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Average kinetic energy of the molecules in a cold liquid less?

  1. Apr 12, 2015 #1
    Why is the average kinetic energy of the molecules a cold liquid less?

    As the temperature of a liquid decreases, the average kinetic energy of its molecules reduce. What is the reason behind this?
     
  2. jcsd
  3. Apr 12, 2015 #2
    Less compared to what?
    And temperature is defined as the average kinetic energy of the particles in a substance. A decrease in temperature is therefore synonymous with a decrease in average kinetic energy.
     
  4. Apr 12, 2015 #3

    DrClaude

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    Really?
     
  5. Apr 12, 2015 #4

    OmCheeto

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    It sounds like you are asking why someone bothered to attempt to define temperature:

    People try to define things all the time. "Why" temperature is defined this way, is probably based on other previous definitions.
    Of course, all definitions are limited, in their scope.
     
  6. Apr 12, 2015 #5
    Yes. Are you joking or are you serious?
     
  7. Apr 13, 2015 #6

    DrClaude

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    I'm serious. I've never seen temperature defined as the average kinetic energy. And think about it: does it make sense for a solid? What about spin temperature? Yes, the average kinetic energy does increase with temperature, but that's a consequence, not a cause.

    Historically, temperature was defined basically as "what a thermometer measures," which is a bit of a circular definition. In modern thermodynamics, temperature is defined through the concept of thermal equilibrium (two objects that can exchange energy are at the same temperature if there is no net flow of energy from one to the other) and quantified by the relation
    $$
    \frac{1}{T} = \frac{dS}{dU}
    $$
     
  8. Apr 13, 2015 #7
    Most introductory physics textbooks have this explanation, right?
    (I didn't set that in bold)
    I'm not an expert in thermodynamics, but isn't it reasonable to loosely define kinetic temperature for a liquid in this manner? I doubt a very formal treatment of entropy is required here o_O
     
  9. Apr 13, 2015 #8

    DrClaude

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    Nope. Just to be sure, I just checked in all those I have on my shelf, and none state that, even the ones in physical chemistry, which tend to be less rigorous :wink:

    That is plainly and simply wrong.

    The way to do it without invoking entropy is to state that temperature is what a thermometer measures, and discuss thermal equilibrium and the 0th law of thermodynamics. You could do all of classical thermodynamics without even invoking atoms or molecules, so the motion of atoms or molecules is not necessary for the concept of temperature. Of course, in modern times, one should introduce early on the concept that an increase in energy leads to an increase in the internal motion of the constituents of matter, but it is wrong to state that this is how one defines temperature.

    Let me take another example to show the problem with this: at the triple point of a substance, you have a solid, a liquid, and a gas phase coexisting at the same temperature. If temperature is defined as the average kinetic energy, is one to conclude that the average kinetic energy is the solid, the liquid, and the gas is the same?
     
  10. Apr 13, 2015 #9

    Vinay080

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    In everyday life, hot or high temperature matter, is defined to have high average kinetic energy (as most of them state here), rather than simply high energy. Why is it defined to have high average kinetic energy rather than high energy?

    Whatever we touch and feel hot, might have high energy but less kinetic energy, it might even continue to have same kinetic energy when we feel it to be cool (here other form of energy might have decreased rather than kinetic energy). What proof do we have to show that temperature or heat is the function of just kinetic energy rather than energy?

    OP seems to mean the same thing.
     
    Last edited: Apr 13, 2015
  11. Apr 13, 2015 #10
    Are you telling me to compare the kinetic and potential energies of ice, water and water vapor?
     
  12. Apr 13, 2015 #11

    DrClaude

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    Not even. When you have a phase transition, there is no relation between energy and temperature, because temperature is constant while the energy of the system increases or decreases depending on the direction of the phase transition.
     
  13. Apr 13, 2015 #12

    DrClaude

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    Once more, temperature is never defined as the average kinetic energy. In everyday life, temperature is what a thermometer measures.

    The simple answer is that it's not. Only for a classical ideal gas is temperature strictly related to kinetic energy.
     
  14. Apr 13, 2015 #13

    DrClaude

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    With all this, I forgot to answer the OP :oops:

    As I mentioned above, temperature is a measure of the tendency of an system to give away or receive energy from another system. When two object of different temperature are in (thermal) contact, the overall exchange in energy will be from the hot object to the cold one, until equilibrium is reached, meaning that there is no overall flow of energy, and the objects are said to be at the same temperature. That energy going to the cold object ends up in all possible ways that object can have energy, including kinetic energy. There is also the equipartition theorem that states how that energy is distributed among the possible ways of storing energy (called degrees of freedom).
     
  15. Apr 13, 2015 #14
    If I were to make a guess, all that I would be able to say is that in a phase transition, there is no transfer of heat among the particles (they are all at the same "temperature"). I can't make much progress on the idea except perhaps say that temperature is dependent upon a property that all particles share during phase transitions. It's almost as if both my hands are tied behind my back if I were to define temperature without using translational kinetic energy.
     
  16. Apr 13, 2015 #15

    DrClaude

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    Imagine you have some water at 0 °C, kept in an isolated container. That water is 3/4 ice and 1/4 liquid water. What happens if you leave the system unperturbed? What happens if you furnish a small amount of heat to the system?

    I don't understand why you are all so intent on defining temperature like that. Yes, there is a relation between temperature and kinetic energy, but there is no way you can build a useful definition of temperature by looking only at kinetic energy.

    You are also missing out on some extremely cool (pun intended) physics, such as adiabatic demagnetization (or magnetic refrigeration) and negative temperatures.
     
  17. Apr 13, 2015 #16
    If no external heat energy is supplied and no work is done on the system, then the proportion of ice and liquid water should remain constant, as an equilibrium between their concentrations is established at 0°C. If heat is added to the system, the potential and kinetic energy of the ##H_2O## particles must increase, so a greater proportion of ice will be converted to liquid (latent heat of fusion). The increase in kinetic energy will obviously not exactly correspond with the amount of thermal energy supplied as the potential energy of the particles increases as well. Nonetheless the molecular energy distribution graph for the particles will shift so that the mean molecular energy increases. (Unless I forgot basic principles of thermodynamics!)
    Where do we go from here?
     
  18. Apr 13, 2015 #17

    DrClaude

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    That's the point I was getting at. You clearly have an increase in kinetic energy, while temperature hasn't changed. Therefore, you definitely can't define temperature as average kinetic energy.
     
  19. Apr 13, 2015 #18
    If I'm not misunderstanding this, then there seems to be a problem with this definition. It means that you fundamentally cannot define the temperature of a system without referring to what is outside the system (the temperature of a completely isolated system is not undefined, but under this definition, it will be). However, there is no such restriction when we describe temperature practically. Temperature is an absolute quantity (at least classically).
     
    Last edited: Apr 13, 2015
  20. Apr 13, 2015 #19

    DrClaude

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    For a stricter, stand alone definition, you need to go further and use the equation I posted in #6.

    It won't be undefined as you can look at what would happen if the system were to be put into contact with another.
     
  21. Apr 13, 2015 #20
    Isn't it circular to define temperature using entropy, when calculating entropy in the first place requires the temperature to be known?
    Sorry but it's taking me some time to wrap my head around this :rolleyes:
     
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