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quantum123
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What is the thermodynamics temperature scale, both in theory and in practice?
Is it really independent of physical substance? Why?
Is it really independent of physical substance? Why?
quantum123 said:What is the thermodynamics temperature scale, both in theory and in practice?
Is it really independent of physical substance? Why?
Gerenuk said:I believe theoretically the (thermodynamic) temperature is found by assuming you have an ideal gas with
[tex]pV\propto T[/tex]
So basically you keep on variable constant and thus find the relative change of the other variables.
Experimentally
https://www.physicsforums.com/showpost.php?p=2270246&postcount=10
The ideal gas law is valid whenever the energy and the density of states are power laws of the momentum.
There might be better experts who can answer this...
Andy Resnick said:I'm not sure what temperature 'is', but thermodynamics uses 'temperature' as a real, positive, number that (somehow) relates to the energy of a system. The specifics of that relationship have not yet been worked out for the general case, only for equilibrium conditions.
Kelvin and Rankine scales are 'absolute' in the sense that the measured temperature is independent of the thermometer- something that was not the case in the early days of thermometry.
In practice, the temperature of an object tells you how hot it is.
Could you please describe the way with which they get a temperature reading, other than using the ideal gas law? But don't forget that it should be a fundamental principle not relying on data parameters which is collected by ideal gas law thermometers.netheril96 said:In practice,it is realized using different ways in different range of temperature.See http://www.bipm.org/utils/common/pdf/its-90/SUPChapter1.pdf
netheril96 said:Even thermodynamic temperature can be negative
See http://en.wikipedia.org/wiki/Negative_temperature
f95toli said:This topic comes up a lot.
The truth is that there IS no good general definition of temperature, which is why the international temperature scale (ITS-90) uses many different "definitions" (depending onf which fixed point is used for a particular range); although there is e.g. no fixed point at very low temperatures (ITS-90 starts at 0.65K).
While this might not be very satisfactory from a philosophical point of view it does have the advantange that it works...
Also, some primary thermometers (e.g. nuclear orientation thermometers and noise thermomenters) are based on statistical principles that are quite difficult to connect to more classical thermodynamics.
Actually that is the most general definition that I support. But the question back to you: So you are given a piece of wood. How would you measure the temperature in practice, now?Mr.Miyagi said:[tex]\frac{\partial S_1}{\partial E_1}=\frac{\partial S_2}{\partial E_2}[/tex],
where the volume of each compartment and the number of particles in each compartment is kept constant. S is the entropy.
Could you mention the method they use to find a temperature (which is not based on ideal gas laws)? How can you guarantee that any other method doesn't cause inconsistencies? For example with a flawed method I could measure the temperature of one body, bring it on contact with another body, measure the other body and suddenly get a different temperature reading.f95toli said:The truth is that there IS no good general definition of temperature, which is why the international temperature scale (ITS-90) uses many different "definitions" (depending onf which fixed point is used for a particular range)
What do you mean by related? Of course if you change one quantity of a system, all others will most likely also change. But note that energy is a function of temperature alone (no p or V dependence) only in the special case that [itex]p\propto T[/itex] at constant V. That special case doesn't necessarily have to hold.Andy Resnick said:Nobody here (IIRC) will disagree that temperature is related to energy.
Gerenuk said:Actually that is the most general definition that I support. But the question back to you: So you are given a piece of wood. How would you measure the temperature in practice, now?
Your definition only shifts the problem to a new (unknown) quantity, the entropy.
The hidden assumption is, that you need someone who has actually solved the problem of measuring temperature before you. And he must have used another method than you, because with your method alone you wouldn't be able to measure temperature.Mr.Miyagi said:But now I'm really curious as to what the objections to this are, since I assume you've considered this as well. I can't find any hidden assumption in this process.
OK, so in the end you are using the ideal gas law definition all the time? You have a definition of temperature with entropy, but you never use it. Instead you bring all you objects in contact with an ideal gas and read off the temperature?Mr.Miyagi said:Well, by sticking a thermometer in it. When the thermometer is in equilibrium with the wood, it gives a certain reading. Then calibrate the thermometer in a controlled experiment, where the value of the partial derivative is known. A gas that can be described as a perfect classical gas would do.
How do you pick which number exactly you want to assign? Who is telling you which test chamber pressure corresponds to which temperature?Mr.Miyagi said:I then assign values to the states of the device.
Why do you think that if you take two different devices, you won't get contradictory results when cross-checking the temperature of the same test object?Mr.Miyagi said:Then, since the ideal gas law is compatible with the definition of temperature as given before, the device could now be used to measure other things.
When do you ever need your quoted definition of temperature? You might as well have said [itex]\partial \sqrt{S}/\partial E=T[/itex] and nothing would be different?Mr.Miyagi said:These other things don't have to follow the ideal gas law, just the second law of entropy.
Step 4) was to bring both bodies in thermal contact (for a long time)Mr.Miyagi said:I still don't see why measurement of body B needs to give the same reading as body A. It seems to me that this should only true if body B does actually have the same temperature as body A.
Gerenuk said:Now say you measure temperature of body A. Then you bring body A in contact with body B. Now you measure temperature of body B. Why are you sure that temperature B will give the same reading?
Sure, but but cannot have two different definitions/properties of temperature. You have to derive all properties from one definition. Other consistency isn't guaranteed.SpectraCat said:Because that is the zeroth law of thermodynamics. If A is in thermal equilibrium with B, and B is in thermal equilibrium with C (the thermometer in this case), then A is also in thermal equilibrium with C. That is the thermodynamic definition of what it means for things to be in thermal equilibrium ... it provides the basis for what temperature *means*, although it does not provide a quantitative scale for measurement.
Mr.Miyagi said:I don't disagree with that. All I tried to say is one can devise a definition of temperature that is "good" and general. I don't see how a definition like dS/dE=1/T isn't good or general. This may very well be the result of my ignorance, which is why I'm participating in the discussion. Perhaps we have a different notion of what goodness and generality is in this case?
Gerenuk said:Sure, but but cannot have two different definitions/properties of temperature. You have to derive all properties from one definition. Other consistency isn't guaranteed.
I think the definition
[tex]\frac{1}{T}=\frac{\partial \ln\Omega}{\partial E}[/tex]
with knowledge about microscopic partition functions is best.
Maybe for some purposes. But wouldn't that screw up all nice physical equations with T in them?SpectraCat said:I don't think I agree with that ... I would rather say that one can have as many independent definitions of temperature as one cares to, provided that they all obey the zeroth law.
I thought about this definition for a while, but it's more tricky than that. It is crucial that you find a reversible engine operating between both sources (otherwise the equation you quote doesn't hold). That could be that catch when such engines between two particular sources don't exist. But I have to think again about it.SpectraCat said:Now take the system for which you want to measure the temperature and set it as the heat source for your ideal Carnot engine, with the cold sink in equilibrium with the triple point of water. Measure the efficiency of the engine, and use the Carnot formula to calculate the temperature of the hot source:
That's not the whole story. Of course I need to calculate [itex]\Omega[/itex] which I can only do for something idealized like an ideal gas. So in the end you can derive all of thermodynamics.SpectraCat said:Yeah .. but that doesn't work for your block of wood example, right?
Gerenuk said:Sure, but but cannot have two different definitions/properties of temperature. You have to derive all properties from one definition. Other consistency isn't guaranteed.
I think the definition
[tex]\frac{1}{T}=\frac{\partial \ln\Omega}{\partial E}[/tex]
with knowledge about microscopic partition functions is best. I just was trying to point out to Miyagi that he uses many hidden assumptions that people have derived before. They can all be resolved, but one has to think about it.
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.Andy Resnick said:They *can't* all be resolved. Your definition works fine for a very limited set of processes (thermostatics), but there is not yet a sound foundation for thermodynamics.
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).Andy Resnick said:Experiments do not use ideal gases; why should you require such a restricted basis for the measurement of temperature?
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.Andy Resnick said:And what happens if you do not know the "microscopic partition functions"- which is the case for almost all real objects?
Gerenuk said:The Carnot definition requires hypothetical engines which do not necessarily exist and no physical device can be confirmed to be strictly Carnot-like.
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.
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).
Or maybe it's rather 1% and you don't know how to write the partition function for 98% of physics.Andy Resnick said:This is really disappointing- you have unilaterally declared 99% of reality to be off-limits of physics.
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.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?
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.