Thermodynamics - do all particles in a system undergo a state transition at exactly the same time?

Question

Basic question:

Why does temperature remain constant during a state change?

The general answer I find in most places is that during a state change, the energy supplied is used to change the potential energy of the molecules in the substance and not the kinetic energy.

On Quora, one user writes "When the solid has been completely melted into the liquid phase, adding more energy will once again increase the average kinetic energy of the particles (make them move faster)."

With this in mind, my main question is this:

Do all particles in the substance undergo the state transition at the same time?

Motivation/thought experiment:

Consider a solid to liquid state transition over an interval $0 \leq t \leq T$.

Individual particles in the substance will break bonds at different points in $[0, T]$.

Heat being input into the system, if distributed randomly, could therefore act to increase the average kinetic energy of these early transitioned particles since their bonds have already broken. Therefore, the average kinetic energy of the system will therefore also change (perhaps slowly, perhaps quickly, I'm not sure) over the interval $[0, T]$, which by definition means the temperature of the system will change.

Instead, however, we observe a very flat line on heating curves at a state transition as if to suggest (as stated by the Quora user), it is only when the whole solid has completely melted does the temperature start to increase.

So what happens to the particles that break their bonds early on in the interval $0 \leq t \leq T$? What dictates that the whole system should be completely melted before the temperature increases?

Hope my question makes some sense.

Answer 1

Do all particles in the substance undergo the state transition at the same time?

Not necessarily. It depends on the temperature difference between the substance and the heat source/sink, as well as the physical characteristics of the substance, such as its surface to volume ratio.

For example, let's say we have an ice cube. If we place it directly onto a hot frying pan it will undergo a change in state from ice to water. Clearly the ice at the surface of the cube will change state from a solid to a liquid before the ice in the interior of the cube does. So the intermolecular bonds of the ice at the surface will break before the intermolecular bonds in the interior. Once those bonds are broken at the surface, additional heating will increase the molecular kinetic energy (raise the temperature of the water produced at the surface).

On the other hand, if the ice cube is subjected to an environment with a temperature slightly greater than the melting point of ice, so that temperature throughout the cube is nearly the same, the intermolecular bonds throughout the cube will be broken at nearly the same time.

Regardless of the timing of the breaking of the bonds, the temperature at which the bonds are broken (temperature at which the change in state occurs) will be the same for the entire cube, 0$^0$C at 1 atm, whether it's placed on a frying pan or in a room at a temperature slightly greater than the melting point. Which is why we call it a constant temperature (of the ice) change in state process.

Hope this helps.

Answer 2

Do all particles in the substance undergo the state transition at the same time?

Since your question mentions melting I assume that when you say "state transition" you mean phase transition.

The phase of a substance is a bulk property which only applies to collections of millions of molecules. It does not make much sense to say "this molecule is in a liquid phase, but that one is in a solid phase". At best you might say "this molecule is more tightly bound to its neighbours than that molecule".

It is also perfectly possible for two or more phases to be coexist in equilibrium, with individual molecules transitioning continuously from the part of the mixture in one phase to the part of the mixture in another phase and back again.