Supercooled water - latent heat of fusion

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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?

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[anorlunda]

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.

 

[mfb]

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).

 

[anorlunda]

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.

 

[Chestermiller]

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

 

[lightarrow]

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.

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

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[Chestermiller]

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

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

Chet

 

[lightarrow]

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

Chet

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

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[Chestermiller]

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})] ?

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

 

[lightarrow]

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

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.

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[Chestermiller]

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 T_i C, where T_i 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)c_wT_i. The internal energy of State 2 is -mH_f calories. So,
(M+m)c_wT_i=-mH_f

So,

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

Chet

 

[lightarrow]

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 T_i C, where T_i 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)c_wT_i. The internal energy of State 2 is -mH_f calories. So,
(M+m)c_wT_i=-mH_f

So,

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

Chet

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.

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[Chestermiller]

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, 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?

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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

 

[lightarrow]

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

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

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[lightarrow]

The internal energy of State 2 is -mH_f

Chestermiller, How do you deduce that?

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[mfb]

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

 

[Chestermiller]

Chestermiller, How do you deduce that?

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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

 

[lightarrow]

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

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

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