[Eternal Student]

Hi.

   I've only just seen that I was mentioned recently....

Eternal Student has answered the question which is different than yours. Where do you think he got it wrong?

    Probably spending time on the forum instead of doing my housework, I would have thought.
However @alancalverd   highlighted something like this:

the reasonable expectation of microscopic random interactions is that some melting and some freezing will occur in each region, it's just that overall there's no net change...

   With the remark.... "Alas, No..."   

    Radioisotope tracers can be used.   I think @Bored chemist  mentioned it some long time ago.     Put tritium enriched  water  ice cubes   into  some liquid water   all kept at  0 deg C   (as best as you can).     Some of the radio-isotope moves out of the ice and into the liquid.    Some melting and also some freezing does seem to be happening at a microscopic level,  it's just that overall or in terms of "net" changes there aren't any.   However, the ice particles are not the same ice particles you started with.
     Temperature tells you very little about the action and interaction of individual particles,   you know they have a range of energies and sometimes interactions occur.   If only you could talk about very local temperatures - then ice will melt if the local liquid is just above 0 deg C.  Meanwhile, on average, some equivalent amount of liquid is locally below 0 deg C and soon joins to the existing solid ice and freezes.  So there's no net change.  However, it's best not to even try and consider "temperature" to be something that is sensibly defined at such small and local levels.   It's preferable to say the  liquid-ice mixture retained a global temperature of 0 deg C but individual groups of particles had a range of energies and some interactions occurred where particles left the ice phase and entered the liquid phase  (and vice versa).
     You (@alancalverd) were technically right to say no heat flowed.   Heat is usually defined as the flow of energy from one place to another (by natural processes like conduction etc.) due to a temperature difference.   The energy didn't flow from a thing with a higher temperature to one of lower temperature,  it wasn't really possible to give that microscopic element a temperature at all.   Energy (not heat) certainly was taken from some particles and given to others though.   Moreover, thermodynamics doesn't say anything about energy never being exchanged between two bodies at the same temperature,  they can exchange, of course they can, just so long as the input equals the output and there is no net exchange of energy.

Best Wishes.

[alancalverd]

However, it's best not to even try and consider "temperature" to be something that is sensibly defined at such small and local levels

...precisely because, like life, it is not a property of individual particles but of a very large ensemble of particles, i.e. a classical mesoscopic "body".

[Eternal Student]

Hi again.

   There's an odd overlap going on here between two related threads.

..precisely because, like life, it is not a property of individual particles but of a very large ensemble of particles, i.e. a classical mesoscopic "body"...

    I think that is going in the right direction for this thread**.   I don't know but I can only imagine the OP is trying to identify the deeper meaning of temperature and, as is often done in science, they are trying for a reductionist approach.    E.g.   What is temperature at the most fundamental level?   It's probably nothing at all, or just an abstract property as suggested in the approach taken by theoretical temperature scales.    It is usefully left as a macroscopic or emergent property that a body can have.

LATE EDITING:  **  Possibly for the other thread called "What is temperature" and I forgot I was writing on this one.   Let's say it's going in the right direction for some of the assumed purpose for some of the related threads.

Best Wishes.

[alancalverd]

Back to your radioactive tracers:

Diffusion and isotope exchange don't imply energy flow, at least to the first order. If we have two gases with identical molecular masses at the same temperature and bring them into contact, they will diffuse together. If there was a significant difference in their molecular masses we might be able to separate them with a centrifuge but we would be adding energy to the system and there's no a priori equipartition so the concept of temperature would not apply to the whole.

Now given that your ice/water mixture is in thermal equilibrium, for every T₂O molecule that moves in one direction we expect on average to see an H₂O moving in the opposite direction at a slightly greater speed but conveying the same amount of kinetic energy, so that the internal k.e. of both ice and water remains constant: that's the definition of thermal equilibrium.

[Eternal Student]

Hi.

   Not sure where this diffusion conversation is going but it seems ok.

    If it's relevant, I think my much earlier comment ( N pages back but recently re-quoted by @hamdani yusuf ) about more ice melting in a region where there is more ice may have been an unreasonable simplification.    At about 0 deg. C, ice only tends to form where there is already a nucleation site.   So where there is more ice we might expect there to be more localised melting but also there may be a similar increase in localised freezing at precisely the same surfaces and locations. 
    There may also be practical stuff and limits on reality, there usually is.

Best Wishes.

[hamdani yusuf (OP)]

It won't.
Because, that would require the transfer of heat to or from teh ice and, because everything is at the same temperature, there is no impetus to drive the energy from one place to another.

Here's the list of possible results.
1. both cups retain their ice/water ratio.
2. both cups increase ice/water ratio
3. both cups decrease ice/water ratio
4. first cup retains while second cup increases ice/water ratio
5. first cup retains while second cup decreases ice/water ratio
6. first cup increases while second cup retains ice/water ratio
7. first cup decreases while second cup retains ice/water ratio
8. first cup increases while second cup decreases ice/water ratio
9. first cup decreases while second cup increases ice/water ratio

Which one is the most likely result?
Do you think that a real life experiment can settle our disagreement?

[Origin]

Which one is the most likely result?
Do you think that a real life experiment can settle our disagreement?

After 7 pages you still don't understand that there will not be a net heat exchange between water and ice at 0 degrees?

[alancalverd]

There may also be practical stuff and limits on reality, there usually is.

Especially with water. Our existence depends on its anomalous properties, and the need for nucleation in boiling and freezing just adds to the fun.

[alancalverd]

Do you think that a real life experiment can settle our disagreement?

No. It will most likely demonstrate all sorts of anomalous behavior peculiar to water, and a whole lot of problems with your method. Sir Lawrence Bragg was a bit of an amateur, having only two Nobel Prizes to his name, but he used to say in his Royal Institution annual  lecture demonstrations that if you get 20% repeatability in a heat experiment you are doing very well. Having mucked about with ice calorimetry at the UK National Physical Laboratory (long after Bragg retired as  Director) I can only concur.

Why not just accept the obvious definitions of elementary thermodynamics that everyone else uses?

[hamdani yusuf (OP)]

No. It will most likely demonstrate all sorts of anomalous behavior peculiar to water, and a whole lot of problems with your method. Sir Lawrence Bragg was a bit of an amateur, having only two Nobel Prizes to his name, but he used to say in his Royal Institution annual  lecture demonstrations that if you get 20% repeatability in a heat experiment you are doing very well. Having mucked about with ice calorimetry at the UK National Physical Laboratory (long after Bragg retired as  Director) I can only concur.

Why not just accept the obvious definitions of elementary thermodynamics that everyone else uses?

Because I want to avoid making false assumptions which can lead to unexpected results. I'm not sure with everyone else. But your suggestion would make science more dogmatic and harder to make corrections from previous errors.

[hamdani yusuf (OP)]

Which one is the most likely result?
Do you think that a real life experiment can settle our disagreement?

After 7 pages you still don't understand that there will not be a net heat exchange between water and ice at 0 degrees?

It shouldn't be hard for you to pick an answer as someone who already understand it? I guess you would pick number 1, but let me know if it's not the case, and why.

[Bored chemist]

Because I want to avoid making false assumptions which can lead to unexpected results.

Thermodynamics has been working for about 200 years.
If there were any "unexpected results" we would have seen them by now.
But we don't.
Which is one reason why we know that the real world obeys the laws of thermodynamics.
The other reason, is that the conservation laws are mathematically proven to be true.
https://en.wikipedia.org/wiki/Noether%27s_theorem

So why do you continue to question something which has been seen to be true and proven can't be false?|
Are you trolling?
The other day you asked me to multiply 0.1 by 3 for you.
You also asked for video evidence that holding a lit match near a piece of charcoal won't set it on fire (and, by implication, said that it would if the ambient temperature was above 40C)

It's becoming increasingly clear that you are either trolling, or you just don't have the background to do science.

[alancalverd]

Because I want to avoid making false assumptions

A definition is not an assumption. Two objects are said to be at the same temperature if no heat can flow from one to the other.

[Bored chemist]

It shouldn't be hard for you to pick an answer as someone who already understand it?

I did pick an answer.

The problem is that you can not , or will not understand what I wrote.

Quote from: hamdani yusuf on 10/05/2022 05:01:00
See the result, if the ratio of ice-water in the cups change.
It won't.

Only 1 of your outcomes is consistent with my reply.

This seems to be part of the problem. When people answer you, you don't understand that they have answered.
That's probably because you refuse to learn science.
Why do you do this?
Why do you make a fool of yourself here, rather than going to something like teh Khan academy and finding out?

[hamdani yusuf (OP)]

 
Thermodynamics has been working for about 200 years.

Astronomy has been working for centuries before Copernicus questioned it.

[hamdani yusuf (OP)]

Because I want to avoid making false assumptions

A definition is not an assumption. Two objects are said to be at the same temperature if no heat can flow from one to the other.

Definitions involve assumptions.
Heat is a form of energy, instead of fluid.
Other forms of energy can be transferred in the absence of temperature difference.
Other forms of energy can be converted to heat and vice versa inside an object.
From the operation of chillers, we know that evaporation reduces temperature while condensation increase temperature of working fluid. By analogy, similar things may happen in melting and freezing.
The point of doing the real experiment is to see if local fluctuation can produce observable effects. It's okay if you are not willing to spend your time and resources to do the experiment, but don't discourage those who are. If you are right, you can use the result to support your position in future debates. On the other hand, if you're wrong, you can stop spreading false information. You have nothing to lose, and everything to gain.

[hamdani yusuf (OP)]

So why do you continue to question something which has been seen to be true and proven can't be false?|
Are you trolling?
The other day you asked me to multiply 0.1 by 3 for you.
You also asked for video evidence that holding a lit match near a piece of charcoal won't set it on fire (and, by implication, said that it would if the ambient temperature was above 40C)

It's becoming increasingly clear that you are either trolling, or you just don't have the background to do science.

My response here is similar to my previous response to Alan.

[hamdani yusuf (OP)]

I did pick an answer.

The problem is that you can not , or will not understand what I wrote.

You quoted my response to Origin.
Perhaps you can learn how to write better.

[hamdani yusuf (OP)]

Only 1 of your outcomes is consistent with my reply.

This seems to be part of the problem. When people answer you, you don't understand that they have answered.
That's probably because you refuse to learn science.
Why do you do this?
Why do you make a fool of yourself here, rather than going to something like teh Khan academy and finding out?

Do you pick the first option, where the ice/water ratio don't change in both cups? Implicit answer may be used as an excuse to evade responsibility for giving a false but definitive answer.

[hamdani yusuf (OP)]

In another thread, a scientific disagreement was finally resolved by making a bet.
 

Let me make the first move. I'm on Veritasium's side. I'll explain my reasoning later.
Will someone give it a try to show their prowess in physics?

It crossed my mind to do similar thing here, although perhaps with lower and affordable stake. Since the thanks seems to be valuable here, perhaps we can bet so members in the losing side give thanks to the posts of members in the winning sides (up to n posts). I think it's an interesting idea.
But then my curiosity creeped and I start doing the experiment I described previously. So it won't be fair to other members.