The entropy of a system undergoing a phase transition increases if the phase transition is towards higher internal energy (e.g., melting) and decreases if the phase transition is towards lower internal energy (e.g., freezing). The change in the entropy of the surroundings is of opposite sign. If the phase transition occurs quasistatically at thermodynamic equilibrium, then the total entropy change of the system plus surroundings combined is zero, otherwise the total entropy change is positive.

Dear Jack, thanks for the very clear response: it is enlightening!

I have to agree with Jack. But may be some qualifications needed which he obviously is aware of . I'm not goirng to consider anything specific about Hopf bifucations since that's  just the simplest bifurcation you can have in a system with 2-d state space with oscillatory/rotational motion like in a 1-d complex wave or in a 2-d rotational motion like e.g, in fluid vorticity etc.

. The first thing i would like to point out is that what he means are probably not inner energy , but free energy. Also I would like to point out that what he is saying is only valid for near equilibrium systems with detailed, stochastic  balance (see below) and linear perturbations and a system with positive energy production , and as i will argue for possible for and possibly only for no variation in initial background entropy- i.e . fixed background in space say. Reason why i am saying this is that I've in one of my many unpublished works defined the meaning of dynamic free energy for fluid mechanics which have  in free energy a dynamic energy v^2 /2  in addition to the internal and subtracted -TS term and a couple of others. Also my fluid model free energy quite natural separates out the natural basic modes such as the gradient longitudinal part , and the vorticity part which notorioullsy has a negative energy part giving rise to instability if you take vort.  energy  away and probably decrease entropy (never checked, but..obvious?)  and even the heat modepart which people does not look at dynamically for some reason i've never understood since they are  assuming adiabaticity or even constant fixed entropy. .  If you now go to a system with also dissipation etc you will find that in addition to the dynamical entropy variations ,there will be also be  heat etc  mode variations in entropy  and this total entropy part might increase,  but may be not due to that:  1. The for sure increase is only valid I now believe in the fluid frame and in the case of the dynamic perturbations  created due to dissipation are in detailed balance which for like

dear Tor,

thanks For the very interesting comment. As we Know from your model, the changes in free energy are not important in brief timescales, but in evolutionary ones. This is an important achievement you made, which sounds different from "classical" schemes, such as, I.e., Friston's free energy, where free energy needs to decrease also in very brief brain timescales

The statement of the question is logically independent from the text. The answer to the question is that entropy changes, if the transition is first order, it doesn't if the transition is second order. 

The entropy does change in a second-order phase transition, but the entropy change is "spread out" over a sizable temperature range. In a first-order phase transition, the entropy change is concentrated into an extremely small temperature range.

No, these statements don't make sense in the context that they have any meaning at all, namely thermodynamics:  the entropy is a continuous function of the temperature, across a second order transition and is a discontinuous function of the temperature at a first order phase transition: there isn't any latent heat, at all,  in the former case, there is latent heat in the latter case. The absence of latent heat is the statement that entropy doesn't change at a second order phase transition, since DQ = TDS and the LHS is the latent heat. 

There is a latent heat in a second-order phase transition, but it is "spread out" over a sizable temperature range, as "extra" heat capacity over this sizable temperature range over and above the heat capacity that would have obtained in the absence of the second-order phase transition. In a first-order phase transition the heat capacity reaches an extremely high value on the order of (N^2)k (compared to normal heat capacities on the order of Nk) over an extremely tiny temperature range on the order of L/[(N^2)k] centered on the temperature T* where the latent heat L is equal to T*dS. This temperature range is on the order of L/[(N^2)k] rather than zero because the total entropy of a material undergoing a first-order phase transition plus its heat bath combined can by Boltzmann's relation between entropy and probability typically fluctuate to a value on the order of k below its maximum value. Thus there is a significant probability that the substance will be frozen if the heat bath is at temperature T* + L/[(N^2)k] or melted if if the heat bath is at temperature T* - L/[(N^2)k].

Entropy: A concept that is not a physical quantity

https://www.researchgate.net/publication/230554936_Entropy_A_concept_that_is_not_a_physical_quantity

Following

The understanding(NOT definition) on heat and work( or say, process quantity) of the standard textbook IS WRONG !

The standard textbook says that

1) heat Q and work W are process quantities. This is right.

2)heat Q and work W(or say, process quantities) can not be the functions of system state variables. However, this is WRONG !

In fact, when the process path is fixed, the process variable is ALSO a state variable.

So,

on the one hand, heat Q and work W are process quantities.

on the other hand, when the process path is fixed, to the fixed process path, heat Q and work W are state quantities.

When we discuss the heat and work, what we discuss is just the heat and work of ONE process path, although it can be any one process path, but it is just ONE given(fixed) process.

All this time, people thought that dQ was not a complete differential, but dQ/T was considered as a complete differential.

However, this is wrong.

The facts are just the opposite: dQ is actually a complete differential, but dQ/T is meaningless.

in ΔQ/T, the relationship of Q and T is the ratio of ΔQ and T or the product of 1/T and ΔQ, so, in ΔQ/T, Q and T can be any relationship.

But in dQ/T, the relationship of Q and T is not the ratio of dQ and T or the product of 1/T and dQ, but the relationship to Find the Original Function of 1/T in dQ/T=1/TdQ.

For we know Q is not a single-valued function of T, (in fact, Q=f(P, V, T) ), so, ΔQ/T can NOT turn into dQ/T, or say, dQ/T is meaningless.

The problem is not whether dQ is meaningful or not here, it is 1/TdQ is meaningless !

Q = f(T, V, P) is a process quantity which varies with path, it has innumerable forms between the same original and terminal states, and has a unique form for fixed reversible process path. When the given path is fixed, Q = f(T, V, P) is the system state variable.

So, dQ=df(T, V, P) is a perfect differential, it is meaningful;

but the integral variable of 1/TdQ is self-contradictory (T and T, V, P), so, 1/TdQ=1/Tdf(T, V, P) is meaningless, that is ∫T 1/TdQ = ∫T 1/Tdf(T, V, P) is not a meaningful integral, or say, it is not a integral at all.

That is, ΔQ/T can NOT turn into dQ/T,

there is not the result∮1/TdQ=0 at all.

The so-called "entropy" does NOT exist at all.