Help, I am terribly confused about entropy. On the one hand, I am taught at school that a substance such as an ice/solid has a lower entropy than its gaseous equivalent and that a process such as solidification reduces entropy, which is justified with the explanatory premise that the solid is more "ordered".

  1. All this seems dogmatic to me, and it somewhat conflicts with my own intuition (probably means I am wrong). For one, if a change of entropy is measured with the equation $\Delta S = q/T$, (btw, can anyone justify this equation / show its derivation?) take solidification - the heat content ($q$) falls so it's exothermic, hence $q$ is a knowable negative value (the enthalpy of fusion, if I am not wrong?) and $T$ is the critical temperature at the given atmospheric pressure - however, how about the entropy change when the temperature changes too - what if the change in the heat content results in a change in temperature - how is that measured?
  2. Another idea that conflicts with my mind is why solids/colder things are said to have lower entropies at all? I understand from a statistical point of view why overall entropy increases, as things progress to chaotic, and this entails that the heat death of the universe will see maximum entropy - or another way to express it: no work will be attainable from the system/universe. However, I am told that colder things have LOWER entropies? I can only intuitively understand entropy from a complete picture - where hot and cold mix to increase overall entropy - how can a part - like the cold area by itself (given it is uniformly cold so that its own potential energy, by itself, is null) even be given an entropy value? The same goes for the hot part - it only makes sense to be if we consider the entropy of the system as a whole when hot and cold are separate or mixed - but obviously, there must be something I don't get.
  3. Could it be that I am confusing different kinds of entropies - btw (final point) if (in Chemistry at least) entropy is defined by change in heat content/temperature (in Kelvins), then how could overall entropy ever decrease as the loss of entropy over there (via loss of heat content) would increase the entropy here (which compensates) and so on. This definition (for me) entails that entropy simply is transformed/passed on like energy. This clearly even conflicts with my own intuitive view of entropy as more of a measure of order, and hence attainable work from a system.

You can see, I have many questions and if you know the answers but can't be bothered, do not hesitate to answer only a fraction of it all, I am grateful for anything. Thank you very much.

  • $\begingroup$ Another idea that conflicts with my mind is why are solids/colder things said to have lower entropies at all? As you mentioned, entropy is a measure of disorderness. Greater the disorderness, greater will be the value of entropy. In case of cooler solid, disorderness will be much lesser than the hotter solids. So, entropy value for cooler solids will be less than that of hotter solids. $\endgroup$
    – Sensebe
    Dec 8, 2013 at 19:03
  • $\begingroup$ Note too that $\frac{\delta Q}{T}$ is exact, therefore the entropy does not depend on the path that you take, even if the process is irreversible. However the entropy of the neighborhood of the system WILL change if the process is irreversible (note too that the entropy of the system will NOT change, but the entropy of the universe will increase). $\endgroup$
    – user40276
    Dec 8, 2013 at 19:37
  • $\begingroup$ I think that in 3 you're confusing entropy with enthalpy. The enthalpy $H=U+PV$, hence $dH=dU+dPV+PdV = \delta Q$ when $dP = 0$. $\endgroup$
    – user40276
    Dec 8, 2013 at 19:39
  • 2
    $\begingroup$ First I started to write a comprehensive answer, but then I thought that most of your ambiguities would resolve by reading the corresponding Wikipedia article on entropy. Have you tried that? $\endgroup$
    – Mostafa
    Dec 8, 2013 at 19:53
  • $\begingroup$ Thanks anyhow, I read some of it, maybe I need to study it. $\endgroup$ Dec 8, 2013 at 20:07

1 Answer 1


It seems you're coming at entropy from a thermodynamics standpoint. This is completely consistent with (and, at the macro scale, equivalent to) the statistical derivation of entropy, but you might find the statistical version more intuitive, if the thermodynamic version is causing you issues.

I warn you, statistical physics is both math-heavy and takes some time to get your head around. But, if you have enough mathematical grounding, the following non-mathematical sentence might help: Increases in entropy correspond to larger phase spaces, which is the same as saying there are more possible ways to arrange the components of the system to give an equivalent macroscopic system. Equivalent here means the same temperature, pressure, and other thermodynamic properties.

When macroscopic things are colder, the system has less total energy available to be spread around the microscopic components. So, there are fewer ways to arrange the system. Think about electrons in energy levels of identical atoms, as an example. If you have a lot of atoms but very little energy, almost every electron has to be in its ground state. If you add some energy, then more electrons can go into excited states. But if one electron could be in an excited state, so could any of the others (though not all at the same time). Since each atom in this example is identical, all those different possible configurations of excited electrons are equally possible (and all give the same macroscopic description of the system), so the system has more entropy. You can imagine how quickly entropy will grow if you go from absolute zero (everything in the ground state - one configuration) to just enough energy to excite one electron (N configurations, where N is the number of atoms), to enough for two excited electrons, and so on.

  • $\begingroup$ Ahh - so for an event like the heat death, the temperature may be lower - but the phase spaces (due to lack of pressure) will be much larger - hence, nevertheless, entropy is high. $\endgroup$ Dec 9, 2013 at 19:42

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