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

[mfb]

Keeping within the narrow question, the interesting/difficult/counterintuitive behavior is that both the ice and drink can cool down when you combine them if both start above the freezing point of the drink.

Only if you take things like cooling from mixing into account, and even then I did not see a convincing situation yet that would prove this. If you do not (e.g. separate ice and drink by a thin plastic sheet), this is not possible. Heat will flow from the warmer to the colder object, heating the colder object.

 

[russ_watters]

Only if you take things like cooling from mixing into account, and even then I did not see a convincing situation yet that would prove this. If you do not (e.g. separate ice and drink by a thin plastic sheet), this is not possible. Heat will flow from the warmer to the colder object, heating the colder object.

I'm not certain what you mean by "cooling from mixing", but yes, if you wrap the ice in plastic it will not be able to go into solution with the drink when it melts, producing different results. For example, if the ice starts at 0C and the drink above 0C, it will have an equilibrium at 0C (assuming sufficient ice).

 

[DrStupid]

I'm not certain what you mean by "cooling from mixing"

I think he means positive enthalpy of mixing.

but yes, if you wrap the ice in plastic it will not be able to go into solution with the drink when it melts, producing different results.

Why should no dilution produce different result than ignoring dilution (as assumed in your post #33)?

 

[Kevin McHugh]

A five year-old doesn't have much chance here, but I'll try to be basic...

This isn't like mixing two liquids together - the concept of an equilibrium temperature somewhere in between doesn't apply. A liquid freezes or a solid melts at one and only one temperature (per a given substance and pressure). If you add heat to ice (put it in a warmer liquid) it will melt at its melting temperature, period. If you remove heat from water, it will freeze at its freezing temperature, period. If those are different temperatures, the lower temperature wins because while the ice can exist as ice at a lower temperature than its freezing point, you can't remove heat from a liquid (by making it melt ice) without dropping its temperature.

This may seem like a bit of a contradiction, but it is actually somewhat similar to the simpler case of a liquid at its boiling point. There is only one boiling point at a given pressure and no matter how much heat you add, that won't change. Conversely, if you lower the pressure above a liquid (say, put it in a vacuum chamber), it will start to boil and that will lower the temperature to get it to the new boiling point.

Phase changes are isothermal and are either exo or endothermic. Water freezes at 0^oC. Which is an endothermic process. The more heat you remove doesn't effect temperature of the phase change, but it does hasten the kinetics. Same as boiling water at 100^oC, the more heat you put in, the faster the water boils, but never exceeds 100^oC. Somebody mentioned freezing point depression. Mixtures of substances will have lowered freezing points, and elevated boiling points.

 

[Daanh]

Keeping within the narrow question, the interesting/difficult/counterintuitive behavior is that both the ice and drink can cool down when you combine them if both start above the freezing point of the drink.

In what possible scenario would they both cool down? Assuming you wouldn't put them both in a deep-freezer...

 

[russ_watters]

Why should no dilution produce different result than ignoring dilution (as assumed in your post #33)?

I'm not referring to dilution there, I'm referring to the existence of the solution (or not) in contact with the ice.

 

[russ_watters]

In what possible scenario would they both cool down? Assuming you wouldn't put them both in a deep-freezer...

I gave an explicit example above:

If you add ice at 0C to a liquid which is at 10C, but has a freezing point of -5C, the final mixture will be both at about -5C (ignoring dilution).

 

[russ_watters]

Phase changes are isothermal and are either exo or endothermic. Water freezes at 0^oC. Which is an endothermic process. The more heat you remove doesn't effect temperature of the phase change, but it does hasten the kinetics. Same as boiling water at 100^oC, the more heat you put in, the faster the water boils, but never exceeds 100^oC. Somebody mentioned freezing point depression. Mixtures of substances will have lowered freezing points, and elevated boiling points.

You're looking at the scenario backwards: when ice melts, the act of melting removes heat and reduces the temperature of the remaining ice.

In our example, heat transfer between the ice and drink combined with the absorbed latent heat of melting result in a lower final temperature.

 

[Daanh]

If you add ice at 0C to a liquid which is at 10C, but has a freezing point of -5C, the final mixture will be both at about -5C (ignoring dilution).

When all the liquid is at zero degrees, what will make the temperature drop more?

 

[russ_watters]

When all the liquid is at zero degrees, what will make the temperature drop more?

The latent heat absorbed by the melting ice.

 

[russ_watters]

While I'm stil not certain we're all talking about the same issue/scenario, it appears to me that there is general disagreement with my explanation/understanding of how this works. So I propose an experiment (which I will do and video tonight):

1. Take an insulated travel mug and fill it 3/4 with ice from my freezer.
2. Fill it further with cold water from my tap, so the ice is just floating. Cover and shake.
3. Measure and record the temperature (with a calibrated RTD probe).
4. Add 1/4 cup (60g) of table salt. Cover and shake.
5. Measure and record the temperature.

My freezer is colder than it needs to be, but not absurdly cold. I'll measure it, but I would guess around -10C. My tap water will vary depending on how long I leave it running, but likely between 10C and 15C. My tap water is very hard, but I use a softener and filter. This will probably leave some dissolved ions in it.

So:
What will the temperature be before and after adding the salt?

 

[DrStupid]

If you add ice at 0C to a liquid which is at 10C, but has a freezing point of -5C, the final mixture will be both at about -5C (ignoring dilution).

Without dilution the minimum temperature is 0 °C.

4. Add 1/4 cup (60g) of table salt.

That's where the dilution comes into play. NaCl has a positive enthalpy of solution and reduces the freezing point of water. Without these effects the temperature cannot drop below 0 °C. As ethanol has a negative enthalpy of solution (according to http://sites.chem.colostate.edu/diverdi/C477/experiments/isothermal%20microcalorimetry/references/j_mol_struct_1993_v300_p539.pdf) that won't work with alcoholic drinks.

 

[Daanh]

The latent heat absorbed by the melting ice.

I see it now. Yes, that could be true. The melting of the ice takes heat from it's surroundings, the remaining ice or the liquid. Latent heat is a tricky thing.

 

[russ_watters]

Without dilution the minimum temperature is 0 °C.
That's where the dilution comes into play. NaCl has a positive enthalpy of solution and reduces the freezing point of water. Without these effects the temperature cannot drop below 0 °C. As ethanol has a negative enthalpy of solution (according to http://sites.chem.colostate.edu/diverdi/C477/experiments/isothermal%20microcalorimetry/references/j_mol_struct_1993_v300_p539.pdf) that won't work with alcoholic drinks.

I think we are using the word "dilution" to refer to different things and the way I designed the experiment may have an impact there. When I say "dilution" I am referring to a change in the concentration of the liquid solution as the ice melts. In cases where the amount of melting ice is small compared to the amount of water, there is very little change in the concentration of the solution and hence very little change in the melting point of that solution. That's what I was referring to when I said I was ignoring dilution.

You are referring to enthalphy of dissolving the solute, which I inadvertently brought into the situation by saying I'd add the salt after the ice and water are mixed. We can go back to the original formulation, which is the ice being added to the solution if that is helpful. Either way, the enthalpy of solution should not come into play here. The main reason I did it is that I wanted to demonstrate that both the ice and water can drop in temperature and that is harder to do if they aren't already stable at 0C.

I'll restate my point/undertanding though: whether we're talking about alcohol or water/salt, the melting of the ice can provide the energy required to cool the liquid below 0C even if both start above 0C.

I'll have to concede though that I had not considered that water and alcohol might behave differently in such a test...

 

[ToddSformo]

I'

Phase changes are isothermal and are either exo or endothermic. Water freezes at 0^oC. Which is an endothermic process. The more heat you remove doesn't effect temperature of the phase change, but it does hasten the kinetics. Same as boiling water at 100^oC, the more heat you put in, the faster the water boils, but never exceeds 100^oC. Somebody mentioned freezing point depression. Mixtures of substances will have lowered freezing points, and elevated boiling points.

I'm not sure if I understand this: correctly: " Water freezes at 0oC. Which is an endothermic process." Freezing (liquid to solid phase change is an exothermic process), while melting is endothermic. Please correct me if I'm wrong.

 

[mfb]

No ice will melt if both ice and drink are at 0 C, at least if we keep ice and drink separate the ice melting and freezing is in equilibrium. Mixing changes that, but there we are back to the mixing thing...

 

[Charles Carter]

Actually, that isn't true(you aren't the only one who said it...): they meet at the freezing temperature of the liquid, regardless of either's starting temperature.

So that means, counter-intuitively, that they can end up colder than either started.

So the only thing that matters here is the freezing temp of the drink (as long as you add enough ice to reach it).

I think you're mistaken.
The temp will always be at some intermediate value at homeostasis. This is not evaporative cooling. A state change will extract heat/energy without a change in temp but that heat can only move to a substance of lower temperature.
The correct answer to the question is 1) ice may be at any temperature below 0 and b) mixing EtOH with water results in a mixture with a lower freezing point than water.

 

[russ_watters]

Ok, new experiment, designed to avoid the issue of the dissolving enthalpy and the issue of dilution as the ice melts. Also, we'll test both water/alcohol and water/salt:

Setup #1: 100 mL of water mixed with 50g of salt
Setup #2: 100 mL of Southern Comfort (50% alcohol by volume)

1. Refrigerate all components of the experiment. Ice is equalizing in an insulated ice bucket in the fridge as well.
2. Measure and record temperatures of both liquids.
3. Add ice to each.
4. Measure and record temperatures again.

At the start, the ice should be right at 0C and the water and alcohol should each be at a few degrees C. What happens when I add the ice?

 

[russ_watters]

No ice will melt if both ice and drink are at 0 C, at least if we keep ice and drink separate the ice melting and freezing is in equilibrium. Mixing changes that, but there we are back to the mixing thing...

In a real drink and in both of the experiments I propose, there will be mixing -- so, what happens to the temperature of the mixtures?

 

[russ_watters]

Ok, new experiment, designed to avoid the issue of the dissolving enthalpy and the issue of dilution as the ice melts. Also, we'll test both water/alcohol and water/salt:

Setup #1: 100 mL of water mixed with 50g of salt
Setup #2: 100 mL of Southern Comfort (50% alcohol by volume)

1. Refrigerate all components of the experiment. Ice is equalizing in an insulated ice bucket in the fridge as well.
2. Measure and record temperatures of both liquids.
3. Add ice to each.
4. Measure and record temperatures again.

At the start, the ice should be right at 0C and the water and alcohol should each be at a few degrees C. What happens when I add the ice?

Test completed. Results:
Starting ice temp: about 0C
Starting Water/Salt Temp: 4.6C
Starting SoCo Temp: 5.0C

7 min Water/Salt Temp: -6.1C
7 min SoCo Temp: -7.6C

Incidentally, now after about 25 minutes, the temperatures are about the same, but the ice in the SoCo is noticeably more melted. You can probably tell, but that's probably because the green mug is substantially lower quality than the black one. Photo after 7 min and 3 videos:

 

[DrClaude]

Nice experiment. Mixing is a key ingredient here. Would you be willing to redo the experiment with the ice kept in plastic bags?

 

[russ_watters]

Nice experiment. Mixing is a key ingredient here. Would you be willing to redo the experiment with the ice kept in plastic bags?

Sure - though I am in agreement with mfb about what the result will be, and it is also a different experiment from what the OP was asking about.

I was wrong about one thing: The freezing point of 50% alcohol is -52C (I didn't look that up ahead of time) and with a starting point 57C higher and a poorly insulated mug, melting and dilution of the alcohol was a significant factor in the final temperature of the mix. The result was a substantial amount of melting and dilution and thus an equilibrium temperature that was not predictable (though I actually didn't try to predict in anyway beyond whether it would be above or below 0C). I may try that part of the experiment again as well, with ice and alcohol in my freezer. I'll by trying to get ice to melt at -20C or lower (as it does on a road when you add salt) and achieve a lower temperature than my freezer.

Note, the key result here was what several people said outright or implied could not happen: the mixture ended up at a temperature lower than either constituent started and lower than 0C.

 

[Chestermiller]

Yes. 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. The excess heat of mixing H^E of alcohol and water is known experimentally as a function of temperature and mole fraction, and this can be used to predict the final temperature, assuming that the system is adiabatic. However, I doubt that this calculation will yield a final temperature anywhere close to the freezing point of the initial (or even the final) water/alcohol mixture.

Chet

 

[russ_watters]

Yes. 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. The excess heat of mixing H^E of alcohol and water is known experimentally as a function of temperature and mole fraction, and this can be used to predict the final temperature, assuming that the system is adiabatic. However, I doubt that this calculation will yield a final temperature anywhere close to the freezing point of the initial (or even the final) water/alcohol mixture.

Chet

No, I don't think so. The heat of fusion of ice is an order of magnitude larger than the mixing enthalpy and as someone else pointed out, the mixing enthalpies of the two example solutions go in opposite directions.

 

[russ_watters]

I need to make a correction:

If ice and drink are at -2C for example, they are in thermal equilibrium (by definition of temperatures). Nothing melts or freezes, assuming the alcohol content is sufficient.

Agreed, and in the real world that can happen since sometimes people keep their alcohol in the same freezer as their ice. The wording of the question in the title implies that isn't the starting condition.

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.

 

[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.