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

[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.

 

[gjonesy]

I assumed the whole of 50c and a steady drop of the same rate, (38 degrees c) based on the earlier calculations. I didn't account for the heat exchange of the solution. I did assume total ice melt and at a higher rate, so unless the liquid that lowers the melting point of water(salt solution/alcohol/ ethylene glycol) is at a steady room temp and not already heated, the heat energy transfer wins and the solution total temp will be higher. That is why I asked the question. Its counter intuitive to think that added heat energy wouldn't factor in.

I did a different conversion and got a different calculation 31.1 c so right around 25c.

 

[russ_watters]

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.

I bought a better mug and verified that insulation matters a lot. The melting rate of the ice (and therefore the internal heat transfer) drops as the temperature drops, which means you need really good insulation to approximate the adiabatic case. In reality, I only achieved -22C in a -18C freezer, despite a freezing point that should have been about -32C.

 

[gjonesy]

I have witnessed instant freezing in beer. 5.5% alcohol as soon as I twisted off the top it froze into a slushy type mixture instantly.

 

[russ_watters]

I did also do more tests, which I will share when I have time. Room temperature alcohol an excess ice at 0C does still yield a final temp below 0C. Both in the real world and the math, if you don't add enough ice (all the ice melts), just add more ice until you have some unmelted in the mixture. 0C is the highest possible temperature of such a mixture.

 

[mfb]

I assumed the whole of 50c and a steady drop of the same rate, (38 degrees c) based on the earlier calculations. I didn't account for the heat exchange of the solution. I did assume total ice melt and at a higher rate, so unless the liquid that lowers the melting point of water(salt solution/alcohol/ ethylene glycol) is at a steady room temp and not already heated, the heat energy transfer wins and the solution total temp will be higher. That is why I asked the question. Its counter intuitive to think that added heat energy wouldn't factor in.

I did a different conversion and got a different calculation 31.1 c so right around 25c.

I don't understand your description at all. There is no time-evolution that has to be taken into account with perfect isolation, room temperature is not relevant at all, and no melting points (apart from ice at 0 C) are relevant.