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

[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.

 

[gjonesy]

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

I didn't assume time as a factor at all, I assumed heat energy transfer. 0c-32f to 50c-122f to 88f-31.1c with a weighted average of 25c or 77.00f Is the math wrong?

Maybe I misunderstood the first calculation. 20c=68f -0c-32f dropping to -18c or -0.4f I understand that alcohol lowers the melting temp of water below its freezing point so it melts and gives up its energy even in a cold environment. But I didn't take equilibrium into account with the first calculation

 

[mfb]

It would help if you could make clear where which numbers come from and why you combine them how. Putting together various numbers without context makes it really hard to follow.

I understand that alcohol lowers the melting temp of water below its freezing point so it melts and gives up its energy even in a cold environment.

But nothing drops below zero in this example. The final temperature is positive (in C).

 

[gjonesy]

The second example I simply converted C temp to F temp and subtracted. 50c=122f subtracting 0c=32f gave a total 31.1c or 88f

But nothing drops below zero in this example. The final temperature is positive (in C).

I know, I was basing my math on the maths already expressed in earlier experiments instead of doing a straight calculation on (my scenario) and assumed that because we were talking about alcohol which melts ice below the freezing point 32f and freezes at -7c or 22f that it would be the same steady drop which was a wrong assumption 50g of ice dropping 20c alcohol 38 degrees. In other words I cheated and got the wrong answer.

so I am agreeing that it would not be 12c

 

[jbriggs444]

The second example I simply converted C temp to F temp and subtracted. 50c=122f subtracting 0c=32f gave a total 31.1c or 88f

You are subtracting 32 degrees Fahrenheit (the temperature of the ice) from 122 degrees Fahrenheit (the temperature of the booze) and getting a delta of 88 degrees Fahrenheit? There's a math error there of two degrees, but we can chalk that up to rounding error.

You then proceed to convert this scale-less number (it's not on the Fahrenheit scale any longer, even though it is measured in degrees Fahrenheit) to degrees Celsius using the "C = 5/9 (F-32)" formula. But that formula is wrong. For a scale-less temperature (which this is), the formula is "degrees C = 5/9 degrees F". So your result which should have obviously been 50 degrees Celsius (50 - 0 = what, again?) is incorrectly reported as 31.1 degrees Celsius.

 

[gjonesy]

You are subtracting 32 degrees Fahrenheit (the temperature of the ice) from 122 degrees Fahrenheit (the temperature of the booze) and getting a delta of 88 degrees Fahrenheit?

I rounded it of in my head to (120f) and did the subtraction on paper Then I looked it up on a internet f to c table.

I used straight subtraction because of the number mfb replied with.( 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)

That's the only way I could get a number close enough to his to make any sense.

The thermal exchange I was guessing would differ with alcohol content + heat.
The limited capacity of the container and the amount of liquid and ice it could contain.

My guestimate (if I hadn't been told it would be easy to calculate) would have been room temperature about 70 degrees Fahrenheit without doing any math.

My reasoning would have been that the liquor is hot about the same (as hot tap water) or (drinkable coffee), and the ice would melt quickly causing the drink to cool rapidly and dilute then reach the ambient temperature of the surrounding air within a short time after. Like a cup of hot coffee with few ice cubes dropped in it.

BTW yeah 50c isn't 120 its 122f 122f-32f=90f I did goof the calculation I do math in my head and sometimes I should just write it out.

 

[gjonesy]

OK since I screwed the pooch on both calculations I'll leave the precise number up to someone else. I converted (incorrectly) 50c x 1.8 +32 = 120 should have been 122f. So with simple math and not complex formulas what's the correct answer?
It can't remain at 50c and there should be a number besides 0 we can use to come up with the correct answer or am I bonkers lol?


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

 

[Chestermiller]

Russ et al,

I completed a calculation with the heat of mixing effect included. The heat of mixing data was somewhat incomplete, and I had to use my best judgment at filling in the blanks. The initial state was a mixture of 44 g EtOH, 56 g water at 20 C, and 150 g ice at 0 C.

As best I could estimate, the final state was a mixture of 44g EtOH, 146 g water, and 60 g ice, all at -14 C.

Chet

 

[Chestermiller]

@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.

I did this calculation including the heat of mixing effect (based on the analysis in post #99).

Initial state:
Mixture of 26.7 gm EtOH, 135.3 gm water @ 50 C
42 gm ice at 0 C

Final state:
Mixture of 26.7 gm EtOH, 177.3 gm water
0 gm ice

Final temperature 27 C

Chet

 

[jerromyjon]

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.

Heineken's FTR I slush twisted teas too Dunkin' Donut's straws are the best, for ice balls of water (when you 'suck' the ice out it escalates...

 

[The-Mad-Lisper]

How about this: freezing is exothermic reaction, latent heat is given off. Melting is an endothermic reaction. So when ice is added to alcohol, it takes energy to melt the ice, and the energy comes from the liquid, in the form of heat, which, in turn, lowers the temperature of the ice-alcohol system?

Not true, endothermic reactions only remove heat from the system of the drink if the contents are evaporating. Since the higher energy molecules are literally flying out of the drink, the average energy of the drink is effectively lowered. Ice melting doesn't produce this effect because the melted molecules are still part of the drink.

 

[Drakkith]

Not true, endothermic reactions only remove heat from the system of the drink if the contents are evaporating. Since the higher energy molecules are literally flying out of the drink, the average energy of the drink is effectively lowered. Ice melting doesn't produce this effect because the melted molecules are still part of the drink.

I'm not quite sure what you're getting at. Are you saying that melting isn't an endothermic process? Or are you just saying that heat isn't removed from the ice-drink system when the ice melts? Or something else?

 

[The-Mad-Lisper]

Or are you just saying that heat isn't removed from the ice-drink system when the ice melts?

Precisely.

 

[AbsoluteZero22]

Can someone explain how this happens?

Equilibrium.

 

[Chestermiller]

Not true, endothermic reactions only remove heat from the system of the drink if the contents are evaporating. Since the higher energy molecules are literally flying out of the drink, the average energy of the drink is effectively lowered. Ice melting doesn't produce this effect because the melted molecules are still part of the drink.

Are you saying that, if the container had a lid and there were negligible head space, the decrease in temperature would not occur (since evaporation could not occur)?

 

[Ronie Bayron]

Can someone explain how this happens?

Final temperature of the alcoholic drink will be fixated at the temperature of the ice whatever it is and it can not surpassed that. Second law of thermodynamics limits the process. Even if you have a glass of alcohol, poured down inthe grounds of Antartica. It can't be any cooler than what is on the ground except if ice and alcohol has some sort of exothermic reaction when comes to contact.