# Water to ice expansion in $1\textrm{mm}^3$ pit - pressure on the pit walls?

[EDITED] by mistake, the subject was regarding 1m^3 instead of 1mm^3. There should be a significant difference between the two...

A 1x1x1mm pit filled with water is frozen at a slow rate (1K/minute).

When the water becomes ice and expands 9%, I assume that it will be pressed out of the pit since it has nowhere else to go. But how much pressure does it apply to the pit walls?

We can assume that the water is impure and will freeze at 0 degrees C. Whether the pit material or the air is first to reach below 0 degrees C cannot be assumed.

As you can imagine, I need to know whether the pit may be damaged.

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This is a hard problem as it depends on the local temperatures in and around the ice forming. Guessing from the geometry of the problem the expanding ice can push out some water at the top and the pressure should be negligible for a pit made out of a metal, crystal or similar material. –  Alexander Jun 21 '12 at 11:07
This is impossible to solve without saying whether the pit will get cold first, or the air. If the water freezes from the top down, it will break the pit, if it freezes from the bottom up, nothing will happen. –  Ron Maimon Jul 5 '12 at 6:54
Also note that your cooling rate is not quite right: the freezing process can (and probably should?) be considered to occur at $0^\circ\textrm{ C}$. The cooling rate that matters is the rate of removal of latent heat, i.e. something with dimensions of power. –  Emilio Pisanty Jul 5 '12 at 13:48
Your diagram that says "Ice Expanded 9%": is that a given expansion along the vertical direction or are you referring to total volumetric expansion? This problem has too many unknowns to solve outright unless the height of the ice protrusion is known. (Also, technically the problem is statically non-determinate without knowing or assuming the hardness of the walls; perhaps assume they are perfectly rigid?) –  nicholas Jul 5 '12 at 20:50
@nicholas. I recognize the point that with perfectly rigid walls, the water would just be pushed upwards. The walls in this case are reinforced plastic (top 0.5mm), rubber o-ring (bottom 0.5mm) and silicon at the bottom. I tried to define the problem as generic as possible, but I can now see that it is impossible to define the problem without including the hardness of the walls. –  Christian Madsen Jul 11 '12 at 18:46

## 4 Answers

1 atm.

The atmosphere is pushing down on the water, holding it the box. When the freezing water expands, it pushes against the atmosphere and wins. Once the pressures are balanced, the volume comes to rest. There is no pressure on the side walls. If there was, the ice would expand if you took it out of the mold.

There is, of course, mg/(1 mm)^2 pressure down simply from the weight of water. That doesn't change when you freeze it.

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"If there was, the ice would expand if you took it out of the mold." Yes, definitely. But why are we sure that that wouldn't happen? –  Emilio Pisanty Jul 5 '12 at 14:07
Since the water expands in all directions, I don't understand how we get 1 atm on the side walls. –  Christian Madsen Jul 11 '12 at 18:26
The water will push in all directions until it comes to equilibrium. If the water is pressing at more than 1 atm, it will expand upwards until it is pressing 1 atm at the walls. I'm assuming a slow, steady, non-interesting freeze. If the top freezes first then all bets are off. –  16BitTons Jul 13 '12 at 1:41

An exact answer of your problem will only come from an experiment. The exact conditions and temperature profiles can make all the difference.

For a slow and homogeneous cooling process the pressure should not exceed atmospheric pressure significantly, as the ratio between volume and surface area is not critical.

The highest measured pressure in frost weathering is 207 MPa, so it can be enormous if the water is enclosed and has only narrow paths to escape. There is a fairly large review article available from 1991 about frost heave in which several factors are discussed, such as the influence of different temperature gradients and of different the crack sizes. A more recent study is unfortunately behind a paywall but you can always ask the authors directly for a copy.

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Cool problem! A little phase transition thermodynamics, though more information will be necessary for a complete answer, e.g. ambient air temperature, container surface temperature, and the temperature range change for each, if applicable. You may have to bust out the old pchem book for this one.

Water will freeze first at the boundary where it's losing energy. Let's look at the scenarios:

1) water freezes from the top first because that's where it's losing energy the fastest. - the thicker the water layer gets, the greater chance you have of doing damage to the container, because the frozen layer is in place while the liquid water beneath is freezing and expanding. It will expand into the path of least resistance; it's likely that as the ice wall becomes thicker, the path of least resistance will be the material of your container, unless it's a very thick and strong material. What's it made out of? How thick is it?

2) water freezes against the container walls first (if the walls are really cold. Is this outside or in a lab?) - if each wall surface is the same temperature, then the water will freeze uniformly and expand against the liquid towards the air barrier. In this case, you're safe.

3) water freezes against the container and air boundaries with a liquid core. This result will be similar to case 1.

For the floor of your container, 1m^3 of Vienna standard mean ocean water has a density of 1000kg/m^3, and this will exert a pressure on it of P = F/A = mg/A = (1000 kg)(9.80665 m s^-2)/(1 m^2) = 9806 N m^-2 = 1.422 psi, from the weight alone. Liquids will exert a pressure on the side walls, but if you've got case 2, ice will not once it's solid.

That's the best I can do with the given information; temperature measurements from the walls and air, as well as information about the container will let us answer your question more concisely.

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in scenario 1, the ice surface would probably not get thicker in the "down" direction, but rather as pressure builds in the remaining liquid water push the ice cap upwards, maintaining atmospheric pressure. in scenario 2 a similar pushing would happen (possibly even allowing water to spill). For 1 and 2, given a sufficient cooling rate, no pressure would build. For scenario 3, in which we treat the cube as an expanding solid, stress could build against the container. –  nicholas Jul 5 '12 at 20:55

The water follows the path of least resistance, which in this case is upwards. I think the result will be that the ice would finish up with a domed of slightly pointed appearance

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