B How can ice cool an alcoholic drink below 0°C?

[Chestermiller]

I need to make a correction:

I let myself be led astray here: this is false.

Yes, it is counter-intuitive to say that ice in a drink is not at thermal equilibrium if both are at -2C, but it is true. In the experiment I did last night, the ice/alcohol mixture passed -2C on its way to an equilibrium temperature of about -8C.

The issue is in how temperature is defined and works. In many problems it is acceptable to assume that the temperature of a subtance is completely uniform, when in reality it is not: it is a bell-curve, centered around its average value, caused by atoms vibrating randomly against each other. In this case, the bell curve matters because that is what is driving the process being investigated.

When ice is at -2C, not all of it is at -2C. Some is at -10C, some at 0C. Some molecules gain enough energy to pop out of the solid and into the liquid. When that happens, energy is removed from the ice (via the latent heat of fusion), lowering its temperature. This is the mechanism that causes two substances that are both at -2C to spontaneously get colder.

I'll demonstrate this tonight as well. It was specifically mentioned earlier that some people keep their aclohol in the freezer. Based on my results from last night, if I take ice and alcohol from a freezer (at about -15C if I remember correctly) and mix them together, I will achieve a temperature perhaps as low as -30C.

The other specific case mentioned by the OP was room temperature alcohol mixed with ice at 0C. I'll try that as well. The result will again be a mixture a little below 0C.

You are saying that, at thermodynamic equilibrium, the temperature of this (rather simple) system is not uniform. Is that correct?

 

[mfb]

The issue is in how temperature is defined and works. In many problems it is acceptable to assume that the temperature of a subtance is completely uniform, when in reality it is not: it is a bell-curve, centered around its average value, caused by atoms vibrating randomly against each other. In this case, the bell curve matters because that is what is driving the process being investigated.

Temperature is defined in the thermodynamic limit, but I know what you mean.

The effect is not directly the mixing enthalpy, but it is related - you reduce the re-freezing rate because the melted water gets mixed with alcohol, we have the entropy change in both.

 

[jbriggs444]

You are saying that, at thermodynamic equilibrium, the temperature of this (rather simple) system is not uniform. Is that correct?

The claim seems to be the converse -- that the temperature is constant throughout but that thermodynamic equilibrium has not been attained.

My take is slightly different (the same thing in different words, no doubt). We do not have a constant temperature throughout. We have a temperature gradient pointing downhill toward a low temperature at the solution/ice interface. Heat flows from both directions toward the interface. At the interface we have a mass flow from ice into solution with the associated latent heat of fusion acting as a heat sink at the current interface position. Equilibrium is not attained until the mass flow rate and the temperature gradients zero out.

 

[Chestermiller]

The claim seems to be the converse -- that the temperature is constant throughout but that thermodynamic equilibrium has not been attained.

My take is slightly different (the same thing in different words, no doubt). We do not have a constant temperature throughout. We have a temperature gradient pointing downhill toward a low temperature at the solution/ice interface. Heat flows from both directions toward the interface. At the interface we have a mass flow from ice into solution with the associated latent heat of fusion acting as a heat sink at the current interface position. Equilibrium is not attained until the mass flow rate and the temperature gradients zero out.

So is it realistic to consider the final thermodynamic equilibrium of the system or not?

 

[Chestermiller]

Russ,

I would like to do some thermodynamics modelling calculations to see if I can match up with your experimental results at the final thermodynamic equilibrium state. I will do the calculations first by (a) neglecting the heat of mixing and then (b) including the heat of mixing. The calculations will be for an adiabatic system. Would you be so kind as to specify an initial state for the system (for ice and liquid mixture separate from one another to begin with), with

Ice: mass and temperature
Liquid: mass, temperature, mass fraction alcohol

Thanks.

Chet

P.S., I have come around to your view that, if there is any ice left at final equilibrium, the temperature of the system will have to be at the freezing point of the final liquid mixture.

 

[russ_watters]

You are saying that, at thermodynamic equilibrium, the temperature of this (rather simple) system is not uniform. Is that correct?

Yes. Broader, ANY similar substance (whether at equilibrium or not - just uniform) has a temperature defined in terms of the average kinetic energy of the molecules. The actual kinetic energies follow a Maxwell-Boltzmann distribution. It is that distribution that allows ice to melt when its measured temperature is below 0C.

When I say "thermodynamic equilibrium", I'm not just saying uniform temperature throughout, I mean no internal heat flow. In this case, the random distribution of energies causes a non random heat flow: from the water to the ice.

 

[russ_watters]

Russ,

I would like to do some thermodynamics modelling calculations to see if I can match up with your experimental results at the final thermodynamic equilibrium state. I will do the calculations first by (a) neglecting the heat of mixing and then (b) including the heat of mixing. The calculations will be for an adiabatic system. Would you be so kind as to specify an initial state for the system (for ice and liquid mixture separate from one another to begin with), with

Ice: mass and temperature
Liquid: mass, temperature, mass fraction alcohol

Thanks.

Let's stick with the OP's main question and merge it with my testing:

...would the drink be able to drop below 0°C if the drink was initially at room temperature, and the ice was initially at 0°C?

I used 100 mL of 50% by volume, but let's go with 100g, and 44% by mass.

Room temp: 20C

The goal is excess ice, but I'll have to guess: 50g.

 

[mfb]

P.S., I have come around to your view that, if there is any ice left at final equilibrium, the temperature of the system will have to be at the freezing point of the final liquid mixture.

I'm not sure about that. The stopping condition is an equal rate of freezing and melting. At the freezing point of the liquid, freezing of the liquid and melting of frozen liquid (!) is in equilibrium. But the frozen liquid melts easier/faster than the ice, and we are looking for an equilibrium of freezing of water molecules onto ice and melting of ice.

 

[DrStupid]

When I say "dilution" I am referring to a change in the concentration of the liquid solution as the ice melts.

Me too, but you are referring to the average concentration and I also to the local concentration. That means without dilution (as I mean it) the local concentration of water would increase around the melting ice or even result in a layer of pure water (a practical example would be ice in oil). In this sense you are assume infinite fast dilution of the melting water. Thus it seems we were just talking cross purposes about the same situation.

It seems clear that the heat of mixing effect, resulting from melting of ice to form pure water which then mixes with the water/alcohol solution to yield an accompanying heat of mixing, is key to this phenomenon.

Definitely not. According to the paper I linked above the heat of mixing of water and ethanol is negative and the mixing therefore exothermic. That means mixing of water/ethanol mixtures with equal temperature but different concentrations should always result in an increased temperature of the resulting mixture (under adiabatic conditions).

The main effect is a shift of then equilibrium between melting and freezing to lower temperatures (or with other words: the freezing-point depression). Ice and pure water are in an equilibrium at 0 °C. The addition of alcohol reduces the freezing-point with the result that the ice is still melting at 0 °C but the solution do not longer freeze at this temperature. The endothermic melting without the exothermic freezing decreases the temperature until there is no ice left or the freezing point of the solution is reached (whatever happens first).

 

[DrStupid]

But the frozen liquid melts easier/faster than the ice

In case of water/ethanol mixtures the "frozen liquid" is pure water ice.

 

[justaman0000]

Can someone explain how this happens?

It doesn't. Unless you could get the ice below the freezing point of pure alcohol. Water freezes at 32 degrees F. Alcohol is much lower.

 

[Chestermiller]

Let's stick with the OP's main question and merge it with my testing:I used 100 mL of 50% by volume, but let's go with 100g, and 44% by mass.

Room temp: 20C

The goal is excess ice, but I'll have to guess: 50g.

What about the temperature of the ice? I'm sure it wasn't 0 C. The freezer in my frig at home is set at 0 F, which is about -20 C. Any idea what temperature for the ice?

 

[justaman0000]

It doesn't. Unless you could get the ice below the freezing point of pure alcohol. Water freezes at 32 degrees F. Alcohol is much lower.

BTW the freezing point of alcohol is under -170 degreez F. So, there is probably no way you are going to get ice that cold.

 

[Chestermiller]

At a temperature where pure ice can be in equilibrium with a (freezing) solution of water/alcohol, the chemical potential of water within the solution is equal to that of the water ice. This means that, if there were a vapor phase in equilibrium with the water/alcohol solution (vapor phase containing only water and alcohol), the equilibrium vapor pressure of the water in the vapor phase (i.e. the partial pressure of the water in the vapor phase) would be equal to the vapor pressure of pure water ice at that (lower) temperature. This means that the equilibrium vapor pressure of water for the water/alcohol solution would be less than that of pure water in equilibrium with ice at 0 C. The net outcome of all this is that (pure) water ice can be in equilibrium with a freezing water/alcohol solution at a lower temperature than 0 C. Of course we already know that because the solution has a freezing point below 0C (and only pure water ice freezes out); this is clearly shown on a water/alcohol phase diagram.

Chet

 

[russ_watters]

What about the temperature of the ice? I'm sure it wasn't 0 C. The freezer in my frig at home is set at 0 F, which is about -20 C. Any idea what temperature for the ice?

0C. It was specified by the OP in what I quoted and is easily achieved with an ice bucket, per my experiment.

I greatly prefer 0C to -20C here because of the persuasive power of both constituents starting warmer than the final mix. It also makes the calculations slightly easier.

 

[Chestermiller]

0C. It was specified by the OP in what I quoted and is easily achieved with an ice bucket, per my experiment.

I greatly prefer 0C to -20C here because of the persuasive power of both constituents starting warmer than the final mix. It also makes the calculations slightly easier.

No problem.

 

[justaman0000]

Sorry,You said alcoholic drink, and I was thinking of pure alcohol. You can freeze an alcoholic drink easily if it has a low alcohol content like beer. It is not the alcohol freezing though but the water that it is distilled with.

 

[gjonesy]

No one's made ice cream?!

I do, or have in the past. I diabetic now so its off limits for the moment. That is until I figure out the cure to diabetes

good old rock salt and crushed ice in a rotary churn!

 

[Chestermiller]

Hi guys. I completed some preliminary calculations for the case that Russ specified:

100 gm alcohol/water solution, 44% alcohol mass percent at 20 C
50 gm ice at 0 C

I ran the calculation for the case of no heat of mixing. This is a pretty easy calculation to do. Here's what I found:
At final equilibrium, essentially all the ice will have melted, and the final temperature would be about -18 C. This would be close to (although slightly above) the freezing point of the final solution, which would contain about 29% ethanol.

I will be running additional calculations with more ice to start with to see if we can end up (modeling-wise) with a two phase system. The next case i will try will be 100 gm ice initially.

If anyone is interested in the model development, I will be glad to flesh out the analysis.

Chet

 

[mfb]

In case of water/ethanol mixtures the "frozen liquid" is pure water ice.

Okay, then I agree with the conclusion.

 

[Chestermiller]

Further Calculations (no heat of mixing):

Initial state:
100 gm alcohol/water solution, 44% alcohol mass percent at 20 C
100 gm ice at 0 C

Final State:
155 gm alcohol/water solution, 28% alcohol at -18 C
45 gm ice at -18 C

So, in the case where I start out with 100 gm ice, the calculations indicate that there will still be ice remaining in equilibrium with the final alcohol/water solution at its freezing point, -18 C and 28% alcohol.

Tomorrow, if I feel like it, I will include the heat of mixing contribution. This is also pretty straightforward to calculate.

Chet

 

[Kevin McHugh]

Has anybody taken into account the heat capacity of the calorimeters? Don't forget, they are in contact with the solutions and are removing heat to come to thermal equilibrium.

 

[Chestermiller]

Has anybody taken into account the heat capacity of the calorimeters? Don't forget, they are in contact with the solutions and are removing heat to come to thermal equilibrium.

I haven't included that in my calculations (yet), but, if anything, they would add heat to the mixture. The calculations I have done so far were just to see what the maximum amount of cooling could be.

 

[russ_watters]

Hi guys. I completed some preliminary calculations for the case that Russ specified:

100 gm alcohol/water solution, 44% alcohol mass percent at 20 C
50 gm ice at 0 C

I ran the calculation for the case of no heat of mixing. This is a pretty easy calculation to do. Here's what I found:
At final equilibrium, essentially all the ice will have melted, and the final temperature would be about -18 C. This would be close to (although slightly above) the freezing point of the final solution, which would contain about 29% ethanol...

If anyone is interested in the model development, I will be glad to flesh out the analysis.

I don't mean to be a pain, but I want to make sure I understand the analysis, so I worked it backwards and here is what I got:
50g ice absorbs 100*334 = 17,200J when it melts
44g of ethanol releases 44*2.44*38=4,079J in dropping from 20C to -18C
106g of water releases 106*4.186*18=7,987J in dropping from 0C to -18C

These should sum to zero, but they don't. What am I missing? Are the specific heats of water and alcohol different when they are mixed?

 

[russ_watters]

I'm not sure about that. The stopping condition is an equal rate of freezing and melting. At the freezing point of the liquid, freezing of the liquid and melting of frozen liquid (!) is in equilibrium.

Agreed.

But the frozen liquid melts easier/faster than the ice, and we are looking for an equilibrium of freezing of water molecules onto ice and melting of ice.

Why? Throughout the dropping temperature, the ice preferentially melts because it can go into solution with the liquid and can't easily come out of solution. What changes when the freezing point of the liquid is reached is that the liquid can no longer hold any more "melted ice". The mixture is saturated and any more dilution would raise the freezing temp and require some to freeze.

 

[russ_watters]

Me too, but you are referring to the average concentration and I also to the local concentration. That means without dilution (as I mean it) the local concentration of water would increase around the melting ice or even result in a layer of pure water (a practical example would be ice in oil). In this sense you are assume infinite fast dilution of the melting water. Thus it seems we were just talking cross purposes about the same situation...

The main effect is a shift of then equilibrium between melting and freezing to lower temperatures (or with other words: the freezing-point depression). Ice and pure water are in an equilibrium at 0 °C. The addition of alcohol reduces the freezing-point with the result that the ice is still melting at 0 °C but the solution do not longer freeze at this temperature. The endothermic melting without the exothermic freezing decreases the temperature until there is no ice left or the freezing point of the solution is reached (whatever happens first).

Agreed.

BTW the freezing point of alcohol is under -170 degreez F. So, there is probably no way you are going to get ice that cold.

Combining these two and analyzing my experiments last night, we detect a flaw that I may not have explicitly detailed but at least was thinking: I hadn't checked the freezing temperature of alcohol (-32C at 50% BV) and so didn't realize just how cold it is. That has (or manifests) two effects on my experiment:

1. There is more ice melting than I realized, more dilution and therefore a larger shift in freezing point than I would have anticipated.
2. With the large temperature differences and (I think) poorly insulated mug, heat transfer with the surroundings is occurring at enough of a rate to matter. Hence, my mixture got nowhere close to its freezing point.

 

[Chestermiller]

I don't mean to be a pain, but I want to make sure I understand the analysis, so I worked it backwards and here is what I got:
50g ice absorbs 100*334 = 17,200J when it melts
44g of ethanol releases 44*2.44*38=4,079J in dropping from 20C to -18C
106g of water releases 106*4.186*18=7,987J in dropping from 0C to -18C

What about the heat released by the 56 g of water in cooling from 20 C to 0 C?

 

[russ_watters]

What about the heat released by the 56 g of water in cooling from 20 C to 0 C?

Yeah, duh. Let's try that again:
50g ice absorbs 100*334 = 17,200J when it melts
50g of water releases 50*4.186*18=3,767 J in dropping from 0C to -18C
44g of ethanol releases 44*2.44*38=4,079J in dropping from 20C to -18C
56g of water releases 56*4.186*38=8,908 J in dropping from 20C to -18C
Sum: 446J --- close enough. Thanks.

 

[russ_watters]

Some references I'm collecting:
https://en.wikipedia.org/wiki/Cooling_bath
https://chemtips.wordpress.com/2015/02/09/methanolwater-mixtures-make-great-cooling-baths/
http://www.larkinweb.co.uk/science/2010/files/Cooling baths.pdf
http://antoine.frostburg.edu/chem/senese/101/solutions/faq/why-salt-melts-ice.shtml <---good one, talking about the mechanism

 

[mfb]

Why? Throughout the dropping temperature, the ice preferentially melts because it can go into solution with the liquid and can't easily come out of solution. What changes when the freezing point of the liquid is reached is that the liquid can no longer hold any more "melted ice". The mixture is saturated and any more dilution would raise the freezing temp and require some to freeze.

You are right. And it does not even matter how exactly the liquid freezes.

To come back to the steel+water example: the don't mix well, some steel will go into solution but the water is saturated long before the freezing temperature is reached. Water can't melt steel beams (scnr).