Background (hydrogen)
In the case of recently liquified hydrogen (which is quite cold of course) it must be re-equilibrated before loading on to a rocket as fuel to avoid a sudden exothermic equilibration of the ratio of the ortho- and para- forms. This is because the nuclear spin degree of freedom (singlet vs triplet) initially remains hot even when the other degrees of freedom of the molecule are cold.
The question What is the ortho/para issue with LH2 as a fuel? quotes Hydrogen Fundamentals on a hydrogen safety website:
Liquid hydrogen (LH2) has the advantage of extreme cleanliness and the more economic type of storage, however, on the expense of a significant energy consumption of about one third of its heat of combustion. Another drawback is the unavoidable loss by boil off which is typical to maintain the cold temperature in the tank. The evaporation rate is even enhanced when ortho hydrogen is stored. The heat liberated during the ortho-para conversion at 20 K is huge with 670 kJ/kg compared to a figure of 446 kJ/kg for the latent heat of vaporization at the same temperature. This represents a safety issue requiring a design of the hydrogen loop which is able to remove the heat of conversion in a safe manner.
Water
If instead I started with extremely cold water ice (we've now switched from $\ce{H2}$ to $\ce{H2O}$) say at liquid helium temperature, and add a enough heat to quickly raise it to 0°C and just melt it to liquid, it would be mostly para- water. This will be out of equilibrium, and the nuclear spins will still want to warm up as well.
What would happen next? Would the water quickly refreeze in a fraction of a second? Or would it take hours or days to quietly re-equilibrate?
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