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

[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.

 

[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).

 

[gjonesy]

I find this discussion very interesting, yet counter intuitive in certain scenarios. Considering the op didn't give any particular circumstance, would these same results hold under different conditions.

Example let's say I have a bottle of brandy 20% alc per volume in the trunk of my car and its sat there all day (on a hot day) and I plan to drink it at the bond fire later that afternoon. It comes out of my trunk at 50c (122 f) and I have just a few 8 oz cups to share with my friends at the bond fire. Could you put enough ice in that size container to bring the alcohol down to 0c? Or would the heat from the alcohol win out given the limited space, melting all the ice before the solution could cold down to 0c?

Just wondering.

 

[Chestermiller]

I find this discussion very interesting, yet counter intuitive in certain scenarios. Considering the op didn't give any particular circumstance, would these same results hold under different conditions.

Example let's say I have a bottle of brandy 20% alc per volume in the trunk of my car and its sat there all day (on a hot day) and I plan to drink it at the bond fire later that afternoon. It comes out of my trunk at 50c (122 f) and I have just a few 8 oz cups to share with my friends at the bond fire. Could you put enough ice in that size container to bring the alcohol down to 0c? Or would the heat from the alcohol win out given the limited space, melting all the ice before the solution could cold down to 0c?

Just wondering.

This can all be precisely calculated.

 

[gjonesy]

This can all be precisely calculated.

so I'll have to do some figuring eh...lol

 

[mfb]

You can always use a lot of ice and just a tiny amount of drink to reach the freezing temperature of the (watered) drink. You might water your drink significantly if it starts too hot. Some ice bath as precooling?
Oz are a weird unit.

 

[jbriggs444]

Oz are a weird unit.

You have to be a wizard to master Oz?

 

[gjonesy]

An 8 oz cup will hold 236.58 ml of liquid - 67.5 ml of that volume for 3 standard ice cubes 42 grams of ice

42g ice
169.08ml of volume left in the cup
20% alcohol at 50c

final temp some where around 12c?

 

[mfb]

A rough estimate (with 0% alcohol content) gives something closer to 25 degrees.

 

[gjonesy]

I converted the 38 degree temp drop from 20c to minus 18c and converted for heat.

what did I miss?

I used russ's experiment as a model since my alcohol is weaker I assumed it would have less of a melting point affect and that the energy from the ice would transfer in the same way 8 grams less ice 8% less alcohol.

 

[Chestermiller]

Including heat of mixing in model:

The excess enthalpy of mixing for an alcohol and water mixture H^E(x_E,T) is usually reported experimentally in J/mole, where x_E is the mole fraction of ethanol in the mixture and T is the temperature. So the enthalpy of a water-alcohol mixture can be expressed as

$$H_{mixture} = (m_wC_w+m_EC_E)T+\left(\frac{m_w}{M_w}+\frac{m_E}{M_E}\right)H^E\left(\frac{m_E/M_E}{\left(\frac{m_w}{M_w}+\frac{m_E}{M_E}\right)},T\right)$$
where the m's are the masses of water and alcohol in the mixture, the M's are the molecular weights, the C's are the heat capacities of the pure liquids, and T is the temperature in degrees C (the reference state for zero enthalpy of both alcohol and water is taken as pure liquid at 0 C). Using the same reference state for the ice, the enthalpy of the ice is given by:
$$H_i=m_i(-H_f+C_iT)$$
where H_f is the heat of fusion of ice at 0 C. So the total enthalpy of the system in the initial state is:
$$H_0=(m_{w0}C_w+m_EC_E)T_{L0}+\left(\frac{m_{w0}}{M_w}+\frac{m_E}{M_E}\right)H^E\left(\frac{m_E/M_E}{\left(\frac{m_{w0}}{M_w}+\frac{m_E}{M_E}\right)},T_{L0}\right)+m_{i0}(-H_f+C_iT_{i0})$$where m_w0 is the mass of water initially in the liquid mixture, m_i0 is the initial mass of ice, T_L0 is the initial temperature of the liquid mixture, and T_i0 is the initial ice temperature. In the final state of the system, the mass of water in the liquid mixture is $m_{wf}=m_{w0}+\Delta m$, the mass of ice is $m_{if}=m_{i0}-\Delta m$, and the final temperature of the water and the ice is T_f, where $\Delta m$ is the mass of ice that has melted. So, in the final state of the system, the enthalpy will be:
$$H_f=(m_{wf}C_w+m_EC_E)T_f+\left(\frac{m_{wf}}{M_w}+\frac{m_E}{M_E}\right)H^E\left(\frac{m_E/M_E}{\left(\frac{m_{wf}}{M_w}+\frac{m_E}{M_E}\right)},T_f\right)+m_{if}(-H_f+C_iT_f)$$
Since the system is adiabatic, the initial and final enthalpies must be equal. In addition, if there is any ice remaining in the final state, the final temperature must be equal to the freezing point of the solution at its final concentration.

 

[mfb]

@gjonesy: ?
The whole ice will melt, and melting ice requires about as much energy as heating the same amount of water by 80 K. That cools the drink by ~20 K, so we have 42g of water at 0 C and 170 g of water at ~30 C. The weighted average is ~24 C. Rough estimate, but the number is not off by 12 K.