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What does high temperature mean?

  1. Jan 11, 2012 #1
    Hello All,
    Suppose i have two identical boxes having sufficient oxygen ( these boxes contains same amount of coal, but one in one box coal is raw while in the other coal is burning ) when i use thermometer, the thermometers reads different readings one is high other is low. Let us suppose that i didn't know what was in the two boxes , what conclusion should i draw by seeing their temperature to be different? how do the two systems different on microscopic level. I mean what has happened two the hotter system that could explain its temperature being higher, note that both the system has same internal energy.
  2. jcsd
  3. Jan 11, 2012 #2


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    Coal and oxygen have potential energy locked up in their molecular bonds. The coal is a solid and so it sits inert on the floor of the box. Low movement, low temperature. By starting a fire, you provide enough energy to start some of the coal molecules combining with the oxygen molecules. The byproduct of the oxidation is carbon dioxide and water. But - the bonds of CO2 and water have lower energy than the initial ingredients, and this excess energy is liberated. It manifests as kinetic energy of the molecules - i.e. they are now in motion, and they bounce around the box, ricocheting off other molecules, making a chain reaction, as well as off the walls (and the thermometer). This is what you are measuring when you measure the temperature.
  4. Jan 11, 2012 #3
    Thanks DaveC426913 for your support.As far as it seems you have no where in your answers written anything wrong about how the temperature has risen. But the gravity of my question is more towards the the state of hotness than the cause of hotness. Let me take a different example i have a block of iron at 80 degree C and a glass of water(liquid) at 30 degree C. I am sure the kinetic energy of water molecules will be higher though the its temperature is found low. So what actually is happening to a body which is at higher temperature.
  5. Jan 11, 2012 #4
    Are you? You're probably thinking of translational kinetic energy, and then you're right. But there is also rotational and more importantly vibrational kinetic energy, and this can easily be higher for a solid, and will also be the case in your example.
  6. Jan 11, 2012 #5


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    HUH ??? You SPECIFICALLY asked for explanation of WHY the temperature goes up by asking

    and Dave gave you exactly that.
  7. Jan 11, 2012 #6
    hmmm is this the thing? so does it mean not only the translational kinetic energy contribute to temperature but other types of kinetic energy(vibrational and rotational) too contribute to temperature? for an ideal gas they how translational kinetic energy depends on temperature(or we may interpret the reverse) they show translational K.E is directly proportional to absolute temperature. what about other types of K.Es?
  8. Jan 11, 2012 #7


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    I may have it wrong but it has been my impression that temperature is just a measure of how much activity there is among atoms and molecules. It has NOTHING to do with how that activity was created.
  9. Jan 11, 2012 #8
    Why do you think the kinetic energy of the water molecules will be higher than the kinetic energy of the iron atoms?

  10. Jan 11, 2012 #9
    Responding phinds, as i already emphasized " Let us suppose that i didn't know what was in the two boxes , what conclusion should i draw by seeing their temperature to be different? how do the two systems different on microscopic level." i don't know what is there in the two systems no information of oxygen or coal. But yes i had chosen a chemistry type example and that response of dave was obvious. Asking straight forward i want know what are the behaviors of sub particles of matter with changing temperature that could explain the meaning of high temperature. Regards
  11. Jan 12, 2012 #10
    Well the ideal gas is a special case: you assume that translational kinetic energy is the only kinetic energy, but in more general gases, the temperature is also "due" to vibrational and rotational kinetic energy. In general, one has per particle and per degree of freedom an energy of [itex]\frac{k_BT}{2}[/itex]. (You can look this is up as the equipartition principle; it's valid in classical mechanics; it is modified in quantum mechanics, but this is inessential to your question so let's ignore that for the moment being.)

    For example, in an ideal gas, you have per particle 3 degrees of freedom, the three directions of translation: translation in x direction, in y direction and in z direction. Hence, according to the above equipartition principle, every particle has energy [itex]3 \times \frac{k_B T}{2}[/itex] and hence the total energy of an ideal gas is [itex]\frac{3}{2} N k_B T[/itex].

    But take your hot solid: there the particles cannot translate (fixed in a lattice), but they can vibrate, and this again in three directions (x,y,z), hence per particle you get, again, an energy of [itex]3 \times \frac{k_B T}{2}[/itex]. I'm not familiar enough with solids to know if they only vibrate, but at least they have this contribution.

    Doesn't this answer your question? If the temperature is different, then this means that per particle and per degree of freedom, the energy is different.
  12. Jan 12, 2012 #11
    ok i understand what - what contribute, here is one more thing i want to clearfy. does raising temperature raises all types of kinetic energies ( a system posses) in the same way. like doubling the absolute temperature doubles each kind of K.Es? or it depend on a particular system and the response(to temperature) of different K.Es is some what complex? Thanks a ton.
  13. Jan 12, 2012 #12
    Again, the equipartition theorem solves our question :) Every particle gets degree of freedom [itex]\frac{k_B T}{2}[/itex], so every degree of freedom will raise an equal amount of energy when you up the temperature.

    The story gets more complex with quantum mechanics: at low temperatures, certain degrees of freedom freeze out, e.g. below a certain temperature a particle in a solid cannot vibrate anymore.
  14. Jan 12, 2012 #13
    dear vodka, ignoring quantum effects, just as in ideal gas we quantify translational kinetic energy average per atom)3/2kt can we say all K.Es to have same values at the same temperature. and doubling temperature will double the other K.Es? Regards
    Last edited: Jan 12, 2012
  15. Jan 12, 2012 #14
    It depends on the number of degrees of the "kinetic energy". Translational kinetic energy has 3 degrees, rotational has 2 degrees, vibrational has 3 degrees. Every degree gets the same amount of energy gain at a certain temperature rise, so you see translational and vibrational gain more energy than rotational kinetic energy for a certain temperature rise.

    Does this answer your question?
    Last edited: Jan 12, 2012
  16. Jan 12, 2012 #15
    yes this perfectly answers. i got it thanks a lot:)
  17. Jan 12, 2012 #16
    your answer has now create one another question in my mind. May i ask it?
  18. Jan 12, 2012 #17
  19. Jan 12, 2012 #18
    does it happen that for a given system with the change in temperature some of its ,degrees of freedoom goes blocked or may increases that could also be an explanation of why the specific heats of substance varies with temperatre?
  20. Jan 12, 2012 #19
    Good question: two things.

    1) You are correct: the freezing out or appearing of degrees of freedom will change the specific heats of substances and this discovery was very important (beginning 20th century) to understand the behaviour of specific heats at low temperatures and I think this is the reason that [itex]C \to 0[/itex] as [itex]T \to 0[/itex] (all degrees of freedom freeze out).

    2) But specific heats can also be temperature dependent for high temperatures, where quantum effects are negligible. Why is this? Well, remember: temperature is a measure of kinetic energy (translational, vibrational, ...) but heat capacity is the change of total energy when you change temperature, not just kinetic energy. And you can imagine that when you heat a substance, the molecules in the solid (for example) start moving so heavily that the structure itself changes (the molecules get new positions; this is extremely evident when you melt something) and this can change the potential energy. The heat capacity also counts changes in potential energy.
  21. Jan 12, 2012 #20
    sorry now my office timings are now over and i have no other access of internet so i will like to keep this matter for tommarow.
  22. Jan 13, 2012 #21
    oh i think i understand the matter, when temperature increases the available molecular motions speed up i.e they increases their energy from the heat(energy to be correct) supplied to them externally for each degree of freedom the rise in K.Es per unit rise in temperature is 1/2k. And as u explained some of the degrees of freedom freezes out while some may appear besides adding energy(generally in form of heat) also utilities in changing the potential energy and this explains the changing behavior of specific heats( ignoring quantum mechanical effects). an increases in temperature increases change all the types of K.Es in the same way but as such all k.Es do not posses the same degrees of freedom therefore translational K.E may change by 3/2k(when dof=3) while rotational K.E may be change by k(when dof=2) and Translational K.Es may change by 3/2k(when dof=3) when the change in temperature is 1K.and thus overall change in all types of K.Es are generally not the same. do i correctly interpret?
  23. Jan 13, 2012 #22


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    It always seems totally backward to me when folks say that a rise in temperature CAUSES an increase in molecular activity, when in reality, temperature is a MEASURE of the molecular activity, so the molecular activity has to go up to cause the temperature to go up, not the other way around.
  24. Jan 13, 2012 #23
    The energy kT/2 is for a quadratic degree of freedom (as many dofs are). Also, keep in mind that the Boltzmann factor k is really just a unit conversion factor. Once upon a time physicists carried around a factor J in their equations to convert BTUs (for heat transfer) to foot-pounds (for mechanical energy). Eventually everyone realized it was pointless to have different units for heat and work, and especially silly to carry around a unit conversion factor in their equations, and J disappeared from textbooks and papers. It may take longer for k to disappear (and it might never) because the colloquial units of temperature are so darn convenient. I've found over the years that many people are confused by k and that it almost always obscures the real physics.
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