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The Meaning of Temperature

  1. Apr 12, 2007 #1
    It is possible to put almost every unit in physics in terms of a product of integer powers of mass (kg), displacement/distance (m), time (s), and charge (q). Here are some examples:
    [tex] \begin{array}{c} speed = m^1 \times s^{-1} \\ acceleration = m^1 \times s^{-2} \\ jerk = m^1 \times s^{-3} \ \\ momentum = kg^1 \times m^1 \times s^{-1} \\ force = kg^1 \times m^1 \times s^{-2} \\ energy = kg^1 \times m^2 \times s^{-2} \\ power = kg^1 \times m^2 \times s^{-3} \\ pressure = kg^1 \times m^{-1} \times s^{-2} \\ current = q^1 \times s^{-1} \\ voltage = kg^1 \times m^2 \times q^{-1} \times s^{-2} \\ resistance = kg^1 \times m^2 \times q^{-2} \times s^{-1} \end{array} [/tex]
    ...and so on. My question is: Is there any way to put units of temperature (degrees) into this format?
     
    Last edited: Apr 12, 2007
  2. jcsd
  3. Apr 12, 2007 #2

    rbj

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    temperature is essentially the amount of unordered kinetic energy per particle per degree of freedom. it's a measure of energy and the conversion factor is the Boltzmann constant. Temperature can be removed as a physical quantity, expressing it in terms of the other base physical quantities (time, mass, length) but so can electric charge (as they do with the cgs system of units). however, whereas i don't consider temperature to be a fundamentally different physical quantity (it's just a measure of energy), i do consider electric charge to be a fundamentally unique or separate dimension of physical "stuff".


    i might recommend Wikipedia at Fundamental unit, Dimensional analysis, Planck units
     
  4. Apr 12, 2007 #3
    How would one express temperature in terms of the other units, then?

    I looked it up on Wikipedia: It defined units of temperature in terms of energy and entropy. Then I looked up entropy: it defined the units in terms of temperature. Neither entry mentioned the fundamental units. (The fundamental units article didn't help, either, because they treat temperature as a fundamental unit in itself!)
     
    Last edited: Apr 12, 2007
  5. Apr 12, 2007 #4
    I personally measure entropy as a unitless quantity, and temperature in the units of energy. However, units don't tell you what something means, necessarily.

    Mehran Kardar's notes for Stat. Mech. I on the MIT CourseWare website have a great discussion of temperature from a microscopic viewpoint. Basically, he shows that temperature can be thought of as a constraint between all the parameters between two systems in thermal equilibrium. In my mind, this is a fruitful way of looking at temperature, since it's a very physically grounded defintion.
     
  6. Apr 12, 2007 #5
    Is there any chance you could put that in somewhat simpler terms...? I know what thermal equilibrium means, and I know what a system is, but I don't quite understand the "constraint between all the parameters" part.
     
  7. Apr 12, 2007 #6
    This is why I use charge as one of my fundamental units instead of the traditional SI unit, amperes.
     
  8. Apr 12, 2007 #7

    ShawnD

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    Temperature sort of is a fundamental unit. The correlation between temperature and energy is sketchy because things can sponge a different amount of energy at different temperatures. That chart says you might be able to change the temperature by 1 degree using 1J of energy at a given temperature, then you might need 1.2J to get the same change at a higher temperature, then 1.5, then 1.6 etc etc.

    You can't directly correlate temperature to an exact amount of energy without knowing what the material is, and that's why I think temperature is a fundamental unit.
     
  9. Apr 12, 2007 #8
    So basically what you're saying is, the correlation between temperature and energy in a given substance is an intrinsic property of the substance, and therefore it should be treated as a fundamental unit?
     
  10. Apr 12, 2007 #9

    ShawnD

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    Yes that is exactly what I am saying. You can put it in terms of entropy and enthalpy but in the end you still need to know what the material is.
     
  11. May 13, 2007 #10
    can you express length unit (m) in terms of the other units???



    the same with temp.
     
  12. May 13, 2007 #11
    Yes, you can. Distance equals velocity times time. Also, temperature equals Heat divided by (mass times specific heat). However, I still consider distance and temperature to be fundamental units.
     
  13. May 13, 2007 #12

    Hootenanny

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    Yes, but what are the units of velocity? That's right meters per second; you can't use a unit to define itself!
    Erm, no it doesn't. That would be change in temperature, not actually temperature.
     
    Last edited: May 13, 2007
  14. May 13, 2007 #13
    That's exactly why I still consider each of those units to be fundamental.
     
  15. May 14, 2007 #14
    Basically, the argument is as follows.

    System A can have its microscopic configuration specified by a set of parameters A_1, A_2, etc. The same can be done for system B and system C. If A and B are in thermal equilibrium, that this indicates that there is a "constraint equation" (same concept as with Lagrangian mechanics) f(A_1, A_2, ... B_1, B_2, ...) = 0.

    He uses this argument to prove the zeroth law of thermodynamics from microscopic considerations, and later proves that this constraint serves the function of temperature, and therefore calls it as such. I can't really think of a "simpler" way of putting this, it's a little abstract to begin with.

    Here is the url for those particular lecture notes: http://ocw.mit.edu/NR/rdonlyres/Physics/8-333Fall-2005/BC713F26-6A2D-499A-B6DB-657804F9C32A/0/lec1.pdf
     
  16. May 14, 2007 #15
    In the modern system of measuring units, time is determined in terms of the frequency of oscillation for a particular element of cesium, and a meter is then defined as the distance light travels in some absurdly short period of time. So technically, the meter is defined in terms of a velocity, but it's a universally constant velocity multiplied by a constant time (in the same inertial frame).
     
  17. May 18, 2007 #16
    I believe the answer to the original post of Izzhov is that
    temperature is unitless; i.e. the degree is really not a fundamental unit but
    just determines the numbers that are used in a particular temperature scale.
    Temperature is a measure of whether there will be a net flow of heat from one
    body to another with which it is in thermal contact. Any
    scale of temperature must satisfy the zeroth law of thermodynamics,
    which effectively says that there will be a net heat flow from a hot
    body to a cooler body. The choice of scale is not determined from this
    law.

    The temperature of a system can be obtained by measuring the product (PV) of a
    fixed amount of any gas in contact with the system at a series of
    pressures and then extrapolating the product to zero pressure. P is the
    pressure and V the volume of the gas. This is
    the ideal-gas scale of temperature (See Giauque:
    Nature, 143, 623, (1939)) and was
    adopted by the International Unions of Physics and Chemistry:

    T=273.1600 lim(P-->0)(PV)T/lim(P-->0)(PV)t.p.

    Here the subscript t.p. refers to the triple point of water,
    The number 273.1600 is chosen to make the degree of temperature the same size
    as that of the centigrad (Celsius) scale. Temperature is related to energy
    via the equation for the ideal gas: PV = nRT, where n is the number of moles
    of gas and R is the gas constant, which is a fundamental constant:
    R = 8.314 J K-1 mol-1.

    The thermodynamic temperature enters into the definition of entropy
    as an integrating factor. It can be shown from standard thermodynamics
    arguments that in fact the thermodynamic temperature may be chosen to be
    identical to the ideal-gas scale of temperature. This scale does not
    depend on the properties of any one substance used in measuring the
    temperature. In particular, the specific heat of a substance plays no
    role in the definition of the temperature scale.
     
  18. May 20, 2007 #17
    Do you agree with people who say that the thermosphere is very hot?
    (I.E this website says, "The actual temperature in the Thermosphere can reach as high as 2000º C"

    But if you take a jar of boiling water at 100C to the thermosphere, I am sure that heat will flow from the jar to the thermosphere and it will become ice.

    I believe that the thermosphere (and outer space in general) has high speed particles because high speed particles can escape the gravitational field of planets and stars. The Maxwell distribution of velocities in a gas predicts that there will always be a small percentage of particles having a velocity more than the escape velocity.

    But these particles rarely collide with each other, and it is meaningless to say that their temperature is high because of their high kinetic energy. Would you agree?
     
  19. May 21, 2007 #18
    Dear Peaceharris,
    Your have posed four questions:
    1) Do I agree that the thermosphere is hot?
    2) Would a jar of boiling water placed in the thermosphere freeze to ice?
    3) Do I agree that particles in outer space have high velocity because
    they are the ones on the tail of the Boltzmann distriubution that
    lies aove the escape velocity?
    4) Is it meaningless to ascribe a high temperature to particles having a
    high velocity?

    Oh dear, this post may be a bit long! As a preface to my answers, let me
    recall that my reply to the original post about the units of temperature
    was couched in the terminology of classical thermodynamics, in which the state
    functions (e.g. energy, entropy, enthalpy, fugacity, free energy, etc.)
    refer to systems in equilibrium. Moreover, I avoided all mention of
    statistical mechanics (which is just a mechanistic interpretation of
    classical thermodynamics) because it was not needed. To answer your questions,
    one needs to consider the distribution functions that arise in statistical
    mechanics, where temperature can be regarded as a parameter in a function
    determining the distribution of some quantity (e.g velocity or energy or
    quantum state etc) among a number of particles. An important remark here is
    that if temperature is so regared, then it is nonsense to ascribe a
    temperature to any particular particle; i.e. we must always bear this in
    mind. So now to my answers.

    1) The thermosphere is not in thermodynamic equilibrium (and never
    will be in my lifetime!), so to speak of a temperature in the usual
    sense is meaningless. So what do people mean by hot (2000 C in this
    case)? According to the text in the web-site, "temperature is a measure
    of how fast particles move" (This is of course not a very good definition.
    It is like saying that money is a measure of the value of cheese.), so
    they must be using the velocity distribution of the particles to define
    a temperature, i.e. they are asking "What temperature would a Boltzmann
    distribution have to have in order to fit the observed
    distribution of velocities?" The answer is apparently T=2000 C. I
    can agree with this kind of parlance but one must realise that this
    is not the temperature. If a system is not in equilibrium, there can
    be as many temperatures as
    there are quantities for which a distribution can be determined, and
    all of these can be different! Even negative temperatures can
    be observed by playing this game! (I can give you an example in
    another post if you are interested.) At equilibrium, all of these
    temperatures would be equal.

    2) Are you so sure that boiling water would freeze in the thermosphere?
    I have no idea, not being an atmospheric physicist but consider this:
    You are on your way to a wedding at 10:00 am dressed in your best
    black suit. An alien specimen collector snaps you up, puts you in
    a big closed jar of air and places you in the thermosphere directly
    above the spot upon which you were walking. Will you freeze? I think
    not---you will probably be fried in the sunshine. And if instead, you
    were plucked from the street on returning from the wedding at 11 pm? Then
    I think you would freeze in the dark shadow of the earth. Maybe this
    latter situation was your point: the "hot" particles would not prevent
    your freezing. True, but not because they are not hot but because there
    simply aren't enough of them---they would have to compete with the
    rate at which the boiling water is losing energy owing to radiative
    cooling (infra-red radiation).

    3) Suppose you are in the thermosphere and you measure the velocity
    of those molecules leaving the earth that reach you. Yes, you would
    find some molecules with high velocity but you would find many many
    more with low velocity, namely those that do not have
    escape velocity but which are fast enough to have reached your
    position (remember that gravity is a very long range force).
    (In order to have reached you, they had to overcome the gravitational
    force and have therefore a smaller velocity than when they started
    their journey)
    If you play the fitting game with your observed distribution, you
    might find a "temperature" which was not too different from the
    original temperature of the molecules nearer the earth. Note that
    many of the molecules reaching you would have zero velocity.

    But there is an even more important consideration: are there any
    other processes that could produce fast particles? I can think of
    one at least. The web-site mentioned that the oxygen and nitrogen
    molecules were good at converting "heat" into velocity. How
    do they do this? One way would be for a molecule to absorb radiation
    from the sun and thus go from its ground state into a highly
    excited electronic state. If the latter were not stable with respect
    to molecular dissociation, the two atoms (e.g two N atoms from
    nitrogen) would then fly apart with a high velocity. I don't know if
    such a process is important in the thermosphere but it is certainly
    well-known in the laboratory and would merit your investigation before
    you place too much faith in your escape velocity theory.

    4) I agree with you. It is meaningless to talk of the temperature of a
    particle (in the context of our discussion here).
     
    Last edited: May 21, 2007
  20. May 22, 2007 #19
    I think we can estimate the final steady state temperature of a sealed jar of water in the thermosphere by making certain assumptions:

    Assume that your jar is spherical with radius r, and it absorbs all energy falling on it, and it is a perfect thermal conductor, and the walls of this jar radiates like a blackbody. The heat absorbed by the jar depends on the heat it gets from the sun and the earth.

    Case 1: Its 12 noon.

    Total power of the sun= 4e26 Watts
    Power per unit area at a distance D from the sun= 4e26/(4*pi*D^2)
    Power incident on the jar from the sun=4e26 * pi * r^2/ (4*pi*D^2)

    The power radiated out from the walls of this jar using Stefan-Boltzmann's law =s * 4 *pi*r^2 * T^4 (where s is Stefan's constant)

    Total power radiated by earth assuming the earth radiates like a black body, with a uniform surface temperature of 300K = Stefan's constant * 4 *pi*R^2 * 300^4 (where R is the radius of the earth)

    Total power radiated by earth per unit area at the thermosphere 100km from the earth's surface (at the thermosphere)=Stefan's constant * 4 *pi*6360000^2 * 300^4 /(4*pi*6460000^2) (I have assumed R=6360000m)

    Power incident on the jar from the earth=Stefan's constant * 4 *pi*6360000^2 * 300^4 *pi * r^2 /(4*pi*6460000^2)

    At steady state, power incident on the jar from sun and earth = power radiated out, so we can solve for T.

    At 12 noon, based on this arithmetic, T is about 301K

    Case 2: Its 12 midnight, there's no power from the sun:
    At steady state, power incident from earth = power radiated out

    At 12 midnight, based on this arithmetic, T is about 210K

    I don't think the energy imparted to the jar from collision of high speed particles to the jar is going to make much difference to this number. In any case, the final temperature of the jar initially with boiling water at 100C will reduce, implying that heat has flowed out. So I think people should stop saying that the thermosphere is hot or is at 2000K, because there will always be a net flow of heat from the jar of boiling water of 100C to the thermosphere.
     
  21. May 22, 2007 #20
    Peaceharris: Yes, I fully agree that the effect of the high speed particles on your temperature estimate would be negligible.
     
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