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

[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.

 

[jbriggs444]

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 endothermic reaction when comes to contact.

Which, if you read this thread, turns out to be the case. Ice melting is an endothermic reaction.

 

[Ronie Bayron]

Which, if you read this thread, turns out to be the case. Ice melting is an endothermic reaction.

I mean the reverse rather. An exothermic chemical reaction perhaps

 

[jbriggs444]

I mean the reverse rather. An exothermic reaction

I do not understand. You meant to say that an alcohol/ice mixture cannot reach a temperature below 0 degrees Celsius unless an exothermic reaction occurs, thereby cooling the system by adding heat?

 

[Ronie Bayron]

I do not understand. You meant to say that an alcohol/ice mixture cannot reach a temperature below 0 degrees Celsius unless an exothermic reaction occurs, thereby cooling the system by adding heat?

I am sure it could be either (Chet is able to explain this better, not my expertise) of the two that would effect that. If and only if alcohol reacts with water.

 

[jbriggs444]

I am sure it could be either (Chet is able to explain this better, not my expertise) of the two that would effect that.

You were right the first time. "Endothermic" is the one that drains heat energy and adds potential energy.

If and only if alcohol reacts with water.

Alcohol and ice do interact. The ice melts into the alcohol, forming (or diluting) an alcohol/water mixture. That interaction is endothermic. It results in a lowering of the temperature of the mixture. If you read this thread from the beginning, that fact has been pointed out at length.

 

[Kyx]

The answer to the second one is no, unless you provided extra energy. This is the second law of thermodynamics.

 

[jbriggs444]

The answer to the second one is no, unless you provided extra energy. This is the second law of thermodynamics.

What are you responding to?

 

[Kyx]

I was replying to this

I am not sure, if that would be a part of the answer then please explain it.

Let me word it this way: if I were making an alcoholic drink and took ice out of my freezer, put a few cubes in the drink and stirred, what factors could cool the drink to below zero?

Second question (if the first doesn't already answer it): 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 meant to mark this as high school competency.

What are you responding to?

I thought I quoted

 

[jbriggs444]

I was replying to this [message #3 in this thread]

Please read the entire thread. The temperature will, in fact, drop below 0 degrees without violating the second law.

 

[Intresting]

You can't. You could apply the principal of supercooling to a unopened bottle of beer or wine which is lower in alcohol with a higher water content. As trying to freeze 40% alcohol with an ice cube, well you just can't

 

[Chestermiller]

You can't. You could apply the principal of supercooling to a unopened bottle of beer or wine which is lower in alcohol with a higher water content. As trying to freeze 40% alcohol with an ice cube, well you just can't

You are aware that this thread is not about freezing alcohol, correct? It is about mixing ice at 0C with a solution of alcohol and water at a temperature > 0C to lower the combined temperature to below 0 C. In this process, no freezing occurs, only melting of ice.

 

[Intresting]

Is the ambient temperature -114 Celsius?, otherwise the question is moot

 

[Chestermiller]

Is the ambient temperature -114 Celsius?, otherwise the question is moot

Let's see your equations. Mine are in post #99 for your consideration. Also, how do you explain Russ Waters' experimental results, in which his final temperature was about -6 C?

 

[Intresting]

Well I haven't seen Russ Waters experiment, but at a guess I would say that with the reduced latent heat a larger proportion of the ice would be able too freeze once nucleation begins, Like I said ambient temperature will also play a roll.

 

[Orodruin]

Well I haven't seen Russ Waters experiment, but at a guess I would say that with the reduced latent heat a larger proportion of the ice would be able too freeze once nucleation begins, Like I said ambient temperature will also play a roll.

Physics is an experimental science. If your prediction does not match experiments, it is wrong. Simply waving away experiments with a comment to the effect of "I have not checked" when the results are right there is not a valid excuse. To summarise it for you:

• Start with ice at 0 °C and a water-salt mixture at > 0 °C.
• Mix the two and observe the temperature of the final mixture.
• Ambient temperature is > 0 °C but plays a minor role for the experiment anyway due to relatively good insulation.
• Resulting mixture is < 0 °C.

Ignoring this experiment and the rest of the information which has already been posted in the thread only makes you appear ignorant.

 

[Intresting]

I thought we had water(ice) 0C and an alcoholic mix at 0C That experiment is not the same. I couldn't find the experiment results, too call another person ignorant is pretty condescending. by the way you should factor ambient temperarure with you're formulas. It would make it a bit more precise and a lot less bland.

 

[Chestermiller]

Well I haven't seen Russ Waters experiment, but at a guess I would say that with the reduced latent heat a larger proportion of the ice would be able too freeze once nucleation begins, Like I said ambient temperature will also play a roll.

Russ' experiments are described in this thread. Apparently, you did not read the entire thread before you began expounding your own theory.

Now I told you that no freezing takes place in this system. So why are you talking about nucleation? No ice freezes (read my lips).

 

[Intresting]

Hence I stand by my first response. The question is moot sir.

 

[Chestermiller]

Hence I stand by my first response. The question is moot sir.

Huh?

 

[Orodruin]

I couldn't find the experiment results, too call another person ignorant is pretty condescending.

Not as condescending as ignoring 6 full pages worth of comments which have already been made in the thread is ignorant.

Hence I stand by my first response. The question is moot sir.

No, your answer is moot because it does not connect to the original topic. Nowhere is anybody claiming that you can freeze the alcohol.

 

[Chestermiller]

Hence I stand by my first response. The question is moot sir.

Let me explain the reality of the situation to you. You have 6 Physics Forums Mentors (Nugatory, Russ Watters, Dr. Claude, Drakith, mfb, Chestermiller) with over 100 years of combined experience and with both experimental results and theoretical modeling calculations to back them up arguing one way, and then there is little ol' you, with not equations, no calculations, no cited literature, and no experiments arguing the opposite way. Do you really think that what you are saying can be given credibility by those reading this thread?

Now, if you have any specific criticisms of what Russ did in his experiments or what I did in my thermodynamic modeling analysis, let's hear them.

 

[Jeff Rosenbury]

I have difficulty believing this thread.

It is a simple problem:

There exists something called energy. It is neither created nor destroyed. (For purposes of this discussion we will neglect mass-energy transitions.)

Adding energy to a system will raise the temperature. Taking it away will lower the temperature (again neglecting irrelevant exceptions).

The phase transition from solid and liquid also takes energy. It gives energy the other way around. This energy is called the heat of fusion.

Water ice freezes at 0º C. A water/alcohol mixture freezes at a much lower temperature.

So adding the mixture to ice melts the ice taking energy which must come from somewhere.

Under some circumstances the only place for it to come from is the water/alcohol mixture. Under other circumstances it could come from other places, but the OP question was, "Can this happen?", i.e. are there any circumstances? The answer is yes.

The heat of fusion of the ice is stolen from the drink, potentially making it colder than 0ºC.

As others have pointed out, ice cream was made this way (using salt, not alcohol) for centuries. This isn't new or disputed science. It's something everyone in the world with ice, salt, and a thermometer can check for themselves.

Get a thermometer. Go to a bar. Buy a drink. Isn't science tasty.

 

[256bits]

This isn't new or disputed science. It's something everyone in the world with ice, salt, and a thermometer can check for themselves.

Get a thermometer. Go to a bar. Buy a drink. Isn't science tasty.

Still, 142 posts ( 143 inclusive ) on the subject.
Freezing point depression is a mis-understood phenomena, both by beginers and the well-heeled.
Probably because it is mostly described as a laboratory demonstration starting from a mixture at a temperature, and then cooling the mixture and noticing the formation of cyrtals forming at a subdued temperature from the pure solvent.

 

[Drakkith]

I have difficulty believing this thread.

What do you mean? The consensus of those doing the math and the experiments in this thread agree with you. The temperature of the mixture will drop.

 

[russ_watters]

What do you mean? The consensus of those doing the math and the experiments in this thread agree with you. The temperature of the mixture will drop.

I think he's expressing surprise at the level of incredulity.

 

[Drakkith]

I think he's expressing surprise at the level of incredulity.

Well butter my biscuit...

 

[Chestermiller]

I have difficulty believing this thread.

It is a simple problem:

There exists something called energy. It is neither created nor destroyed. (For purposes of this discussion we will neglect mass-energy transitions.)

Adding energy to a system will raise the temperature. Taking it away will lower the temperature (again neglecting irrelevant exceptions).

The phase transition from solid and liquid also takes energy. It gives energy the other way around. This energy is called the heat of fusion.

Water ice freezes at 0º C. A water/alcohol mixture freezes at a much lower temperature.

So adding the mixture to ice melts the ice taking energy which must come from somewhere.

Under some circumstances the only place for it to come from is the water/alcohol mixture. Under other circumstances it could come from other places, but the OP question was, "Can this happen?", i.e. are there any circumstances? The answer is yes.

The heat of fusion of the ice is stolen from the drink, potentially making it colder than 0ºC.

As others have pointed out, ice cream was made this way (using salt, not alcohol) for centuries. This isn't new or disputed science. It's something everyone in the world with ice, salt, and a thermometer can check for themselves.

Get a thermometer. Go to a bar. Buy a drink. Isn't science tasty.

Jeff,

I'm impressed by your perceptiveness, and your ability to recognize the solution to this "obvious" (to you) problem. Unfortunately, it was not so obvious to naive me (and many other mere mortals). At least at the beginning, it wasn't obvious to me that, as the ice melted, its temperature would be dropping.

When you made your knowledgeable pronouncement, were you aware that solutions of ethanol and water are highly non-ideal, and that the enthalpy of these solutions is a complicated non-linear function of the mole fraction of ethanol and temperature? Had you ever seen a diagram of the excess molar enthalpy H^E vs temperature and mole fraction, and had you taken the behavior shown in these diagrams into account quantitatively when you made your pronouncement?

 

[Bystander]

Two words: "Apple jack."

 

[Jeff Rosenbury]

Jeff,

I'm impressed by your perceptiveness, and your ability to recognize the solution to this "obvious" (to you) problem. Unfortunately, it was not so obvious to naive me (and many other mere mortals). At least at the beginning, it wasn't obvious to me that, as the ice melted, its temperature would be dropping.

When you made your knowledgeable pronouncement, were you aware that solutions of ethanol and water are highly non-ideal, and that the enthalpy of these solutions is a complicated non-linear function of the mole fraction of ethanol and temperature? Had you ever seen a diagram of the excess molar enthalpy H^E vs temperature and mole fraction, and had you taken the behavior shown in these diagrams into account quantitatively when you made your pronouncement?

I'm sorry, I woke up in a bad mood. I should have been more polite. Perhaps I should have done the experiment before posting (though posting while drunk has other problems).

 

[Bystander]

it was not so obvious to naive me (and many other mere mortals).

Two words: "Apple jack."

Okay, maybe it is one word ... Chet, chemical engineers are supposed to know these things.