6
$\begingroup$

Trying to grasp Entropy not from the combinatorics point of view (I understand the micro-macro states thing pretty good now), but from the phenomenological point of view.

So Entropy is:

What entropy measures is how much energy is spread out in a process/T OR how spread out the initial energy of a system becomes in that system (at constant temperature).

Now that makes some sence and it almost sinks in my mind. But there is one small thing I can't comprehend. The units of entropy.

If we say that 10 Joules of energy got dispersed in 10 qubic meters of volume then the measure of such dispersion would be Energy/Volume.

But in the case of Entropy we have Energy/Temperature. Why is it so? I mean do we measure of the spread in the "temperature"? How many Joules got spreaded in each degree of T? Why the temperature? Why not the volume?

Is Temperature being treated here as a kind of a universal "volume"? I now that math checks out and stuff... I'm talking here about intuition though.

P.S. I feel that dividing by temperature here has something to do with heat capacity of the substance. The more the heat capacity the more there is "volume" for the energy to spread inside the substance...

$\endgroup$
  • 1
    $\begingroup$ This issue is thoroughly discussed in this other Physics Stack Exchange post. $\endgroup$ – DanielSank May 16 '17 at 14:28
  • 1
    $\begingroup$ To second @DanielSank 's comment, entropy may be defined to be dimensionless. There is nothing fundamental about its units. $\endgroup$ – Deep May 17 '17 at 4:38
2
$\begingroup$

Entropy change is related to reversibility. In layman's term it is related to the degree of disorder introduced to the system, relative to its initial disorder. Now recall that temperature is actually a measure of molecular (or particle) agitation. Greater the temperature, grater the agitation. On the other hand when energy is transferred as heat it means there are disordered works done at the molecular level. Since we are not able to keep track of all these molecular works we add them up all and call it heat. It is disordered in the sense that the molecules which received heat move randomly.

The unit of entropy is unit of heat by unit of temperature. It actually measures how much energy in form of disordered motion is delivered to the system (heat) taking into account how much disordered motion the system already has (temperature). A small quantity of the first (heat) into a system which already has a lot of disorder (high temperature) will not make much difference. Hence it shall represent a small entropy change.

$\endgroup$
  • $\begingroup$ "taking into account how much disordered motion the system already has" that's my question. If we start from T =0 kelvin then adding couple of random "things" we in result should create "couple of random things" but not INFINITY as in the case 10joule/0 kelvin. Just don't understand it. Yes there was no "disorder" in 0 kelvin but adding something should render to "adding something" and not blowing every thing up. Looks like this "disorder measure" (as you put it) is non additive or something... if so.. why it is not additive? Does it make no sence to have additivity here? $\endgroup$ – coobit May 16 '17 at 14:01
  • $\begingroup$ At the absolute zero you do not have any disorder. Hence adding any small disorder to the system is the same as infinitely increasing its disorder. $\endgroup$ – Diracology May 16 '17 at 14:10
  • $\begingroup$ First: Why we measure disorder as a relative thing (relative to previouse state)? Second: Why the base for this is Temperature and not space(volume) for example? Why it seems that temperature is a universal "space" here? $\endgroup$ – coobit May 16 '17 at 14:13
  • $\begingroup$ The point is that we don't create such a fundamental thing as entropy to use it as a representation of relative disorder. We just seek for an intuition about its given units. You can define some quantity which measure absolute disorder but that will not be entropy. $\endgroup$ – Diracology May 16 '17 at 14:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.