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Theoretically speaking, what is the state of water at bang on 0°C - not any lower or higher?

Any lower would make it a solid whereas any higher would make it a liquid. But what about bang on 0°C?

Thanks in advance

Edit: I understand that other factors are involved, such as pressure and temperature which would shift the equilibrium, but in a 'theoretical perspective', what would occur - assuming that all of the particles are at exactly the same temperature with the same kinetic energy?

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    $\begingroup$ What is "water bang"? $\endgroup$
    – Steeven
    May 3, 2015 at 11:47
  • $\begingroup$ @Steeven It's "water" + "bang on" $\endgroup$
    – Turbo
    May 3, 2015 at 11:48
  • $\begingroup$ @Steeven "bang on" is idiomatic language for "at exactly" or "at precisely" $\endgroup$
    – innisfree
    May 3, 2015 at 11:49
  • $\begingroup$ Ahh, never heard that before :) $\endgroup$
    – Steeven
    May 3, 2015 at 11:55
  • $\begingroup$ Related question with answer: physics.stackexchange.com/questions/295794/…. $\endgroup$ Mar 8, 2018 at 21:34

5 Answers 5

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At the transition point between two phases, both states are thermodynamically (meta)stable. The actual composition, however, is kinetically determined and will depend on the history of how you got to 0$^\circ$C and how long you wait.

For instance, if pure water (no impurities and in a container that does not induce heterogeneous nucleation) is cooled slowly enough then it will remain liquid at 0$^\circ$C for a very long time (until a critical ice nucleus spontaneously forms). In fact, it can reach as low as -10$^\circ$C without freezing (i.e. it becomes supercooled).

If you had an infinitely large block of ice (i.e. no surfaces) and warmed it up to 0$^\circ$C then it would also remain solid for a very long time (again, until a spontaneous critical nucleation event). With surfaces, however, the surfaces will melt and the system will slowly move towards a state consisting of both liquid and ice.

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  • $\begingroup$ Thank you for your answer, but I don't quite grasp your use of the term 'history'? $\endgroup$
    – Turbo
    May 3, 2015 at 11:59
  • $\begingroup$ @Leuchte For example, whether you start with ice and warm it up to 0$^\circ$C, or start with liquid water and cool it down to 0$^\circ$C, you will end up with a different composition. $\endgroup$
    – lemon
    May 3, 2015 at 12:00
  • $\begingroup$ So either way, what would occur? Pardon my misunderstanding $\endgroup$
    – Turbo
    May 3, 2015 at 12:01
  • $\begingroup$ @Leuchte I have updated my answer. $\endgroup$
    – lemon
    May 3, 2015 at 12:15
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You need more information to tell the state. Actually, exactly two more values, the pressure and the volume. With these three you have a fixed point in the phase diagram:

enter image description here

But even if you have a fixed point in the diagram, you can still reach a two-phase sitation. Then e.g. water and ice exists at the same time in fitting fractions of the total mass.

What happens at an equilibrium point/line is not a good question. Because, that line is infinitely thin. You can never be exactly on the line.

A usual phase diagram often seen is in 2D, where only pressure and temperature are considered. This is a good illustration, and the $T/p$ graph is to the left:

enter image description here

Giving a diagram looking like this:

enter image description here

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  • $\begingroup$ Thanks for your detailed response. I am aware of the fact that you can never be exactly on the line, but theoretically speaking, what would occur? $\endgroup$
    – Turbo
    May 3, 2015 at 11:47
  • $\begingroup$ @Leuchte Well, theoretically it is impossible. IF it ever happened, then I would assume that you keep the state you came from in the split-second before reaching the line. $\endgroup$
    – Steeven
    May 3, 2015 at 11:48
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I'm assuming 1 atm pressure. It will be a mix of solid and liquid. It takes a certain amount of thermal energy to change the state of water. Until that amount of energy is reached, it will be a mixture of solid and liquid, both at 0°C. With more energy, a higher portion will be water. Once all of it is liquid, any further energy addition will raise its temperature. Once all is solid, energy removal will lower the temperature.

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  • $\begingroup$ "It will be a mix of solid and liquid" Wrong. See my answer. $\endgroup$
    – lemon
    May 3, 2015 at 11:53
  • $\begingroup$ This answer is wrong. If the water is already liquid and at 0°C, it wouldn't necessarily take energy to change the state of water. The final state will not be one of a mix of solid and liquid. $\endgroup$ Mar 8, 2018 at 21:27
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Temperature defines the amount of energy and vice versa. However, supplying or taking small amounts of energy away from water strictly at zero degrees Celsius, is not going to change the temperature! First, this energy will be used to make transition between liquid and solid form, until you get just one kind of form, either all liquid or all solid. Only after that the temperature will change. In other words, to say that the temperature is zero, is not sufficient to define the phase of water. It could be any! Solid, liquid or a mix.

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  • $\begingroup$ This additional, and may be unknown, amount of energy will define the phase of water. It is called "the history" by the user Lemon. For more information see the definition of the phase change energy. $\endgroup$
    – Alexey
    Mar 8, 2018 at 20:41
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H2O at 0°C is Ice. There is a considerable gap between Ice and Water. After 0°C if you increase the temperature by 0.1°C, that is at 273.1K the equlibrium state occures. This state is called the 'Triple point of water'. This is where water, ice and surprisingly water vapour. After this state if you increase the temp by any amount H20 becomes water. Hope this maybe sufficient.

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