Take the 2-minute tour ×

This question already has an answer here:

We know that density of ice reduces by about 8% during freezing, this means it expands to have little higher volume. But if I fill water in a container (entire volume) which has very low coefficient of thermal expansion at 0 $^\circ$C, will the space constraints stop water from freezing?

share|improve this question

marked as duplicate by nemu, Brandon Enright, John Rennie, jinawee, Waffle's Crazy Peanut Dec 18 '13 at 17:29


Does the space constraint of a cast-iron engine block stop a pure water coolant from freezing? –  User58220 Dec 19 '13 at 0:12
I worked with a professor who used to do the exploding cast iron ball full of water demo. Energetic and surprising even though you know what is coming. –  dmckee Dec 19 '13 at 4:40

2 Answers 2

If you force your liquid in a container you will increase the pressure during the freezing process. This will lower the freezing temperature and thus will indeed "stop the water from freezing". However, if you cool down further, probably at some point a different crystal structure (or the same with lower inter-atomic distances) will form and thus the water will still freeze, just with higher density.

While I do not know this effect from water I know for sure this happens to other materials if the volume is limited.

share|improve this answer

What you need to do is take into account the thermodynamic properties of freezing water into ice and the elastic properties of the ice. You will likely not be forming different types of ice, but at high (for ice) temperatures you will probably have a two phase system of ice and water where the strain that would form from freezing prevents some of the water from freezing. As the temperature lowers you will have just ice but compressed ice.

For thermo with no stress/strain you look at the free energy: $$F=U-TS$$ but when you incorperate stress/strain there is an additional term like: $$F=U-TS-\sigma d\epsilon$$ Which incorperates the energy of compressing the solid.

share|improve this answer
