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Entropy question?

  1. Jun 15, 2011 #1
    Does an ice cube have more entropy than room temp water? I would intuitively think it doesn't because its atoms are more organized in a lattice vibrating like a solid. Conversely water would be more spread out and disorganized. However, the equation

    s = q / t

    states that the higher the temperature something has, the lower the entropy. So an ice cube would be colder than water making it have more entropy.
     
  2. jcsd
  3. Jun 15, 2011 #2
    i think entropy us the measure of disorder an thus it increases when increasing temp and so also in this one the water has greater entropy than ice.
     
  4. Jun 15, 2011 #3

    Drakkith

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    S is the CHANGE in entropy based on amount of heat added (Q) over the temp.
     
  5. Jun 15, 2011 #4

    Mapes

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    Where did you see this equation? It isn't quite correct. Instead, [itex]\Delta S=Q/T[/itex] holds for reversible systems at constant temperature; that is, the addition of thermal energy Q causes a change in entropy [itex]\Delta S[/itex] for a system at temperature T.

    The entropy of a system increases with temperature: [itex]dS/dT=C/T[/itex], where C is the heat capacity (which is positive for typical materials).
     
  6. Jun 15, 2011 #5
    How can the temperature stay constant if your adding it taking heat away? Also is T the temperaute of the water you drop the ice cube in or the ice cube itself
     
  7. Jun 15, 2011 #6

    Drakkith

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    The temperature isn't staying constant. It is simply the temp that the object was initially at before you added heat.

    I believe it is both. You would use a seperate equation for each one. In one you would have a Q and in the other you would have -Q. But I don't know for sure, that's just a guess.
     
  8. Jun 15, 2011 #7

    jambaugh

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    To understand entropy you must understand that when we speak of "a system" with specific macroscopic state variables we really mean a class of systems. So the "system" in this case is somewhat of an abstraction. The entropy then is a measure of how broad one's definition of "the system" is. Entropy is the negative logarithm of the number (or measure in a continuum case) of --in principle-- empirically distinct physical systems satisfy the conditions for this class we call "the system".

    This is why entropy is both "a measurement of our knowledge" and yet physically meaningful and hence calculable for laboratory systems.

    So start with "the system" being the class of n water molecules contained within a fixed volume and having a fixed total energy. Pick a distribution for the velocities of each molecule. You can then count actual distinct cases this system class may represent and this gives you the entropy. Change the distribution and you get a different entropy. Observe that there is a distribution yielding maximum entropy (because it has the most distinct cases) and it will also be the most probable given no other information.

    See how the entropy changes as one varies the total energy and you get the (reciprocal) temperature. Note that you can possibly have a negative temperature if e.g. increasing energy actually decreases entropy. There is also the in-between case where increasing energy increases entropy in a constant fashion, thus both go up while temperature is constant. This occurs e.g. during a phase change like when ice is melting.

    dE = TdS

    Now the equation you wrote: s = q/t when applicable would mean that for fixed energy q, as the entropy increases the temperature decreases, or that for fixed entropy as energy increases so does temperature, or for fixed temperature entropy increases with energy.

    Your intuition about the water is correct since the (constant T) phase change as ice melts (since also T is positive) with dQ = TdS and since we observe the need to increase energy to melt the ice, we then know melting increases entropy as well.
     
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