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I've leant that entropy is a state of randomness, and that solids have a more structured form, therefore having less entropy.

However, I saw a YouTube comment stating the following:

a liquid NOT ALWAYS means higher entropy than a solid it depends...of the context for example, in the south pole, ice means higher entropy, because Mother Nature sets the equilibrium for liquid water to become ice.

Is there justification for this statement? Is it true that even in a more ordered substance like the ice, there is more entropy?

DarkLightA
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    "in the south pole, ice means higher entropy,.." Its true in following sense:- At south pole temperature of surroundings is below zero degrees i.e. <273K. Now consider ice and water in equilibrium at 273K. If water can lose some of its heat to the surrounding to form ice, then the entropy of total system ice+water+surrounding will increase. So the exact statement should be : "in south pole formation of ice is entropically favored". – user10001 Jan 30 '13 at 19:39
  • In the South Pole or ANY other place as well, where ice is melting into water... There is nothing special about the South Pole, nothing more than in my glass of Coca-Cola with ice. – Eduardo Guerras Valera Jan 31 '13 at 10:12

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Let's consider the following situation. Suppose we have an ice block of mass $m$ sitting at $T=0^\circ\,\mathrm C$ in a container. To melt the ice, we need to heat it up, and the exact amount of heat we need is the so-called "latent heat of fusion" of the ice, and is given by $$ Q=mL $$ where $L$ is called the specific latent heat and is specific to the melting substance. The change in entropy of the system during the phase change is, in this case, given by the heat absorbed by the ice divided by its temperature (note here that temperature should be written in Kelvin for the following to be valid which is why we're not dividing by zero) $$ \Delta S = \frac{Q}{T}=\frac{mL}{T} $$ which is positive. This shows that the entropy of an amount of ice at $0^\circ\,\mathrm C$ is less than the entropy of the same amount (mass) of water at $0^\circ\,\mathrm C$.

I'm not sure what the YouTube comment is referring to.

For more info, see this and this.

joshphysics
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    The youtube commenter is probably equating "entropy always increases" with "since the water froze (i.e. went from solid to liquid) it's entropy must have gone up" with considering that the water was exchanging heat with something else while that happened. Not an uncommon misunderstanding. – dmckee --- ex-moderator kitten Jan 30 '13 at 19:06
  • Alright, makes sense! I suppose then that in an isolated system with water and ice, the liquid loses entropy to heat the ice to equilibrium, but the ice gains more entropy than the liquid lost, thereby increasing the total entropy? Also, when it's not about melting, what is the value of Q and how is it found? – DarkLightA Jan 30 '13 at 19:09
  • @dmckee Oh that's interesting; I've never encountered that particular misunderstanding, but I can definitely see how that could be confusing to someone learning about entropy for the first time and hearing the phrase "entropy always increases." – joshphysics Jan 30 '13 at 19:09
  • I truly appreciate you taking the time to help. Also.. (Sorry, just trying to understand this) would a cube ice spontaneously forming in an above-0 degree system of water at equilibrium decrease the total entropy? – DarkLightA Jan 30 '13 at 19:13
  • Hang on, is it maybe not possible to report entropy like that? Is it only a measure of total "randomness" of a system? Or can you say that the ice has a lower entropy than the water? – DarkLightA Jan 30 '13 at 19:30
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The youtube commenter was wrong. The spontaneity of reaction (including phase change from ice to water) is not just determined by entropy otherwise petrol would spontaneously turn into water and carbon dioxide without the need to burn

To determine if a reaction can happen spontaneously the Gibbs free energy ($\Delta$G) needs to be $<0$ which is determined by the change in enthalpy ($\Delta$H) as well as entropy ($\Delta$S).

$\Delta$G = $\Delta$H - T$\Delta$S

Although the entropy of liquid water is ALWAYS higher than ice the melting of ice is an endothermic reaction (i.e. it requires energy input). This results in a positive enthalpy ($\Delta$H) so in order for the ice to melt the entropy ($\Delta$S) and the temperature (T) have to be big enough to overcome the positive enthapy change and make $\Delta$G < 0. It's probably obvious at this point that T needs to be >0 deg C (273.2 K) in order for this to happen

Chris Kirk
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we know that every spontaneous process will in the direction of increasing entropy . At poles water spontaneously converted into ice . so total entropy (entropy (ice+water+surrounding)) will increase

Yasar Np
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