[Eternal Student (OP)]

Hi.
   I apologize for the title, it had to be very compressed to fit in the space available.   Hopefully it makes sense.

Typically when you transfer heat to a substance, its temperature increases and also the entropy of that substance (or the entropy of the system which is just that substance) increases.   

Are there any counter-examples?

    Well, yes, I think so.   An ideal gas, starting from an initial state with pressure and volume P_i and V_i can be subjected to compression (to a volume V_f  < V_i ) and then some surplus heat can be removed while maintaining the volume at V_f.   It should be possible to reach a higher final temperature but lower final entropy.
    You can determine this by looking at the Sackur-Tetrode equation for the entropy of an ideal gas.   (Reference: http://hyperphysics.phy-astr.gsu.edu/hbase/Therm/entropgas.html  ).   I can show the mathematics if required.

     There are probably other examples but you can hopefully see that the situation or the combination of changes you make, can be as important as the substance you choose.
    Now if we insist that all changes are restricted to maintaining a constant volume, we have the question that I was really hoping to ask:   Is there a substance where the temperature can be increased but the entropy of that substance decreases and all changes are done at constant volume?
   (It doesn't seem possible to me, using a thermodynamic definition of entropy,  just because  C_v = specific heat capacity at constant volume is always positive for every substance I know about.   I'm asking because I don't know for certain if there is a substance with C_v < 0.    Also, using a statistical mechanics definition of entropy you don't seem to be asking for too much:   If there was a substance where you can increase the total energy content from E_i  to  E_f    but the number of microstates availble for total energy E_f is less than the number of microstates available for total energy E_i    -  then you would seem to be on to something.)

Best Wishes.

[chiralSPO]

I think you need to be careful about specifying whether you mean temperature or thermal energy.

There are plenty of systems one can imagine (or even observe!) in which increasing energy of the system results in there being fewer available micro-states, and therefore lower entropy. The first example that springs to my mind is the electronic structure of a transition metal ion.

Screen Shot 2022-12-14 at 8.57.56 PM.png (27.42 kB . 964x342 - viewed 3942 times)

If some light is absorbed, the resulting system has both more energy and less entropy than it did before (it's not a closed system).

More broadly, we can imagine any system made up of bits that can be in two states (one high energy, one low). Imagine all the bits start out in the low energy state, and you start feeding energy into the system. Initially the entropy increases with increasing temperature. But once the 50/50 state is reached, any additional energy going in is going to decrease the entropy of the system, until finally it is 100% energized, and has regained its initial low entropy.

Think this isn't realistic? It's how lasers work.

Also, going back to my initial remark about temperature: I never said anything about the temperature of the system of bits, only that you are adding energy. One reasonable definition of Temperature is: T = ∂U/∂S (constant volume). So ... if the energy is increasing, and the entropy decreasing, then the system has a negative temperature (ie below absolute zero) 

[Eternal Student (OP)]

Hi and thanks.

   Lasers and population inversion was exactly what I had in the back of my mind.

This sentence

One reasonable definition of energy is: T = ∂U/∂S (constant volume).

   might be a typing error.   That's a perfectly good thermodynamic definition of temperature.

Obviously I'm going to have to take a bit longer to consider everything.  Thank you very much for your time and it's exactly the sort of thing I was looking for.

Best Wishes.

[chiralSPO]

oof! thanks for the correction--I set myself up poorly for that.

Me: "be careful about temperature and energy"
also me: "energy is defined as T =...." *facepalm*

[Eternal Student (OP)]

Hi.

    I've had a bit of time to think.   As you ( @chiralSPO ) implied:  It's very much a case of needing to keep our definitions of temperature straight (and it does become a little dull then).

    The situation with a simple two-energy level system does illustrate a situation where temperature can increase (in this case it is increasing by becoming less negative) while the entropy decreases.   As you ( @chiralSPO ) implied it's clear that energy is always being added to the system as more particles are driven to the higher energy state,  however that wouldn't guarantee that the temperature was increasing.   
    For a typical substance like an ideal gas held at constant volume, it's apparent that increasing the internal energy will increase the temperature.   For a more arbitrary substance and where statistical temperature is used, you really do have to directly check the gradient  ∂S/∂E.   A diagram will save 1000 words here:

Suppose the entropy, S, of a system of N particles (each particle having only two energy levels) varied with E, the total energy of the system, as follows:


Untitled.jpg (49.3 kB . 1152x648 - viewed 2409 times)
   An increase in the the total energy of the system would not produce an increase in temperature once you've entered the region with most of the particles in the higher energy state.  The temperature is negative in that region but perfectly constant.

  Fortunately the entropy of the simple 2-energy level system doesn't look like that,  it has this sort of shape:


entropy2.png (24.19 kB . 882x680 - viewed 2489 times)
   Where the gradient =  1/T   is  monotonically decreasing with E.

  With that established:   The simple 2-energy level system does have the usual relationship where increasing its energy remains synonymous with increasing its temperature.   The note of caution in your ( @chiralSPO ) first sentence is then substantially laid to rest:

I think you need to be careful about specifying whether you mean temperature or thermal energy.

   More generally, the simple 2-energy level system ticks all the boxes that I was asking for in the original post.  So, well done and thank you.   You can have the best answer award.
-------
     I don't think I need to ask anything else.   If you restrict attention to just non-negative temperatures, then I'm now fairly sure that the answer rests entirely upon finding a substance with specific heat capacity, C_V < 0.
     While I don't know of any substance with that property, in any situation (e.g. in any range of temperature, pressure or other conditions),  if someone else does -  Please let me know.

Best Wishes.

[evan_au]

Is there a substance where entropy decreases at higher temperature?

Perhaps Rubik's Cube?

It is not a 2-state system, but a 20 state system, with the state representing the number of moves required to get it back to a fully-ordered system.

And the number of microstates (ie probability of finding the system in that state) decreases as you move to state 19 or 20.
- But up to state 18, the number of microstates increases if you add a random turn.

Does this fulfill the definition above?

In a cube with 19 or 20 moves to solve, inserting extra energy (random moves) decreases the entropy (the number of moves to solve).


Rubiks_Cube_Moves.png (30.34 kB . 471x587 - viewed 2544 times)
Table from: https://www.quora.com/How-many-permutations-does-a-Rubiks-Cube-have

This came up in https://www.thenakedscientists.com/forum/index.php?topic=85923

[Eternal Student (OP)]

Hi.

   Yes, maybe Rubik's cube is OK.   It's a little artificial thinking of the number of turns to solve as a temperature, or the number of turns applied as an energy input.   For example, if you put one Rubik's cube next to another one then the more turned one doesn't start un-turning while the other cube turns more until they reach an equilibrium temperature.
  None-the-less the principle is being demonstrated quite graphically and it is interesting.

Best Wishes.

[Petrochemicals]

?

[chiralSPO]

I believe it would be impossible for strictly positive temperatures to satisfy the constraints in the OP, if we want to talk about thermodynamic equilibria (I haven't done the math,  but I'm pretty sure one would quickly run afoul of the second law of thermodynamics.)

But, I think petrochemicals post shows an interesting solution when one also accounts for kinetic barriers.

One could imagine a chemical reaction (or just change of state, like melting/freezing) which is thermodynamically spontaneous at a given temperature, T₁ (ΔG < 0), with ΔH < 0 and ΔS <0). If there is an activation barrier such that the rate of reaction is imperceivably slow at T₁, but progresses at a higher T, such that ΔG, ΔH, and ΔS are all negative, then one would observe a system that becomes more ordered as it's temperature increases.

Supercooled water is a good example of this, but perhaps there are even better examples, such as supercooled sodium thiosulfate or sodium acetate solutions. (commonly used for instant heat packs. they crystallize and get hot after the crystallization is initiated--typically done mechanically, but I'm sure increasing temp would work too, if done right).

[Eternal Student (OP)]

Hi.

   Yes, OK, I see that.  It does seem to hinge around whether a supercooled liquid, or @chiralSPO 's suggestion of any chemical reaction that is only proceeding imperceptibly slowly,  really is in an equilibrium state.   Certainly in the case of a slow reaction, it isn't.

   While we do measure the temperature of things on an ordinary and everyday basis whether they have reached an equilibrium state or not  (just by sticking in a thermometer etc),  on a more formal basis we can't actually do that.  Any attempt to do so is only obtaining an empirical estimate of temperature.
   Temperature, Entropy (and similar functions of state) are only properly defined for a system at equilibrium.   

   I'm going to have to think about a supercooled fluid.  I guess that genuinely is in some equilibrium state as long as you keep it in a smooth container etc.  I'll have to think about that.

   Even assuming the supercooled fluid is in a valid equilibrium state,  the tacit assumption you have both made is that the entropy of a solid is lower than the entropy of a liquid.   That's only generally true and based on the notion that particles in a solid can't move around as freely as particles in a liquid.  A solid is generally more ordered and has less modes of supporting internal energy.
     The question is - can a supercooled fluid really support as many microstates as the fluid at a more conventional temperature?   I'm thinking of stuff like Bose-condensation where certain modes of supporting energy are simply frozen out at low temperatures.

Diagram from Wikipedia showing a generic version of the change in C_V which can be explained classically as a loss of modes of supporting energy at various temperatures:


   The main point is that a supercooled fluid may already be in a situation where the higher energy rotation and vibration states have been frozen out.   Overall then it's entropy could be much the same as the entropy of the solid ice.
   
   Anyway, I like the suggestion and thank you both for your time.

Best Wishes.   

Late Editing:  Fixed some spelling, removed some surplus discussion.

[chiralSPO]

It does seem to hinge around whether a supercooled liquid, or @chiralSPO 's suggestion of any chemical reaction that is only proceeding imperceptibly slowly,  really is in an equilibrium state.   Certainly in the case of a slow reaction, it isn't.

I wouldn't call either of these cases as being in an equilibrium state.  I suppose we could say that an uninitiated heat pack could be in thermal equilibrium with another system (ie when it's on the shelf, it should have the same temperature as the surrounding air, and will roughly track the temperature of the air as it rises and falls during the day, absorbing and releasing heat to "re-equilibrate" as long as the energy involved is not enough to disrupt the metastable state and kickstart the transition.)

As far as "very slow" vs 0 reaction rate goes... Because molecules are discrete there can be reactions that are not strictly impossible (even thermodynamically favorable), but are still so improbable (at a given temperature) that even with moles of interacting molecules the molecular reactions would not take place even once in millions of years (or any arbitrary timeframe). But increase the temperature by a couple hundred degrees, and suddenly it's a fast reaction (temperature is usually a term in the exponent of the rate equation).

[Petrochemicals]

The problem with the supercooled stuff is that the temperature follows the state, it is a cause there of, not a concequence.