its conditions are disrupted by an external event. The steady coexistence of material phases is just one example of an equilibrium state.

So, for example, if you have a chemically pure system consisting of two phases not in equilibrium (think about filling up bottle of water originally dry, and closing it leaving a space over the water line, allowing evaporation) and realizing that it can be modelled as a closed system at constant ambient temperature an pressure (the system's boundaries are given by the atmosphere) you certainly can calculate its final state by minimizing the system's free energy, which as an extensive property its equal to the sum of its subsystem's free energies:

$G_{sys}=G_{w}+G_{v}\Rightarrow \displaystyle\frac{dG_{sys}}{dt}=\displaystyle\frac{dG_{w}}{dt}+\displaystyle\frac{dG_{v}}{dt}\leq0$

Without further information, there isn't too much we can say about the evolution of the system. Here we are in a non-equlibrium situation (that is, the place for transport phenomena-related stuff) and so we have a time response of the system, which is dynamically changing its properties. That means that there is a net mass flux from the liquid phase to the vapour phase as long as the evaporation continues.

Nevertheless, given enough time this system will reach a steady, non-changing state, easily recognizable by the fixed liquid level. At this point the free energy of the system will be at a minimum, and become independent of time. So:

$\left. \displaystyle\frac{dG_{sys}}{dt} \right|_{eq}=0$

Even though we know the system has reached equlibrium (and there won't be any more spontaneus stuff happening any more) we can study a reversible evolution, that in this case involves a macroscopically undetectable shift of the system's extensive properties while keeping at equilibrium. Recalling the liquid and vapour phases free energies:

$\left.\displaystyle\frac{dG_{w}}{dt}\right|_{eq}+\left.\displaystyle\frac{dG_{v}}{dt}\right|_{eq}=0\Rightarrow\left.\displaystyle\frac{dG_{w}}{dt}\right|_{eq}=-\left.\displaystyle\frac{dG_{v}}{dt}\right|_{eq}$

Now, we can take a step further and express this reversible process in terms of the mass of each phase and specific properties:

$\left.\displaystyle\frac{dG_{w}}{dt}\right|_{eq}=-\left.\displaystyle\frac{dG_{v}}{dt}\right|_{eq}\Rightarrow \left.\displaystyle\frac{d(M\cdot \widehat{g})_{w}}{dt}\right|_{eq}=-\left.\displaystyle\frac{d(M\cdot \widehat{g})_{v}}{dt}\right|_{eq}\\\left.\displaystyle \widehat{g}_{w} \cdot \frac{dM_{w}}{dt}\right|_{eq}-\left.\displaystyle M_w\cdot\frac{d \widehat{g}_{w}}{dt}\right|_{eq}=-\left.\displaystyle \widehat{g}_{v} \cdot \frac{dM_{v}}{dt}\right|_{eq}-\left.\displaystyle M_v\cdot\frac{d \widehat{g}_{v}}{dt}\right|_{eq}$

Just as $G$ is an extensive property, $\widehat{g}$ is an intensive one, like density, viscosity or refraction index. Now both $\widehat{g}_{w}$ and $\widehat{g}_{v}$ are fixed because of macroscopical equlibrium, but that is nothing but a fancy way of saying that the two phases are (reversibly) exchanging matter between them, as the rest of the terms drop out. Also, the fact that the system is closed means that the increment in one phase's mass can only be the result of a decrease in the other:

$\left.\displaystyle \widehat{g}_{w} \cdot \frac{dM_{w}}{dt}\right|_{eq}=-\left.\displaystyle \widehat{g}_{v} \cdot \frac{dM_{v}}{dt}\right|_{eq}\\\left.\displaystyle \frac{dM_{w}}{dt}\right|_{eq}=-\left.\displaystyle \frac{dM_{v}}{dt}\right|_{eq} \Rightarrow (\widehat{g}_{w}-\widehat{g}_{v})\cdot\left.\displaystyle \frac{dM_{w}}{dt}\right|_{eq}=0$

Finally, as the reversible exchange rate can exhibit a non-unique set of values (and in principle, setting it at zero would be a contradiction with this line of reasoning), the conclusion is that the equilibrium specific free energies are equal for both phases:

$\widehat{g}_{w}=\widehat{g}_{v}$

This is the very condition that two phases at equilibrium must verify.

But what about latent heat and temperature?

Firstly, "latent heat" is the informal (engineering-like) name given to the specific enthalpy change in a system which undergoes a phase transformation. So what we are really looking for is a relation between enthalpy change and temperature, and this is easily deduced from the previous reasoning.

Expressing the specific free energy in terms of enthalpy and entropy at equilibrium:

$\widehat{h}_{w}-T_{eq}\cdot\widehat{s}_{w} =\widehat{h}_{v}-T_{eq}\cdot\widehat{s}_{v}\\(\widehat{h}_{v}-\widehat{h}_{w})=\lambda=T_{eq}\cdot(\widehat{s}_{v}-\widehat{s}_{w})$

Making use of an appropiate Maxwell relation, and realizing that (for pure substances) the pressure at phase equilibrium its only a function of temperature:

$\displaystyle\frac{{\partial \widehat{s}}}{{\partial \widehat{v}}}=\displaystyle\frac{{\partial P}}{{\partial T}}=\left.\displaystyle\frac{{dP}}{{dT}}\right|_{eq}\Rightarrow \int\partial \widehat{s}=(\widehat{s}_{v}-\widehat{s}_{w})=\int\left.\displaystyle\frac{{dP}}{{dT}}\right|_{eq}\partial \widehat{v}=\left.\displaystyle\frac{{dP}}{{dT}}\right|_{eq}\cdot(\widehat{v}_{v}-\widehat{v}_{w})$

Then:

$\left.\displaystyle\frac{{dP}}{{dT}}\right|_{eq}=\displaystyle\frac{\lambda}{T_{eq}\cdot(\widehat{v}_{v}-\widehat{v}_{w})}$

This is the Clausius–Clapeyron relation, and it expresses the slope of the single component, phase equilibrium states pressure curve.

EDIT: As latent heat is actually the change in the specific enthalpy of the substance when going from one phase to another, and specific enthalpy is an intensive property (which, by the Gibbs-Duhem equation, gets uniquely determined for a pure substance by it's temperature and pressure -and at equilibrium, pressure is a scalar function of saturation temperature- so it's actually a function of temperature only) then yes, latent heat is correlated to the temperature of the phase transition. But that's not the end of the story: the Clausius-Clapeyron relation tells you a way to compute that relation without making use of a calorimeter (i.e. measuring directly the latent heat of the phase transition). You just have to measure saturation pressures, temperatures and densities at equilibrium.

• this doesn't answer the question really--- can you change the transition temperature (by fiddling with parameters, like atomic masses--- like changing the isotope) without changing the latent heat. Jul 27, 2012 at 7:22
• @RonMaimon: If I understood the question, the OP was asking for a relation between latent heat and temperature in a phase transition (I'm actually transcribing the title) and that's what I tried to explain. I think that points right to the Clausius-Clapeyron relation (at least for pure substances, which is enough generality, pedagogically talking), and introducing it needs a previous insight into thermodynamic equilibria (and that's why I bothered to give the long introduction).
– Mono
Jul 27, 2012 at 15:42
• The OP was asking "All else being equal, is the latent heat of a transition proportional to the temperature?" His intuition is that if you have a low-temperature or zero temperature transition, it should have less latent heat because the energy in each molecule is something like kT. This intuition is not great, and it's obliquely related to Clausius Clayperon, but now that I think about this more I see you are right, this is a good answer +1. Jul 27, 2012 at 17:08
• Well, then it reduces to interpreting the OP's original intuition, and if you are right, then I should have started by asking him to express his question more clearly. I assumed tacitly it wasn't necessary to explain that latent heat is a difference, a "jump" between two states, and that temperature is a measure of an absolute property (i.e. mean molecular kinetic energy), so it makes no sense comparing directly a relative to an absolute parameter in the sense you state. Thanks for the feedback.
– Mono
Jul 27, 2012 at 17:30
• I think I don't entirely understand what do you exactly refer to when you say: " if the difference between the two phases is purely energetic, like breaking atomic bonds, then the free energy doesn't depend on T". Every classical equilibrium phase transition in the sense I described here implies both a difference in energy (enthalpy) and entropy between the phases; the temperature of the transition is just the quotient between this two differences. And by definition, there's no difference in the free energies of the phases at equilibrium (you may say chemical potentials, in general).
– Mono
Jul 27, 2012 at 23:53
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