What is the thermodynamics temperature scale?

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SUMMARY

The thermodynamics temperature scale is fundamentally independent of physical substances, as established by the efficiency of the Carnot engine and the Second Law of Thermodynamics. The ideal gas law, represented by the equation pV ∝ T, serves as a practical approximation but does not capture the theoretical underpinnings of temperature. The International Temperature Scale (ITS-90) utilizes various fixed points, such as the triple point of water at 273.16K, to define temperature across different ranges. Negative temperatures can occur in specific systems, such as paramagnetic solids, where the relationship between entropy and energy leads to unconventional temperature definitions.

PREREQUISITES
  • Understanding of the Carnot engine and its efficiency principles
  • Familiarity with the ideal gas law (pV = nRT)
  • Knowledge of the International Temperature Scale (ITS-90)
  • Basic concepts of entropy and thermodynamic equilibrium
NEXT STEPS
  • Research the principles of the Carnot engine and its implications for thermodynamic temperature
  • Study the derivation and applications of the ideal gas law in thermodynamics
  • Explore the International Temperature Scale (ITS-90) and its various fixed points
  • Investigate the concept of negative temperatures and their occurrence in specific physical systems
USEFUL FOR

Students of thermodynamics, physicists, and engineers interested in the theoretical and practical aspects of temperature measurement and its implications in various scientific fields.

  • #31
Gerenuk said:
They can be resolved and work for all of thermdynamics, but I'm going to write down the derivation again. If you insist I can try to dig out a part of the derivation that I wrote in another thread.


Because no other way is possible. In another thread I asked before, and the only answer was the ideal gas method (link that I gave earlier in this thread).

The Carnot definition requires hypothetical engines which do not necessarily exist and no physical device can be confirmed to be strictly Carnot-like.


Then you cannot make any statement about the system. It's just some arbitrary system with completely arbitrary dynamics. Of course then you cannot say anything about it.

This is really disappointing- you have unilaterally declared 99% of reality to be off-limits of physics.
 
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  • #32
Gerenuk said:
It's easy to come closer to an ideal gas than any possible Carnot-like engine. Please tell what you propose instead. Because I'm merely saying that there is no better method than ideal gas (to my knowledge and to the last discussions).

Let me back up a bit here: "..no better method than ideal gas"

I don't understand what you are saying- no better method *for what*? thermodynamics? statistical mechanics? Something else?
 
  • #33
Andy Resnick said:
This is really disappointing- you have unilaterally declared 99% of reality to be off-limits of physics.
Or maybe it's rather 1% and you don't know how to write the partition function for 98% of physics.
 
  • #34
Andy Resnick said:
Let me back up a bit here: "..no better method than ideal gas"
I don't understand what you are saying- no better method *for what*? thermodynamics? statistical mechanics? Something else?
A better method for measuring the temperature. Just describe one. But one which doesn't use ideal gases and no Carnot engine. Especially Carnot engines don't exist.
 
  • #35
Andy Resnick said:
It's not complete- one or two simple examples should suffice to show that.

Blackbody radiation is defined as a photon gas in thermal equilibrium with a reservoir at some temperature T. Say blackbody radiation is emitted, and then passes through a spectral filter- the spectral filtering can be as broad or as narrow as you wish. In fact, we could simply pass it through a polarizer and separate out the polarization components. The energy is well-defined, the change in entropy is well-defined, but the photon gas no longer has a temperature- even if all the optical components are at the same (initial) temperature.

Same for monochromatic radiation- a well defined energy, a well-defined entropy, but it cannot be assigned a temperature.

Thank you for the counter-examples. But when you say the entropy and the energy are known, are they not related? Is that the problem?

I'll try to read up a bit more.
 
  • #36
Gerenuk said:
A better method for measuring the temperature. Just describe one. But one which doesn't use ideal gases and no Carnot engine. Especially Carnot engines don't exist.

What's wrong with the mercury thermometer I have in my lab? It seems to work well enough... I also have one of these:

http://www.fishersci.com/wps/portal/PRODUCTDETAIL?prodcutdetail=%27prod%27&productId=1585059&catalogId=29104&matchedCatNo=150782C||150778||150781&pos=5&catCode=RE_SC&endecaSearchQuery=%23store%3DScientific%23N%3D0%23rpp%3D15&fromCat=yes&keepSessionSearchOutPut=true&fromSearch=Y&searchKey=digital||thermometer||thermometers&highlightProductsItemsFlag=Y

which I think uses a thermocouple, and it seems to work ok as well. At least, the two thermometers agree with each other when they are both in an incubator.
 
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  • #37
Mr.Miyagi said:
Thank you for the counter-examples. But when you say the entropy and the energy are known, are they not related? Is that the problem?

I'll try to read up a bit more.

Well for the case of a laser, the light has a very low entropy. But the energy can be fairly arbitrary- just change the intensity. There is no unique way to assign a value of temperature.

This is a good article:

How hot is radiation?
Christopher Essex, Dallas C. Kennedy, and R. Stephen Berry
Am. J. Phys. 71 969 (2003)
 
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  • #38
"How hot is radiation? "
Yes, can I have a copy?
 
  • #39
Andy Resnick said:
What's wrong with the mercury thermometer I have in my lab? It seems to work well enough... I also have one of these:
Because you require an ideal gas experiment to calibrate those thermometers in the first place. So in the end again the only definition that's useful is an ideal gas definition and the zeroth law.
 
  • #40
If I had asked about the time scale and clocks instead, would the discussion be similar?
 
  • #41
quantum123 said:
If I had asked about the time scale and clocks instead, would the discussion be similar?
No, that's more well known. Time is defined by a specified atomic process (whichever is the most current candidate...). Length is the product of time multiplied with a fixed velocity.
 
  • #42
Gerenuk said:
Because you require an ideal gas experiment to calibrate those thermometers in the first place.

Gas thermometers are only used for some parts of the temperature scale. The lowest temperature part is defined using the melting curve of He3. For some parts the melting/freezing points of various metals as well as tripple points are used as fixed points to calibrate a platinum resistor. For very high temperatures fixed points are used together with radiation thermometry.
There are also other primary (primary means that it does not need to be calibrated) thermometers that -although not officially parts of ITS-90- are often used; the most common being nuclear orientation thermometers (used below about 0.3K, I use one in my lab) and various types of noise thermometers (as well as Coloumb blockade thermometers etc).
 
  • #43
To measure absolute temperature experimentally, one can start from the fact that absolute temperature is an integrating factor for heat, dS=dQ/T, at least for reversible changes, where T is a function of the empirical temperature theta (e.g. the height of mercury in a thermometer). Details can be found e.g. in the book by Buchdahl "Concepts of classical thermodynamics", Cambridge UP, 2009 - a very carefull axiomatic treatise on thermodynamics. Experimentally, one has to measure basically heat capacities dU/d theta and compressibilities.
 
  • #44
Gerenuk said:
Because you require an ideal gas experiment to calibrate those thermometers in the first place. So in the end again the only definition that's useful is an ideal gas definition and the zeroth law.

Oh- calibration. I thought you said *measure*.

I would calibrate my thermometers with a triple-point device. Water, as you know, is not an ideal gas at the triple point.

The ideal gas scale of temperature is only used to set the energy equivalent of a change of temperature of 1 degree. To that extent, it is an arbitrary scale.
 
  • #45
f95toli said:
Gas thermometers are only used for some parts of the temperature scale. The lowest temperature part is defined using the melting curve of He3. For some parts the melting/freezing points of various metals as well as tripple points are used as fixed points to calibrate a platinum resistor. For very high temperatures fixed points are used together with radiation thermometry.
Thanks. I didn't know that yet.
OK, but to use the melting curve, you need to know it's theoretical behaviour, right? In fact you always need to know the theoretical behaviour which you will only will get from the microscopic statmech definition. So in this case the statmech definition is consistently used?!
An triple points are single points only where you still don't know how to assign good temperature values between them.

f95toli said:
There are also other primary (primary means that it does not need to be calibrated) thermometers that -although not officially parts of ITS-90- are often used; the most common being nuclear orientation thermometers (used below about 0.3K, I use one in my lab) and various types of noise thermometers (as well as Coloumb blockade thermometers etc).
Can you quickly name the theoretical concepts which stand behind all these methods? I assume they used the statmech definition to predict temperature and compare it to the measure curves?!

Andy Resnick said:
The ideal gas scale of temperature is only used to set the energy equivalent of a change of temperature of 1 degree. To that extent, it is an arbitrary scale.
Of course. And that's the topic here. So in the end the ideal gas scale is applied all the time! That little triple point scaling is just a detail. This ideal gas definition is necessary to set all the marks on your thermometers. The first degree might be arbitrary, but all others are predicted by the ideal gas.

DrDu said:
To measure absolute temperature experimentally, one can start from the fact that absolute temperature is an integrating factor for heat, dS=dQ/T, at least for reversible changes
I think it's highly impractical or even impossible to find a real reversible engine that transfers heat. But apart from that, I agree. If perfect, cyclically operating engines exists which are also able to completely reverse their operation, then temperature ratios can be defined.
 
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  • #46
Gerenuk said:
<snip>
Of course. And that's the topic here. So in the end the ideal gas scale is applied all the time! That little triple point scaling is just a detail. This ideal gas definition is necessary to set all the marks on your thermometers. The first degree might be arbitrary, but all others are predicted by the ideal gas.

<snip>

That's not at all what I meant. The choice of an ideal-gas scale sets the energy equivalent of a 1-degree *change*. In fact, an ideal gas is defined by the choice of temperature scale, not the other way around.
 
  • #47
Andy Resnick said:
That's not at all what I meant. The choice of an ideal-gas scale sets the energy equivalent of a 1-degree *change*. In fact, an ideal gas is defined by the choice of temperature scale, not the other way around.
No. An ideal gas obeys
pV\propto T
So it's a straight line in the graph. No need for personal definitions.
It's a physical law by itself. If you define temperature what way you like, then the ideal gases won't agree with you. You won't get a straight line the the graph.
 
  • #48
Sigh. It's an unfortunate state of affairs when a non-existent idealization is considered the correct basis of reality. Whither viscosity?

All I am willing to do now is recommend you read up on the foundations of thermodynamics. I haven't read Buchdahl's book (recommended above) but the 'peek inside' seems to indicate it's decent. personally, I recommend Truesdell's "Rational Thermodynamics".
 
  • #49
Gerenuk said:
Can you quickly name the theoretical concepts which stand behind all these methods? I assume they used the statmech definition to predict temperature and compare it to the measure curves?!

The basic idea behind Nuclear orientation thermometry is that the pattern of gamma rays emitted from nuclei depends on the alignment of the nuclei (well, the spin). This is implemented using a Co-60 crystal which is bolted to whatever you want to measure the temperature of. Co is a ferromagnet meaning the spins tend to line up in the direction of the intrinsic B-field at low temperatures.
At 0K the emission pattern would ideally look like a d-wave meaning there are nodes in the pattern,but as the temperature increases the spins become more "randomized" meaning the emission from an ensemble is uniform.
The temperature is measured by using a gamma ray detector positioned in a node direction; you get the temperature by calculating the ration N(T)/N(high T), where N means number of counts. "high T" means any temperature high enough to have randomized the spins (in reality everything over about 600 mK or so). This ratio plus a bunch of constants is all you need to calculate the absolute temperature. This is a very reliable method, and if you buy a thermometer calibrated for temperature below about 300 mK this is how it was calibrated, many low-temperature labs (including mine) have nuclear orientation setups (very handy for trouble-shooting since one can rule out problems with thermometery).

The simplest form of noise thermometry is to simply to measure the thermal noise across a resistor in a known BW. This is not a very good method but works in principle.
Most "real" noise thermometers are based on the "escape from a potential well" concept, i.e. what you basically measure is the escape probability from a potential which then can be related to the absolute temperature using the Boltzmann factor (or FD or BE factors).
The tricky part here is of course that you are always measuring an effective temperature, i.e. the thermodynamic temperature+any other sources of noise. Hence, the temperature measured using e.g. an electrical noise thermometer will always be higher (and unless you are careful MUCH higher, I've seen temperature as high as 1K in system thermalized to a 30mK bath) than the phonon temperature.
 
  • #50
Thanks for the details. I'll save that contribution to my files :)

So finally, both methods rely on the statmech theory and the prediction by the Boltzmann distribution?
 
  • #51
I found finally some time to go through my books: A more practical procedure to determine the absolute temperature than trying to run a carnot machine as reversible as possible is described in Peter T. Landsberg, Thermodynamics and Statistical Mechanics, Dover, NY, 1990, Chapter 6.1. "Empirical and absolute temperature scales". With T being absolute temperature and t being an empirical temperature, the important result (his formula 6.2) is
\ln\frac{T_1}{T_0}=\int_{t_0}^{t_1}\frac{(\partial p/\partial t)_<br /> V}{p+(\partial U/\partial V)_t} dt
The right side does depend only on the empirical temperature (e.g. the length of a mercury filament in a thermometer), but neither on absolute Temperature nor entropy.
The reference Temperature T_0 being fixed arbitrarily, the absolute temperature corresponding to any temperature t may be calculated once the two derivatives have been measured for the intermediate temperature range.
 
  • #52
I have found some similar equation. One problem is that you need to be able to go from the reference state to the final state with a reversible engine (i.e. constant entropy). That might be troublesome?!

Also note that the derivative in the denominator should be at constant pressure. So you cannot calculate it while running the Carnot engine. To make a change at constant pressure you would destroy the reversibility. Can one even use the definition in that case?
 
  • #53
No, you don't need a reversible engine!
You only have to measure the quantities in the integral for the range of temperature you are integrating over. But this are all standard calorimetric measurements. More importantly, all functions in the integral only depend on the states but are not path dependent.
E.g., to measure the differential dU/dV at fixed t, you could measure the Heat and Work needed to expand some substance at constant temperature. While heat and work may depend on the path taken on their own, their sum will not.
PS: the derivative in the denominator is at fixed temperature t, not fixed pressure p.
 
  • #54
The whole integral depends on path and has to be taken at constant entropy (strictly reversible process). The derivative dU/dV needs to be taken at constant pressure and no other variable. I've done the calculation myself. Maybe you can check the book again.
Meanwhile I try to figure out if your equation is also correct, but I highly doubt it.

EDIT: I have 5 equations of this type and I found you are indeed correct.
 
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  • #55
But note that the integral has to be taken at constant volume only, i.e. your change of state variables necessarily needs to go along constant volume.
So you can only find temperature ratios between states of equal volume.

Also dU/dV is taken at constant temperature, which does not lie on the constant volume curve on which the integral has to be. So I'm not sure how to do this in practice?!

EDIT: Actually I agree, that there isn't really a path of the integral. But the derivatives which are taken along different routes (constant volume or constant temperature) have to be taken care of. Is that doable?
 
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  • #56
I once read something about the experiments at NIST (if I remember correctly) which lead to the definition of the ITS-90 temperature scale. Seemed to have been a fascinatingly complex experiment. Lamentably, such experiments do not get the attention as e.g. the largest short-circuit experiment at CERN.
I am not an experimentalist, but I don't see any principal problems in evaluating the integral we were talking about experimentally. E.g. for a (real) gas, the pressure and the derivatives dp/dt and dU/dV vary smoothly with V and t. So you could measure them on a grid of points V and t, and interpolate between these points. Then you numerically integrate your formula and you are done.
 
  • #57
DrDu said:
I am not an experimentalist, but I don't see any principal problems in evaluating the integral we were talking about experimentally. E.g. for a (real) gas, the pressure and the derivatives dp/dt and dU/dV vary smoothly with V and t. So you could measure them on a grid of points V and t, and interpolate between these points. Then you numerically integrate your formula and you are done.
Interesting. Actually you could make constant t steps to calculate dU/dV followed by constant V steps to calculate dp/dt. For the first kind, the variable t doesn't change, which does the trick.

As I said I have different such equations (including the one I wanted first), but I wasn't aware of how practical your particular equation is. Thanks for pointing that out.

Now I can go back to derivation and look where it comes from. Because temperature seems to pop out of nothing for a completely arbitrary system!
 
  • #58
As I already said, it is a consequence of the fact that the inverse of absolute temperature is defined so as to be an integrating factor for dQ in a reversible process, i.e. dS=dQ / T(t). I'll sketch the derivation: For a reversible process, dQ=dU+pdV, now express dU=(\partial U/\partial t)_V dt +(\partial U/\partial V)_t dV. Use also that \partial/\partial t=dT/dt \partial /\partial T. Hence dS can be written as dS=Adt+BdV. For dS to be a total differential, \partial A/\partial V=\partial B/\partial t. Solve this condition for T(t).
 
  • #59
I mean that's all just letters. You could take this definition and apply it to a herd of bisons. V being the number of bisons. E being the total age of the bisons. Then you can define any social dynamics between groups of them and calculate things like "temperature"!
Two interacting groups of bisons will have the same parameter "temperature".

That's what I find surprising. It's all just maths with hardly any assumptions and in fact no assumption that relate it to any particular physics.
 
  • #60
The modern formulation of the second law, named after Caratheodory, says that not all states in the neighbourhood of a given state can be reached from it, without the exchange of heat. To reach these states, the system has to loose heat. This shows that heat or some function of it, should be a function of state, which we name entropy. Heat is not a function of state, whence we need an integrating factor. That it has to be a function of empirical temperature is easy to show, the rest is mathematics. Both the books by Buchdahl and by Landsberg, which I already cited elaborate this view in detail (although I even prefer the original article by Caratheodory). That the efficiency of a thermodynamic cycle is limited by the absolute temperatures of the reservoirs follows as a corollary.
However, from classical thermodynamics it is not clear, why for an ideal gas T=pV/nR.
 

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