Cooling a drink with salty ice?

[ice109]

So everyone knows about this method of making ice cream, cooling drinks, w/e. You add salt to ice and it lowers the freezing/melting point and this cools things below 0C. I understand why the freezing point is lowered and all that but I don't understand why this cools things down faster. So here's the thought experiment.

You have ice at some temperature below 0, say -10C and you have two coke cans at 80F. divide the ice up into two piles and pour salt on one of them. What happens? Of course all this is a closed system ideal system with perfect thermal conductivity. we can consider the more realistic system later.

The control experiment will come to some thermal equilibrium as dictated by $$m_1c_1 \Delta T=-m_2c_2 \Delta T$$ but that's not important right now.

The experimental system will come to some thermal equilibrium at a lower temperature? Knee jerk reaction is to say yes. After a little thought i still say yes. I think that because the aqeuous salt solution has a lower freezing point it you have the drink pouring heat into the high heat capacity lower temp water for longer because of the sooner transition to it, water, from ice. So the temperature of the water rises at the same rate ( maybe not ) as the control experiment but it starts rising from a lower temperature.

I think the experimental system reaches thermal equilibrium faster because of Newton's law of cooling, the rate of change is proportional to the temp. differential, but I don't know how correct that DE is. Does it even apply when two bodies are changing temperatures or just a changing body and an ambient temperature?

In addition to all this salt water apparently has a lower heat capacity than regular water but maybe not by much, I can't find it's specific heat.

Supposing the heat capacities for both types of water are the same is my assessment of where they will reach thermal equilibrium accurate? I do not think it matters at which temperature the ice transition to water because that will be the same amount of heat lost by the drink no matter where along the time line it is. I think that since the transition from ice to water in the experimental system happens sooner, the drink is losing heat to the higher capacity water for longer.

I mean this has got to be true or else I wouldn't be able to make ice cream using this method.

If someone could put up some math as to how I could describe the behavior of this system I would be much obliged.

 

[cesiumfrog]

The ice is far from absolute zero, hence frequently molecules on the surface will randomly have enough kinetic energy to escape (moving to liquid), overcoming a potential barrier, using up an amount of energy named the latent heat of fusion. The molecules left behind had less energy on average, and so the melting process effectively cools the ice.

At thermal equilibrium (zero Celsius) this process is still happening but molecules from the surrounding water are also freezing onto the ice at the same rate (converting potential to thermal energy, so that the ratio of ice remains constant and no net heat is exchanged with the environment).

If you add an excess of salt, this dissolves into the surrounding water, resulting in brine (which has a freezing point of maybe -20C). Consequently, while some ice continues randomly melting, the water or brine is not freezing at the same rate. The temperature therefore keeps decreasing.

The bulk ice keeps melting (continually depleting thermal energy) until thermal equilibrium is reached again (which is at the freezing temperature of the brine solution). So provided you have an excess of salt and ice, you can cool anything else (in thermal contact) down to about -20C.

 

[ice109]

The ice is far from absolute zero, hence frequently molecules on the surface will randomly have enough kinetic energy to escape (moving to liquid), overcoming a potential barrier, using up an amount of energy named the latent heat of fusion. The molecules left behind had less energy on average, and so the melting process effectively cools the ice. The bulk ice keeps melting (continually depleting thermal energy) until thermal equilibrium is reached again (which is at the freezing temperature of the brine solution). So provided you have an excess of salt and ice, you can cool anything else (in thermal contact) down to about -20C.

two things

the two bolded statements confuse me. in melting the ice depletes its own thermal energy and hence gets colder? yea i don't understand that, if i have a block of ice and nothing else it will melt on its own, if it is above zero?

and you didn't answer my question, you just gave me a rehash of why solutions have lower freezing points which i already mentioned that i knew. not that its not good for me to read.

 

[cesiumfrog]

rehash? No. I have *not* explained why solutions have lower freezing points. But instead (assuming brine does have a lower freezing point) I've explained why the presence of salt causes the ice (initially at 0C) to actually cool to an even lower temperature.

The part you say confuses you seems to be the microscopic explanation of why the process of melting can cause a decrease in temperature.

Consider a 3-molecule crystal of dry ice. The molecules are attracted to each other (basically by electrostatic forces), and cannot escape (sublimate, to gas phase) without a certain amount of energy (think "escape velocity").

However, it's not as though they have zero energy: being above absolute zero, the molecules are constantly vibrating (seemingly randomly) against each other. Now, while all this thermal (vibrational/kinetic) energy is evenly distributed between the three molecules, none has enough energy to break loose from the electrostatic potential, and they remain as a solid. But if you wait long enough, then randomly (just by chance, every now and then) one or two molecules will bump in just such a way as to transfer most of their energy to another molecule. Then for a moment we have two molecules with only a small amount of vibration, and one molecule with a large amount. That third molecule now has enough energy to escape, so it climbs out of the potential barrier (loosing kinetic energy as it does so, but having enough left over to escape and become part of the surrounding gas nonetheless). And since it departed with so much energy, the left over two-molecule solid is at a lower temperature (they are still vibrating less), at least until the solid absorbs more energy from the surroundings (lowering the temperature of the surroundings).

It should be well known that solidification/freezing and condensation both cause an increase in temperature whilst melting and vaporisation cause a decrease. This is the reason why the air warms up just when it starts to rain, and why sweating makes your body cooler.

 

[ice109]

sinip

yes yes this is all clear to me, I'm sorry to be unreceptive, but i am looking for opinion on my analysis of the system mentioned in the OP. there are three question in it

Will the experimental system will come to some thermal equilibrium at a lower temperature?

and

Does Newton's law of cooling even apply when two bodies are changing temperatures or just a changing body and an ambient temperature?

and

What is the math to describe two bodies changing simultaneously.

 

[Integral]

I am having trouble understanding the point of the first post.

Suppose you have ice at -10C, if there is no liquid in the system then adding salt will do nothing, the ice is at -10C.

A dry bath with normal ice cubes not have very good thermal contact with the can, the ice will only touch with limited surface area. The best thermal contact will be when there is sufficient liquid to form a bath so the cans can be immersed. A non salt water bath will be at 0C minimum, adding salt creates bath which will drop to the temperature of the ice (-10 in our case). Yes Newton's law of cooling applies, yes, due to the larger initial temperature differential the drink cans will cool to drinking temperature faster in the colder bath. A difference in specific heat of salt water and pure water would only matter if there was only just enough water to cool the beverage, we can safely assume that that is not a limitation on the system, that is, there is excess thermal capacity in either liquid system

 

[ice109]

I am having trouble understanding the point of the first post.

Suppose you have ice at -10C, if there is no liquid in the system then adding salt will do nothing, the ice is at -10C.

A dry bath with normal ice cubes not have very good thermal contact with the can, the ice will only touch with limited surface area. The best thermal contact will be when there is sufficient liquid to form a bath so the cans can be immersed. A non salt water bath will be at 0C minimum, adding salt creates bath which will drop to the temperature of the ice (-10 in our case). Yes Newton's law of cooling applies, yes, due to the larger initial temperature differential the drink cans will cool to drinking temperature faster in the colder bath. A difference in specific heat of salt water and pure water would only matter if there was only just enough water to cool the beverage, we can safely assume that that is not a limitation on the system, that is, there is excess thermal capacity in either liquid system

the central question of the first post is whether the system with the salty ice will reach thermal equilibrium at a lower temperature, the system is the salty ice and the can.

 

[cesiumfrog]

Suppose you have ice at -10C, if there is no liquid in the system then adding salt will do nothing, the ice is at -10C.

This isn't correct, you haven't considered the possibility of an endothermic process.

 

[ice109]

This isn't correct, you haven't considered the possibility of an endothermic process.

:grumpy: you guys derailing my thread, but these are interesting questions I am not too upset. cesium you seem to implying that an isolated piece of ice will melt

 

[cesiumfrog]

Not an insulated piece of pure isolated ice, no.

Can you clarify again exactly what the question is?

 

[ice109]

Not an insulated piece of pure isolated ice, no.

Can you clarify again exactly what the question is?

lol

the central question of the first post is whether the system with the salty ice will reach thermal equilibrium at a lower temperature than the system with regular ice and a can, the system is the salty ice and the can.

min post length

 

[cesiumfrog]

yes it will. (Otherwise, the related method of making ice cream would not reach sufficiently low temperatures to freeze the mixture.)

 

[Integral]

This isn't correct, you haven't considered the possibility of an endothermic process.

Ok, so the final temp of the ice would depend on the amount of salt added and the amount of ice present. The final temp will be between -10C and -20C. That will just serve to cool the beverage cans faster.

According to Newton's law of cooling larger temperature differentials cause faster changes in temperature. As the can approaches the temperature of the bath, the rate of change of temperature will drop. This means in a water/ice bath the drink temperature will slowly approach 0C, while in a brine mixture there will still be a significant temperature differential as the beverage reaches 0C, thus the temp will continue to fall until it is near the brine temperature, well below 0C.

BTW:
Mythbusters did this, the brine mixture is much faster at cooling cans to good drinking temperature. IIRC it took only 5min to have a cold drink with brine, signifantly more for ice water.