Why can a block of ice at $0^\circ \rm C$ and water at $0^\circ \rm C$ coexist in an insulated bucket without changes either way in the amount of ice?


The textbook I read mentions nothing of state transformations of matter; so if this phenomenon can be explained without it, I'd appreciate it very much.


This question is posed in chapter 6 ("Dynamics of Heat") of "Understanding Physics" (Cassidy, Holton, Rutherford). The book is free to download (legally) here.

  • $\begingroup$ Could you add some details on how this seems surprising to you? If you mix ice and water both at $0ºC$, what did you expect to happen? They can't exchange heat, since the temperature is the same, so it seems obvious to me that nothing special would happen. $\endgroup$
    – Wood
    Oct 19, 2016 at 6:57

3 Answers 3


It is indeed a very active scene under this picture. Frequent transformation from ICE to WATER and WATER to ICE. But the total amount cannot change since the temperature is kept the same.If you look at this thing at the microscope level, you should only see the vibrating molecules. Ice molecules are more tightly bound than the liquid water molecules. If it is indeed at zero degree, then there could be no amount of energy being transferred to this mix. What you actually see is that for a small volume (suppose this is the ice form) , the molecules are vibrating. If suddenly one molecule at the boundary of this small region absorbs a bit energy from the others around, then the chemical bounds around this molecule become looser. You may regard this process as the transformation of the form ICE into WATER. But, since the energy is from some other molecules', for these molecules losing energy they may change from WATER to ICE.

  • $\begingroup$ It seems as if statistically there is an equal amount of water-to-ice and ice-to-water, therefore there is no net gain/loss of ice. Does this mean that the amount of ice could noticeably increase or decrease, but is unlikely? $\endgroup$
    – Fine Man
    Oct 19, 2016 at 17:37
  • $\begingroup$ Strictly speaking, you cannot distinguish water from ice at the critical point. This is a point where things can show their differences just because of the difference of the group behavior. To be more clear, think of a group of water molecules. You define it to be water because you look at them as a whole but not as a single molecule. $\endgroup$ Oct 22, 2016 at 2:19
  • $\begingroup$ Then I can take some energy from each of these chemical bounds and this group water is going to the form of ice but not exactly. When this group reaches the critical point, we randomly begin to see some part of this group looks like the form of ice. If I then take more energy from the chemical bounds, more and more parts tend to look like ice. But this is just our human view from the macroscopic sense. $\endgroup$ Oct 22, 2016 at 2:20
  • $\begingroup$ @Sir Jony If you want to know whether there will be a noticeable increase of ice, I would say it depends. The molecules are vibrating frequently, and 0 degree is actually a very hot condition for these molecules. Therefore they frequently exchange their energies with each other so that in our time scale we always see and measure an average value. If you are fast enough and you can take a picture of this at a really short instant, you will probably see, ha, there are so "many ices" than the average value. $\endgroup$ Oct 22, 2016 at 2:27

Plain and simple: at any phase transition, both states (in this case, water and ice) are in thermodynamic equilibrium, meaning there will be no spontaneous transition from one to the other (none of them is preferred at that temperature), but you still need to transfer energy to turn from one to the other: you need to heat to turn ice into water and vice-versa. So the amount of ice in the mixture stays constant unless you add or remove energy.

  • $\begingroup$ But at the microscopic level (as @ZHANGJuenjie mentioned), wouldn't there be "hot" (because heat is the average kinetic energy of the particles making up a body) water molecules smashing into the ice and breaking some "ice" molecules off the cube? Thus, the amount of ice won't stay in equilibrium? $\endgroup$
    – Fine Man
    Oct 19, 2016 at 17:41
  • $\begingroup$ Yes, but just as many slow water molecules would hit the ice and get stuck on it. That's what a detailed equilibrium means. Thermal motion is a random process - in equilibrium, it averages out. You always have a whole spectrum of velocities (and the AVERAGE kinetic energy corresponds to temperature) - if the fast ones deposit energy to break a slow molecule off ice, the average kinetic energy would slightly go down, and if a slow molecule sticks to ice, the average kinetic energy would go up. When you look closely enough, things fluctuate, but tha average stays the same. $\endgroup$
    – orion
    Oct 20, 2016 at 7:04

The statement "insulated bucket" tells one that there is no heat transfer between the ice and water and the outside world.
This means that the overall average number of hydrogen bonds made between the water molecules cannot change as it takes energy to break a bond.

The statement that both the ice and the water are at the same temperature means that if they come together they will be in thermal equilibrium which means that there will be no net heat transfer between them. On the molecular scale this means that on average when liquid water molecules collide with ice molecules there is no net transfer of kinetic energy between them.

So you have ice with each water molecule having four hydrogen bonds and liquid with on average each water molecule having slightly less than four hydrogen bonds coexisting together.
There can be an interchange as to which hydrogen bonds are actually made or broken but overall the number stays the same.

The interesting thing is that you allowed some heat into the bucket then as long as there was ice present the temperature would stay at $0^\circ \rm C$ with some of the ice turning into water.


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