The H2O can be held at 0 deg C while adding or removing heat, and this will affect the relative amount of liquid or solid. The OP is not clear on the conditions. Liquid water has 80 calories per gram more heat than ice. If you start with 100% liquid and remove 80 calories per gram, at equilibrium, it will be 100% ice while remaining at 0 deg C. If you start with 100% liquid and remove no heat, equilibrium will still be 100% liquid. If you remove some heat, but less than 80 calories per gram, it will become a mixture of liquid and ice.
In the case of a having a mixture, the ice would probably end up as a big crystal rather than slush.
I am ignoring the additional complication that the H2O is in a glass that is inside a chamber of gas, since OP asks for spherical cow type assumptions.
The weight of the water/ice on itself has not been addressed. If the pressure at the surface of the water/ice is exactly one atmosphere, the pressure at the bottom of the container will be higher, resulting in a lower freezing point at the bottom. Back of the envelope calculations put the freezing point at a depth of 10 cm of water to be roughly .00075 deg C lower than at the surface.
Consider this experiment. Take equal masses of water and ice at approximately 0 deg C and mix them in a cylindrical flask to a depth of 10 cm, then put the flask into a perfectly insulated isolation chamber made by Spherical Cow Inc. After a period of days,weeks, or months, I would suppose that you would wind up with a layer of nearly solid ice in the (roughly) top half and mostly water in the bottom half, due to the pressure gradient.
If you read Crafter_Man’slink from the NIST, they took hydrostatic pressure into account in determining the ice point. I take that to mean the ice point is the temperature at which the *top layer *of water will freeze, but the underlying water will not. So if we perform the OP’s experiment at this temperature, I think all of the ice cube will melt except a thin layer. Starting with a glass of water, it will likely supercool and stay liquid, but if you drop a snowflake into it as a nucleation site, the top layer will freeze.
Btw that link also says to mix up the (ultra-pure) shaved ice and water into a homogeneous slurry, and to keep adding ice + pumping water out of the bottom precisely so you do not get a layer of water on the bottom. So the (real) ice-point measurement is not quite the same as melting ice cubes in a sealed box.
Assume we’re doing this in space. Or, whatever you need to answer the question I’m trying to get to:
If you hold a block of ice or a glass of water at exactly the melting/freezing point, will it stay at its current state, transition to the other, or form a slurry?
The reason I think it will be a slurry is that, through some random movements, you’ll always have ice crystals breaking apart to become water and water molecules sticking together to become ice. At the exact temperature, neither is preferred, so both will exist however you started.
I disagree with Chronos’s statement about oscillation because I don’t see why all of the molecules would go one way or the other. However, he’s a lot smarter than I am when it comes to the sciences, so I’m happy to be shown the error of my ways.
The oscillation, if it happened, would be due to the operation of the thermostat. Let’s say that it’s set up to add heat if the temperature is ever below -0.001º, and remove heat if the temperature is ever above 0.001º. In that case, the thermostat wouldn’t switch until after all the water had changed phase. If you started off with a mix of liquid and solid, and the chamber was gradually gaining heat, the temperature would stay constant at 0 until all the ice melted, then it would increase until it reached the thermostat’s high set point, and then the chamber would start losing heat, and so it’d go back down to 0 and then start re-freezing, and once it’d completely re-frozen, it’d start decreasing the temperature again until it reached the low set point, and so on.
Oops. I missed the point that you said it was at one atmosphere. That means you have air in there too.
I think the end point depends on how much water and how much ice you start with. If you keep your walls adiabatic, the total energy in the closed system can’t change, so to a pretty close approximation the ratio of ice to water can’t change.
Adding magic to a science question always does the exact opposite from clearing things up. We had questions that needed to be answered if the original question could be answered. And now, you’re not only not answering those questions, you’re declaring that they cannot be answered. And thus, the original question can’t be answered, either.
But I’ll take a stab at it. If the temperature is being held constant by magic, then all of the water spontaneously turns into cotton candy.
Ok, let me try one more time. Maybe the answer is, this is unknowable.
If I put a block of ice or a glass of water in a chamber being kept at exactly 0 degrees, will the ice melt or the water freeze or something in between? Or, since no chamber can be kept at that temperature perfectly in real life, the answer is, depends on the chamber? Is 0 degrees a discontinuity where the state of the water is in principle unknown?
Are you assuming water melts/freezes at precisely 0 °C?
Are you assuming ever gas molecule inside the chamber is precisely 0 °C?
Are you assuming the chamber walls are precisely 0 °C?
Are you assuming no air convection currents, and that the air is perfectly still?
Are you assuming the isotopic composition of the water is VSMOW, or something else?
Are you assuming the liquid water has no vapor pressure?
Are you assuming the liquid water has no convection currents, the water is perfectly still, and has no internal pressure gradient?
Are you assuming the block of ice has no vapor pressure, and no internal pressure gradient?
Are you assuming there’s no such thing as supercooled liquid water?
And so on, and so on…
If you make enough assumptions, you can create a simple, mathematical model where the liquid water would get closer and closer to 0 °C without actually “getting” there. Same goes for the ice. But it’s an uninteresting model, and not really worthy of an in-depth discussion.
I think the OP wants the basic physic interpretation, not to be further confused. More information other than temperature and pressure needs to be specified to determine the relative amounts of solid and liquid. Liquid water at 0 deg C and atmospheric pressure has a certain density and specific internal energy. Ice at 0 deg C and atmospheric pressure also has a particular density and specific internal energy, but those are different than the values for liquid water. When you have a mixture of liquid and solid at 0 deg C and atmosphere pressure, you need to know the total amount of internal energy (or some other thermodynamic information) to figure out the relative amounts of liquid and solid.
That is why you need to know how this is being maintained at this temperature and pressure. If no heat is being added or removed (or, equivalently, volume is held constant), the relative amounts of liquid and solid will not change. If you start with 100% liquid and increase the volume a little (i.e., remove some heat) while maintaining the temperature and pressure, you will get some ice to form.
If you have a 50-50 mixture of liquid and ice, and maintain the temperature and pressure while adding or removing no heat, it will remain 50-50. Since the molecules are constantly moving around, which molecule is part of a liquid or solid phase will change with time. That is the basic thermodynamics interpretation. Beyond that you have to worry about super cooling and crystal formation related to surface energy effects. With time, I think fairly stable crystals will form.