No temperature change during phase change - except in my soup?

[bartovan]

I was just heating some frozen soup on my electric stove, and then remembered that supposedly as long as there is some ice left the system stays at 0°C until all the ice is melted, and only then does the temperature rise (if heat is continued to be added).

I looked it up and this page (https://chem.libretexts.org/Courses/Valley_City_State_University/Chem_122/Chapter_2%3A_Phase_Equilibria/2.3%3A_Heating_Curves - sorry, couldn't find guidelines on how to properly post links) phrases it like this: "the temperature of a system does not change during a phase change. In this example, as long as even a tiny amount of ice is present, the temperature of the system remains at 0°C during the melting process". See graph below between B and C.

Only, my soup doesn't act like this. There's a big lump of ice in the middle, and 3 cm of melted soup bubbling and boiling around it. How does this work really?

[studiot]

I don't see any soup in your idealisation diagram.

Water is not soup (at least not in Somerset)

The stipulations you refer to only apply to a single pure substance.

Mixtures will act differently.

 

[bartovan]

Well soup is at least in majority water, in- and outside of Somerset, so it wouldn't be all that weird if it would act like water.

So you think it's only due to not being a pure substance? And how much impurity is acceptable then? Tap water? Distilled water - outside of a dust free lab?

I might try with tap water and see what that gives...

[John Cuthber]

The problem is that you have not stirred the soup adequately.

The requirement for a more or less fixed temperature while it melts is that the ice and water are in thermal equilibrium.

 

That requirement is not met here.

[bartovan]

OK, sounds realistic, although I don't think it's mentioned in the phase change description that the mixture has to be stirred.

The page mentioned before says: [as long as there's ice in the mixture] "the added heat is being used exclusively to overcome the attractive forces that hold the more condensed phase together". No mention of stirring anywhere.

Still I wonder, if you have a swimming pool full of melting ice, with a heat source below (let's say a heated floor), and you're almost at the end of the melting process, with 1g of ice in 20000 liters of water, even if you stir like mad, I can't imagine the whole swimming pool remaining at 0°C until the last gram of ice melts?

So maybe it's actually completely theoretical and idealistic, and not to be taken as such in any real world setting?

[swansont]

22 minutes ago, bartovan said:

OK, sounds realistic, although I don't think it's mentioned in the phase change description that the mixture has to be stirred.

The page mentioned before says: [as long as there's ice in the mixture] "the added heat is being used exclusively to overcome the attractive forces that hold the more condensed phase together". No mention of stirring anywhere.

Still I wonder, if you have a swimming pool full of melting ice, with a heat source below (let's say a heated floor), and you're almost at the end of the melting process, with 1g of ice in 20000 liters of water, even if you stir like mad, I can't imagine the whole swimming pool remaining at 0°C until the last gram of ice melts?

So maybe it's actually completely theoretical and idealistic, and not to be taken as such in any real world setting?

It’s assumed to be close to equilibrium. That’s what the stirring would do.

[Sensei]

52 minutes ago, bartovan said:

Well soup is at least in majority water, in- and outside of Somerset, so it wouldn't be all that weird if it would act like water.

Soup often contains oil/fat (soups here are usually made on boiled chicken parts base). Water density is 1 g/cm^3, oil/fat density is lower than water, so it floats at the top of surface, and changes idealized uniform liquid calculations.

[studiot]

1 hour ago, bartovan said:

Well soup is at least in majority water, in- and outside of Somerset,

Even the Archbishop of Canterbury is over 70% water.
Do you expect him to melt in the same way as a large mug of ice?

Let us examine your conditions  more carefully.

1) Your soup will not melt/freeze at 0 degrees C because of a phenomenon known as depression of freezing point. This is due to both material (salts) dissolved in it and particles of food suspended in it.

2) Lumps of vegetable will act quite differently providing a thermal reservoir either of cold or heat with local effects in the mixture, no matter how well you stir.

3) Particles of suspended food will considerably affect the heat transfer through the mug of soup. This is why you should thaw slowly. The different constituents thaw at different rates.
anyone who has frozen and then thawed a bottle of milk to quickly will have experienced this differential thawing of the components of the milk.

 

So John and swansont are quite right to point out the lack of equilibrium and stirring (although you can't sit until nearly thawed).
And the more inhomogeneous the soup, the more slowly you need to thaw and heat.

 

As a matter of interest we thawed out and then heated some chunky chicken and vegetable soup, in the microwave, tonight.
This behaved exactly as described, and we have to run the microwave at half power to get a satisfactory thaw and reheat.
 

Edited November 9, 2019 by studiot

[J.C.MacSwell]

I saw an ice chunk float by in the harbour last February. I felt bad for everyone heading south looking for a warm swim in the ocean.

Silly thought of course but any time temperature is in transition the assumption of equilibrium is at best an approximate one.

[mathematic]

The description only makes sense for small volume, where everything is in close proximity. ..  Extreme counterexample - the ocean

[bartovan]

Thank you for all your answers. J.C.MacSwell makes exactly the point I'm trying to make: according to the theory, 1g of ice in the ocean would make the entire ocean remain at 0°C, which it doesn't of course. And I do believe that in an ocean of pure water this would be exactly the same.

I have been thinking it over some more and thought of two extremes, as a thought experiment.

1) Extreme case 1: The heat is applied in one single point at the bottom. In that case it seems inevitable to me that, according to where exactly one measures, different temperatures will be measured. Especially in the case of a very large container with only a tiny fraction of ice remaining (floating at the top of course, to make things worse), when one would measure very close to the heating point, the temperature would be significantly higher than close to the ice. Importantly, no amount of stirring would remedy this (think of the extreme of 1g of ice in 20 000 liters of water, or 1g of ice in the ocean).

So, in case 1, you simply have different temperatures in different areas, even if you stir, so the diagram doesn't really apply in a literal sense. However, it does make sense, see below point 3.

2) Extreme case 2: The heat is applied perfectly uniformly (seems impossible in the real world, but it's a thought experiment). (Maybe the microwave experiment of studiot comes close.) So we have a huge chunk of ice at say -50°C and start applying heat in a perfectly uniform manner (meaning: we add kinetic energy to every molecule at exactly the same rate per molecule). In that case the whole block of ice will turn to water in one and the same instant (in other words: at every point, the molecules will reach the necessary kinetic energy to get loose from the solid state), and then, being completely water, will immediately continue to rise in temperature. (Since the phase change happens in one single point in time).

In this case, the diagram doesn't apply either. You go in a straight, rising line from -50°C to 0°C, then in one single instant (a point in time) you have the phase change, and you continue with a straight rising line. No horizontal part between B and C (in the diagram I posted at the beginning).

3) So to come back to case 1, I think what the diagram and the theory mean, is that the system, as a whole, remains at 0°C during the phase change in terms of the sum of kinetic energies. Temperature is actually kinetic energy, right? (at least in one of the multiple frameworks for understanding temperature). So during the phase change, some areas will have a rather high kinetic energy (close to the heating point, where they will be excited more), say corresponding to 10°C; while other areas will have a low kinetic energy, corresponding to 0°C or even less in the ice; still others somewhere in between, etc. Probably, during the phase change, the total sum of these kinetic energies is equivalent to the kinetic energy corresponding to 0°C.

Does that make sense?

(I propose we leave soup and bishops out of the discussion, since the crux of the question is not really in the "purity" part - see the example of a bit of ice in a gigantic quantity of water).

[studiot]

3 hours ago, bartovan said:

Does that make sense?

(I propose we leave soup and bishops out of the discussion, since the crux of the question is not really in the "purity" part - see the example of a bit of ice in a gigantic quantity of water).

No it does not make sense and you introduced soup at the start of this discussion.

It does not make sense because you are making the classic mistake of applying a theory designed for one situation to an entirely inappriopriate one.

Rather than going abroad to find fancy elaborations on what is a simple subject, why not look to your own University of Cambridge?

Here is the griff prepared, taught and examined in UK schools on the subject.

https://isaacphysics.org/concepts/cc_cooling_curves

You need to look at the second part of this webpage where the question is applied to mixtures.

Note the bowl of soup you heated is not as large as an ocean but is the sort of situation discussed in the theory you are trying to apply.

Further it should not be compared with an ocean but with a similar sized bowl of pure water, where you will find that the water remains at freezing point until all the ice has melted.

This experiment used to be conducted in school science lessons because there is a great deal more that can be learned from it.

[Endy0816]

4 hours ago, bartovan said:

Thank you for all your answers. J.C.MacSwell makes exactly the point I'm trying to make: according to the theory, 1g of ice in the ocean would make the entire ocean remain at 0°C, which it doesn't of course. And I do believe that in an ocean of pure water this would be exactly the same.

I have been thinking it over some more and thought of two extremes, as a thought experiment.

1) Extreme case 1: The heat is applied in one single point at the bottom. In that case it seems inevitable to me that, according to where exactly one measures, different temperatures will be measured. Especially in the case of a very large container with only a tiny fraction of ice remaining (floating at the top of course, to make things worse), when one would measure very close to the heating point, the temperature would be significantly higher than close to the ice. Importantly, no amount of stirring would remedy this (think of the extreme of 1g of ice in 20 000 liters of water, or 1g of ice in the ocean).

So, in case 1, you simply have different temperatures in different areas, even if you stir, so the diagram doesn't really apply in a literal sense. However, it does make sense, see below point 3.

2) Extreme case 2: The heat is applied perfectly uniformly (seems impossible in the real world, but it's a thought experiment). (Maybe the microwave experiment of studiot comes close.) So we have a huge chunk of ice at say -50°C and start applying heat in a perfectly uniform manner (meaning: we add kinetic energy to every molecule at exactly the same rate per molecule). In that case the whole block of ice will turn to water in one and the same instant (in other words: at every point, the molecules will reach the necessary kinetic energy to get loose from the solid state), and then, being completely water, will immediately continue to rise in temperature. (Since the phase change happens in one single point in time).

In this case, the diagram doesn't apply either. You go in a straight, rising line from -50°C to 0°C, then in one single instant (a point in time) you have the phase change, and you continue with a straight rising line. No horizontal part between B and C (in the diagram I posted at the beginning).

3) So to come back to case 1, I think what the diagram and the theory mean, is that the system, as a whole, remains at 0°C during the phase change in terms of the sum of kinetic energies. Temperature is actually kinetic energy, right? (at least in one of the multiple frameworks for understanding temperature). So during the phase change, some areas will have a rather high kinetic energy (close to the heating point, where they will be excited more), say corresponding to 10°C; while other areas will have a low kinetic energy, corresponding to 0°C or even less in the ice; still others somewhere in between, etc. Probably, during the phase change, the total sum of these kinetic energies is equivalent to the kinetic energy corresponding to 0°C.

Does that make sense?

(I propose we leave soup and bishops out of the discussion, since the crux of the question is not really in the "purity" part - see the example of a bit of ice in a gigantic quantity of water).

 

They're talking about an idealized isothermal process.

Quote

 

An isothermal process is a change of a system, in which the temperature remains constant: ΔT =0. This typically occurs when a system is in contact with an outside thermal reservoir (heat bath), and the change in the system will occur slowly enough to allow the system to continue to adjust to the temperature of the reservoir through heat exchange.

Quote

Phase changes, such as melting or evaporation, are also isothermal processes when, as is usually the case, they occur at constant pressure.

https://en.m.wikipedia.org/wiki/Isothermal_process

Fairly certain it couldn't ever perfectly occur in reality. Always some losses.

I would recommend taking a look at the ideal Carnot cycle.  Easier to consider it in that scenario Imo.

[Strange]

4 hours ago, bartovan said:

according to the theory, 1g of ice in the ocean would make the entire ocean remain at 0°C, which it doesn't of course.

That is not what any theory says. You can obviously have parts of a system at different temperatures.

 

[bartovan]

3 minutes ago, Strange said:

That is not what any theory says. You can obviously have parts of a system at different temperatures.

Exactly. But then the question is, what do they mean by "the temperature of the system"? And how would you measure that temperature? (Obviously not by sticking a thermometer somewhere, because that would be the temperature in one point).

"Thus the temperature of a system does not change during a phase change. In this example, as long as even a tiny amount of ice is present, the temperature of the system remains at 0°C during the melting process (...)."
(https://chem.libretexts.org/Courses/Valley_City_State_University/Chem_122/Chapter_2%3A_Phase_Equilibria/2.3%3A_Heating_Curves)

10 minutes ago, Endy0816 said:

They're talking about an idealized isothermal process.

Well yes and no. They do take it to the kitchen:

"The rate at which heat is added does not affect the temperature of the ice/water or water/steam mixture because the added heat is being used exclusively to overcome the attractive forces that hold the more condensed phase together. Many cooks think that food will cook faster if the heat is turned up higher so that the water boils more rapidly. Instead, the pot of water will boil to dryness sooner, but the temperature of the water does not depend on how vigorously it boils."
(https://chem.libretexts.org/Courses/Valley_City_State_University/Chem_122/Chapter_2%3A_Phase_Equilibria/2.3%3A_Heating_Curves)

[Strange]

3 minutes ago, bartovan said:

Exactly. But then the question is, what do they mean by "the temperature of the system"? And how would you measure that temperature? (Obviously not by sticking a thermometer somewhere, because that would be the temperature in one point).

As it is only true for a system in equilibrium, you can measure the temperature anywhere because, by definition, it is the same everywhere.

[studiot]

25 minutes ago, bartovan said:

Well yes and no. They do take it to the kitchen:

"The rate at which heat is added does not affect the temperature of the ice/water or water/steam mixture because the added heat is being used exclusively to overcome the attractive forces that hold the more condensed phase together. Many cooks think that food will cook faster if the heat is turned up higher so that the water boils more rapidly. Instead, the pot of water will boil to dryness sooner, but the temperature of the water does not depend on how vigorously it boils."

 

Why are you so stubbornly misapplying the theory?

 

Adding salt to the water increases the boiling temperature.
Putting the lid on increases the boiling temperature and reduces the evaporation rate.
 

Keep it simple

[swansont]

7 hours ago, bartovan said:

Thank you for all your answers. J.C.MacSwell makes exactly the point I'm trying to make: according to the theory, 1g of ice in the ocean would make the entire ocean remain at 0°C, which it doesn't of course. And I do believe that in an ocean of pure water this would be exactly the same.

That's a misapplication of the theory. Your link seems to assume a little bit of background, but even so it refers to "the mixture" which implies a system which is fairly homogeneous (the contents as well as the heat source or sink). A situation like a single chunk of ice in a swimming pool doesn't qualify.

One of the more dangerous things in science is applying a model while ignoring the assumptions that went into deriving it. It's how we get nonsense like "according to physics, bumblebees can't fly" or the confusion you have shown here.

 

7 hours ago, bartovan said:

 

I have been thinking it over some more and thought of two extremes, as a thought experiment.

1) Extreme case 1: The heat is applied in one single point at the bottom.

Localized application of heat would violate the notion of a homogeneous system.

7 hours ago, bartovan said:

 

2) Extreme case 2: The heat is applied perfectly uniformly (seems impossible in the real world, but it's a thought experiment). (Maybe the microwave experiment of studiot comes close.) So we have a huge chunk of ice at say -50°C and start applying heat in a perfectly uniform manner (meaning: we add kinetic energy to every molecule at exactly the same rate per molecule). In that case the whole block of ice will turn to water in one and the same instant (in other words: at every point, the molecules will reach the necessary kinetic energy to get loose from the solid state), and then, being completely water, will immediately continue to rise in temperature. (Since the phase change happens in one single point in time).

The interior of the ice could remain colder. Melting only happens at the surface.

 

 

[Endy0816]

3 hours ago, bartovan said:

Well yes and no. They do take it to the kitchen:

"The rate at which heat is added does not affect the temperature of the ice/water or water/steam mixture because the added heat is being used exclusively to overcome the attractive forces that hold the more condensed phase together. Many cooks think that food will cook faster if the heat is turned up higher so that the water boils more rapidly. Instead, the pot of water will boil to dryness sooner, but the temperature of the water does not depend on how vigorously it boils."
(https://chem.libretexts.org/Courses/Valley_City_State_University/Chem_122/Chapter_2%3A_Phase_Equilibria/2.3%3A_Heating_Curves)

Probably not how I would explain it.

If you consider the phase-change diagram above instead, you can more easily see how you can change from one phase to another and still keep temperature/pressure constant, without issue.

SwansonT brings up an excellent point about not confusing a model with reality. Highly recommend considering that.

Edited November 10, 2019 by Endy0816

[studiot]

3 hours ago, bartovan said:

Exactly. But then the question is, what do they mean by "the temperature of the system"? And how would you measure that temperature? (Obviously not by sticking a thermometer somewhere, because that would be the temperature in one point).

In fairness to the OP he has hit upon a question to so often trips folks up.

The 'system' doesn't have a temperature unless that is the same at every point in the system.

So most of the equations in thermodynamics do not apply to systems to which you cannot assign a temperature.

In such cases you need to break the system down into smaller subsystems (as many as necessary) to apply thermodynamics.

Just sometimes you can obtain an average temperature and use this as 'the temperature of the system', but you have to be very careful when and how you do this.

[bartovan]

2 hours ago, Endy0816 said:

If you consider the phase-change diagram above instead, you can more easily see how you can change from one phase to another and still keep temperature/pressure constant, without issue.

Interesting diagram, it clearly shows the relation between temperature and pressure, but how does that shed a light on the constant temperature of the system during phase change? Since there's no time dimension? It only shows that with constant pressure, if you raise the temperature from 0 K (or whatever the X origin is), at some point you'll go from solid to liquid, not how long this transition will take?

Concerning this:

2 hours ago, swansont said:

The interior of the ice could remain colder. Melting only happens at the surface.

That's not consistent with the premise of the second thought experiment, where heat would be applied perfectly uniformly, thus also at the core of the ice. If you state as a premise (as you do) that the interior of the ice remains colder, you return to the other thought experiment, where different temperatures exist, and thus to the problem of defining/measuring the temperature "of the system", as Studiot also mentioned.

[swansont]

6 hours ago, bartovan said:

That's not consistent with the premise of the second thought experiment, where heat would be applied perfectly uniformly, thus also at the core of the ice. If you state as a premise (as you do) that the interior of the ice remains colder, you return to the other thought experiment, where different temperatures exist, and thus to the problem of defining/measuring the temperature "of the system", as Studiot also mentioned.

It was not clear to me that this is what you meant by uniform. As you note, it’s not physical. 

In that case, yes, the melt would happen all at once. But there would be no situation where you had a water-ice mixture, so it doesn’t seem relevant 

[Endy0816]

8 hours ago, bartovan said:

Interesting diagram, it clearly shows the relation between temperature and pressure, but how does that shed a light on the constant temperature of the system during phase change? Since there's no time dimension? It only shows that with constant pressure, if you raise the temperature from 0 K (or whatever the X origin is), at some point you'll go from solid to liquid, not how long this transition will take?

In physics, temperature is a measure of the average kinetic energy of particles in a system. The individual particles won't necessarily all have the exact same amount of kinetic energy.

No phase change takes a set amount of time. It is this whole semi-random process.

What's really happening is incredibly complex so we have to simplify it so we can make equations and models of the behavior.