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This is my first time posting here, so bear with me. It was stated (in this video: https://youtu.be/7jT5rbE69ho?t=325) (Watch about the next fifteen seconds, until 5:40) that an increase in entropy leads to a decrease in heat (or temperature or absorbs energy - unsure of the correct terminology). (If I'm understanding correctly.) Now, I've never taken a thermodynamics class or anything past high school chemistry, but this is a little confusing to me and I'm hoping to get an explanation. What follows below is my best attempt at understanding why. It is most likely mis-guided, so feel free to correct at will.

  1. I understand that entropy can be partially understood as disorder. When the Helium-3 enters the Helium-4, that constitutes an increase in entropy.

  2. The change in entropy is defined as Heat/Temperature (see below)

    Rudolf Clausius invented the idea of entropy in such a way that the change in entropy is the ratio of the heat exchanged in any process and the absolute temperature at which that heat is exchanged. That is, he defined the change in entropy DS of an object which either absorbs or gives off heat Q at some temperature T as simply the ratio Q/T.

  3. I then figured that maybe the Q-T ratio (entropy) increased by the total energy (of the helium system) increasing, requiring it to pull heat from somewhere

Miscellaneous thought: My above reasoning confused me because I thought it required energy to decrease entropy locally (add order to a system), so it seems strange that a system increasing in entropy (like my helium example) would absorb energy.

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I consider watching any video a waste of time, so I'll be judging from your words alone. (Anyway, your question is essentially self-consistent, which is good.)

  1. Yes, entropy is a measure of disorder (sort of).
  2. No, $dS={\delta Q\over T}$ only for reversible processes.
  3. And no, entropy is not the Q-T ratio for two reasons. First, because the above formula deals with infinitesimal increments and not with finite quantities. Second, because like I just said, that formula is not universally applicable.

All in all, the increase of entropy alone is not necessarily accompanied by the heat flow to the system. It may be so, but it may be otherwise. You may imagine pretty much any kind of possible outcome:

  • You dissolve some $\ce{NH4NO3}$ in water (which surely increases entropy), and the beaker gets palpably cold. The system starts sucking heat from the surroundings, just like you anticipated.
  • You mix your $\rm^3He$ with $\rm^4He$, and... nothing happens (except the increase in entropy, of course, but that's a thing you can't see or touch). There are no thermal effects to speak about, and hence no heat flow.
  • You burn a candle in a closed vessel (supposedly, containing enough air), and guess what? Entropy increased, but heat is flowing outside.

As for your last remark: true, it takes some energy to decrease entropy locally (say, in your fridge). But in doing so, we inevitably increase entropy elsewhere, so the overall balance is positive.

So it goes.

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    $\begingroup$ Good answer. Just to avoid confusion, the second point in the latter list is approximately true. A small heat of dilutation (endothermic) exists, and has been utilised since the 1960s. doi: 10.1007/978-1-4899-6443-4_133 $\endgroup$ – Linear Christmas Dec 28 '17 at 21:44
  • $\begingroup$ Very interesting. I appreciate the quick response. I am giving the video the benefit of the doubt, just because I'm certainly no expert on it. So, why is it that the increase in entropy causes a drop in temperature? Again, thanks - very helpful $\endgroup$ – Zara Dec 28 '17 at 23:55
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    $\begingroup$ Increase in entropy does not necessarily cause a drop in temperature (see my examples 2, 3). In the cases when it seemingly does (like example 1), this is due to some endothermic reaction, which is made spontaneous by the positive entropy gain ($\Delta H>0$, but $\Delta G=\Delta H-T\Delta S<0$). $\endgroup$ – Ivan Neretin Dec 29 '17 at 4:29

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