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Difference between Temperature and Internal Energy

  1. Sep 19, 2011 #1
    Hi :)

    Uhh, this question is quite simple but I'm still confused between these two terms...

    Am I right to say that internal energy is the sum of kinetic energy and potential energy of all the particles in a system? Meaning the two factors are the number of particles, and the energy of each individual particle.

    And that temperature is the average kinetic energy (how about potential energy?) of each particle?

    Then, in that case, heat energy is transferred from the system with higher temperature to one with lower temperature, but not the one with higher internal energy to the one with lower internal energy, correct?

    And this transfer of heat energy occurs until both reach thermal equilibrium.

    So does thermal equilibrium then refer to temperature only, and not internal energy?


    Thank you xD
     
  2. jcsd
  3. Sep 19, 2011 #2
    Potential Energy does not have role while calculating temperature of bodies if thats the case Temperature of a body should rise when it is moved to higher altitude.

    Heat Energy (More correctly Kinetic energy) transfers from higher Temperature bodies to lower ones.

    And finally Thermal Equilibrium refers to Temperature only. Net Heat flow is zero when both bodies reaches thermal equilibrium i.e they attain same temperature.
     
  4. Sep 19, 2011 #3
    Thanks for your reply :)

    I kinda understand now that temperature only refers to kinetic energy, though just as a side note, the "potential energy" is was referring to in my question was that of chemical bonds between atoms in a molecule.
     
  5. Sep 19, 2011 #4

    Philip Wood

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    Your definition of internal energy is fine, and of course you're right to interpret potential energy as that due to inter-particle forces. Gases at low density (such as air at 'ordinary' temperatures and pressures) behave very much like an ideal gas, whose internal energy is wholly kinetic (though for molecules with more than one atom there is rotational KE to be considered). [I've oversimplified a little.]

    For an ideal gas the temperature is indeed proportional to the mean particle KE. For many other systems this is also approximately true, but the proper microscopic (i.e. particle-based) interpretation of temperature uses the concepts of statistical mechanics. See Ken G's post (below)!
     
    Last edited: Sep 19, 2011
  6. Sep 19, 2011 #5

    Ken G

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    That picture often works, but it isn't really what temperature "is." What the temperature of a large system in equilibrium "is" is a measure of the scale of the heat that the system "covets", by which I mean, the amount of heat you'd have to add to the system to get it to increase the number of states it has access to by a factor of e. That might sound like a lot to get your head around, but it is really the crux of temperature-- systems at high temperature are somewhat ambivalent to gaining or losing a unit amount of heat, because it does not much affect the number of states they have access to. For an ideal gas, for which each particle is essentially an independent system that has access to states based on its own kinetic energy, that directly translates into the average kinetic energy of each particle-- you "impress" the whole system if you can "impress" any one of its particles (increase the number of states one particle has by a factor of e and you increase the number of states the whole system has access to by that same factor), and the way you "impress" a particle with kinetic energy of order kT is by giving it more kinetic energy at that same order.

    However, in systems where you cannot treat each particle as having access to states based on its own kinetic energy, this simple picture breaks down. A great example of this is a white dwarf in astronomy-- here you have degenerate electrons who are acting in an extremely interdependent way, such that each electron can have a huge kinetic energy, but a very small number of allowed states. Then if you add even a small energy to one electron, it might increase the states it has access to by a large factor, which means its T is very low even though its kinetic energy is very high.

    Also, you mentioned potential energy, so you can see that if you have potential energy around, you will need to analyze what happens to that potential energy when you add heat. If a given particle has kinetic energy K and potential energy P, and if any heat you add gets shared between them, then you have to analyze how changes in both K and P will alter the number of states the particle has access to. There are, for example, a number of states available for each range in P, as well as each range in K, so they all have to be included to figure out what is the energy scale that will "impress" the particle (which is what kT means).

    Yes, thermal equilibrium means a fixed T everywhere-- all the systems in contact must "covet" the same scale of heat changes.
     
  7. Sep 19, 2011 #6
    Internal energy is the sum of all energies within the system. That includes not only kinetic and potential energy of particles but also their mass–energy equivalence, vacuum energy, dark energy and so on. But you may limit your calculation to all kinds of energies that are not constant.

    Temperature is no energy. It is just a function of state that characterizes a thermodynamic equilibrium. This function is correlated to the kinetic energy (Maxwell–Boltzmann distribution) but also to the potential energy (Boltzmann distribution). But for an ideal gas you may limit it to the mean kinetic energy.

    Yes, this is correct.
     
  8. Sep 19, 2011 #7
    I cannot speak for all physical usages of the term "internal energy", but in kinetic gas theory the term is used to differentiate between the "internal" kinetic energies of molecular rotation and molecular vibration/libration and the "external" kinetic energy of molecular translation.

    Temperature is seen as a measure only of the mean kinetic energy of translation.
     
  9. Sep 19, 2011 #8

    Ken G

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    I'm not sure what you mean by that, diatomic molecules at typical temperatures have a mean kinetic energy of 5kT/2 instead of 3kT/2, the extra kT being in rotations. So in that case, we can use T to characterize both the mean kinetic energy in translations (3kT/2), and also the mean kinetic energy in rotations (kT). In neither case is that what T really is though-- T will always relate to the scale of the added heat we would need to appreciably alter the number of states available to the system, regardless of whether those states are translational, rotational, or configurational, so T is always influenced by the full context.
     
  10. Sep 19, 2011 #9
    Take a mixture of monatomic, diatomic, and polyatomic gas molecules such as make up our atmosphere. The temperature of each of these component gases will be independent of how many atoms make up its molecule. The temperature of each of these gases will be a function of the mean kinetic energy of translation of each gas and those individual gas temperatures will all be the same.
     
  11. Sep 20, 2011 #10

    Ken G

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    The temperatures will all be the same, yes, and so I can use that fact to connect their translational kinetic energies or their rotational kinetic energies (for the diatomics). I still don't see any sense to which the translational modes are the only ones that involve temperature-- all that is happening is kT/2 of energy goes into every quadratic degree of freedom, and translational and rotational modes are both quadratic degrees of freedom (they have energies that depend on some type of velocity squared). It just happens that all the molecules have 3 degrees of translational freedom, so they all get 3kT/2 because of that. If I restrict some of them to move 2 dimensionally, they'll get kT instead.
     
  12. Sep 20, 2011 #11
    If the temperature of a gas having rotational and\or vibrational kinetic energies is exactly the same as the temperature of a gas having neither, then it appears to me that the thermal sensor is sensing and measuring only the translational kinetic energies.

    Let us take the standard statistical mechanical definition of gas temperature:
    T = mv^2/3k.​
    Here, T is the temperature in kelvins, m is the mass in kilograms, v is the square root of the mean of the squares of the velocities along the molecule's true paths in meters per second, and k is Boltzmann's Constant. Note that there is no allowance or accommodation for any internal energy. Such energies are irrelevant to the measurement of gas temperatures.
     
  13. Sep 20, 2011 #12

    Ken G

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    The temperature is the same because they are in thermal equilibrium with each other, it doesn't make any difference what types of degrees of freedom they have available to put energy into. I could have some sort of Ising spin configuration that can only put energy into spin-spin interactions, and those will come to the same T also if they are in good thermal contact with your system. Any two systems that increase their number of available states by the same factor when I add a given heat Q are at the same temperature, no other information is needed to say they are at the same T.
    That is only true in the special case of a nonrelativistic nondegenerate ideal gas moving in 3 dimensions. There are many other types of systems that admit to a perfectly normal statistical mechanical meaning for T that doesn't equal that. But more to the point, for the diatomic nearly ideal gas in the air around us right now, I can equally well write T = Iw^2/2k, where I is the moment of inertia and w^2 is the root-mean-squared angular velocity, and this expression is every bit as much the T of that gas as your expression is. Being able to write the expression does not imply that T is only sensitive to v^2 and nothing else.
    Not so, if I measure the root-mean-square w^2 of the air around me, and I know the moment of inertia, I have the T without any reference to translational motions. Many roads lead to T in thermodynamic equilibrium.
     
  14. Sep 21, 2011 #13
    So far I vote for Philip Wood and Ken G as having the best responses.

    My 2¢: In the general case, you have to distinguish between bulk KE & PE, and microscopic KE & PE. So if you have a container of gas in your lab and you raise it off the table and put in on a high shelf, you have increased the bulk PE but not altered the thermodynamic properties, which are only associated with the microscopic KE & PE. Usually the term "internal energy" refers to the total microscopic KE & PE. Raising the temperature of a system will increase the internal energy by an amount that depends on (or defines) the specific heat of the system, c = du/dT. In that sense, temperature reflects all the contributions to microscopic KE and PE.

    Many of the formulas that are being bandied about here are for ideal gases, which have zero potential energy. Keep in mind that solids definitely have a PE due to the lattice potential -- each atom in a crystal, for example, experiences a restoring force due to overlap from its neighbor. In essence this explains the http://en.wikipedia.org/wiki/Dulong%E2%80%93Petit_law" [Broken] (which says that crystalline solids have a molar specific heat equal to 3R, which has contributions of 1/2 for each of the three translational kinetic and three potential energy degrees of freedom). But fundamentally, thermal equilibrium means that temperature is uniform, not internal energy.
     
    Last edited by a moderator: May 5, 2017
  15. Sep 21, 2011 #14
    Ken,

    You are absolutely right in the entirety of your post #12.

    Virtually all of my physics work has been in the various fields of atmospheric physics. There, I ignore the internal molecular energies when it comes to thinking of atmospheric temperatures because some atmospheric molecules have them and some don't. That doesn't mean that they don't exist and are not important.

    I use them all the time in dealing with the specific heats of different air masses and in considerations of absorption and emission spectra for atmospheric gases.

    Mea culpa! I am going to have to learn to switch into the analytic mode when faced with disagreement instead of leaping immediately into the argumentative mode.
     
  16. Sep 21, 2011 #15

    Ken G

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    I think I see what you mean though-- you mean that the concept of temperature can be invoked even when all you want to do is analyze the translational motions, you don't need to consider anything else if your interest is the translations and the temperature stays fixed.
     
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