Supercooled water - latent heat of fusion

In summary, during water solidification at 1 atm, about 80 cal/g of heat are released due to the ice specific heat being less than 1 cal/(g °C). This means that if this heat were completely transferred to the ice, its temperature would increase by more than 80°C. However, the question arises of how supercooled water can solidify so quickly and how the heat is immediately exchanged with the environment. It is possible for water to freeze suddenly even at -1°C, but only a small amount of supercooling would result in a small amount of ice being formed. The equation used to calculate this is (m/M) = (Cw/(Cw-Ci)) * [1-exp
  • #1
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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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  • #2
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
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
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
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
Chestermiller said:
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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  • #7
lightarrow said:
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
 
  • #8
Chestermiller said:
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:

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

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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  • #9
lightarrow said:
Thanks Chestermiller.
Did you compute it using an equation similar to:

[tex] \frac{m}{M} = \frac{C_w}{(C_w-C_i)} * [ 1- exp(-\frac{(C_w-C_i)(T_1-T_0)}{H_f})] ?[/tex]
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
Chestermiller said:
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):
[tex]H_f⋅dm = m⋅c_i⋅dT + (M−m)⋅c_w⋅dT[/tex]
[tex]\frac{dm(T)}{dT} + \frac{(c_w - c_i)}{H_f}⋅m(T) - \frac{M⋅c_w}{H_f} = 0[/tex]
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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  • #11
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
 
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  • #12
Chestermiller said:
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
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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  • #13
lightarrow said:
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
 
  • #14
Chestermiller said:
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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  • #15
Chestermiller said:
The internal energy of State 2 is -mHf
Chestermiller, How do you deduce that?

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  • #16
This is the heat we get out if we freeze water of mass m at 0°C.
 
  • #17
lightarrow said:
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
 
  • #18
Chestermiller said:
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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What is supercooled water?

Supercooled water is water that is in a liquid state at a temperature below its normal freezing point. This occurs when the water is pure and free from impurities or disturbances that can trigger freezing.

What is the latent heat of fusion for water?

The latent heat of fusion for water is the amount of energy required to change one gram of water from a solid state to a liquid state at its freezing point, which is 0 degrees Celsius or 32 degrees Fahrenheit. This value is approximately 334 joules per gram.

Why is supercooled water important?

Supercooled water is important in many scientific fields, such as meteorology, climatology, and material science. It also has practical applications, such as preserving perishable items during transportation and creating clear ice for ice sculptures.

What causes water to become supercooled?

Water becomes supercooled when it is in a pure and undisturbed state and is cooled below its normal freezing point without any impurities or disturbances to trigger the formation of ice crystals. This can occur in extremely cold temperatures or in the absence of nucleation sites, such as scratches or dust particles.

Can supercooled water spontaneously freeze?

Yes, supercooled water can spontaneously freeze if it is disturbed or if it comes into contact with a nucleation site, such as a dust particle or a scratch. This is known as the Mpemba effect, where warmer water can freeze faster than colder water due to the absence of impurities in the warmer water.

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