# Supercooled water - latent heat of fusion

1. Aug 14, 2015

### lightarrow

During water solidification at 1 atm, about 80 cal/g of heat are released; since ice specific heat is less than 1 cal/(g °C) it means that if this heat would be totally transferred to the ice, its temperature should increase of more than 80°C.

How then it is that very fast solidification from supercooled water can happen? How it's possible that the heat is immediately exchanged with the Environment?

--
lightarrow

2. Aug 14, 2015

### Staff: Mentor

I like your question. I can't answer it, but the Oimplication is that the internal energy of the supercooled liquid is about the same as the ice. Oherwise, there would have to be a massive heat transfer to ambient, and the transition could not be so rapid.

3. Aug 14, 2015

### Staff: Mentor

As far as I know, not all water freezes. Enough to get a solid structure, but you have to cool it more to get massive ice everywhere.

It doesn't have to be more than 80 degrees - you also have to consider all the energy that got extracted while cooling down the water (while being liquid).

4. Aug 14, 2015

### Staff: Mentor

Here is one of hundreds of youtube videos demonstrating "instant ice" I believe that is what the OP has in mind, but he's interested in the energy aspect.

5. Aug 14, 2015

### Staff: Mentor

How much is the supercooling to begin with? Once you know that, you can easily calculate the final state of the system, assuming no heat is exchanged with the surroundings.

Chet

6. Aug 19, 2015

### lightarrow

Actually I have no idea. I wonder if water can freeze suddenly even if T = -1°C, for example.

--
lightarrow

7. Aug 19, 2015

### Staff: Mentor

In my judgement, yes, but, with that small amount of supercooling, only about 1/80 of the water would freeze.

Chet

8. Aug 24, 2015

### lightarrow

Thanks Chestermiller.
Did you compute it using an equation similar to:

$$\frac{m}{M} = \frac{C_w}{(C_w-C_i)} * [ 1- exp(-\frac{(C_w-C_i)(T_1-T_0)}{H_f})] ?$$

m = mass of ice formed;
M = water mass + ice mass ;
C_w = water specific heat;
C_i = ice specific heat;
H_f = standard entalpy of fusion;
T_0 = initial temperature of supercooled water;
T_1 = final temperature of freezed water

--
lightarrow

Last edited: Aug 24, 2015
9. Aug 24, 2015

### Staff: Mentor

No. The equation I ended up with was simpler than this (although it does match this result in the limit of a small amount of supercooling). Please show us how you derived this equation.

Chet

10. Aug 25, 2015

### lightarrow

Assuming the energy released by the freezing of a mass dm of ice is totally absorbed by water+ice without any exchange with the environment, the energy goes in increased temperature of the already frozen ice (of mass m) and of the remaining liquid water (of mass M−m):
$$H_f⋅dm = m⋅c_i⋅dT + (M−m)⋅c_w⋅dT$$
$$\frac{dm(T)}{dT} + \frac{(c_w - c_i)}{H_f}⋅m(T) - \frac{M⋅c_w}{H_f} = 0$$
which is a first order diff. eq. in the unknown m(T) with the initial condition m(T_0) = 0, the solution of which, for a generic temperature T_1: T_0 < T_1 < 0°C, is the one I've written.

--
lightarrow

Last edited: Aug 25, 2015
11. Aug 25, 2015

### Staff: Mentor

This is a much simpler problem than the differential analysis you've presented. I haven't checked it over, because it doesn't match the results of my analysis below.

We have a finite change that takes place between two discrete states:

State 1: (M+m) grams of supercooled liquid water at temperature Ti C, where Ti is negative.
State 2: M grams of liquid water at 0 C in equilibrium with m grams of ice at 0 C.

Because the system is isolated, the change in internal energy between states 1 and 2 is zero. If we take the basis of internal energy liquid water at 0 C, the internal energy of State 1 is (M+m)cwTi. The internal energy of State 2 is -mHf calories. So,
(M+m)cwTi=-mHf

So,

$$\frac{m}{M+m}=\frac{-c_wT_i}{H_f}$$

Chet

12. Aug 26, 2015

### lightarrow

I see. But there is something which doesn't convince me in your analysis (not that I'm totally sure of mine, anyway): you start from (M+m) grams of liquid water at 0°C (I know that this is not the state 1 that you wrote, but you write "If we take the basis of internal energy liquid water at 0 C" ), then you supercool it and you end up with M grams of liquid water plus m grams of ice, so this final state is not the same as the initial one; how can you say that Delta(U) = 0 from these two?

Edit:
Said in another way: State 1 and State 2 are different so their internal energies shouldn't be different? I know that you say the system is isolated, but state 1 is not an equilibrium state.

--
lightarrow

Last edited: Aug 26, 2015
13. Aug 26, 2015

### Staff: Mentor

In state 1, it had already been subcooled to less than 0 C. I don't have to account for how it got to this state. In state 2, it is at 0C, with some ice present and the rest liquid water.

Chet

14. Aug 26, 2015

### lightarrow

Sorry, I was editing my post while you were already answering it.

--
lightarrow

15. Aug 26, 2015

### lightarrow

Chestermiller, How do you deduce that?

--
lightarrow

16. Aug 26, 2015

### Staff: Mentor

This is the heat we get out if we freeze water of mass m at 0°C.

17. Aug 26, 2015

### Staff: Mentor

We are taking liquid water at 0 C as the datum for zero internal energy U. So the internal energy of the M grams of liquid water in state 2 is 0. The internal energy of the m grams of ice in state 2 is 0 minus the heat of fusion of m grams of liquid water to ice (the specific internal energy of ice is equal to the specific internal energy of liquid water minus the heat of fusion).

Chet

18. Aug 28, 2015

### lightarrow

Thanks, I misunderstood the way you computed it. Now I have to find what is wrong in my reasoning.
Regards,

--
lightarrow