entropy of system + environment is increasing
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Replying to @RyanDavidReece
System = the whole thing. Like the contents of your freezer. Not specifically the H2O molecules. The surrounding air + water looks disorganized. Then suddenly, order. Yet entropy has increased. That's the point.
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Replying to @fchollet
then your OP is less mysterious, because the disorder of the total system has increased, it's just not localized in flakes to which we attend.
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Replying to @RyanDavidReece
No, you're still missing the point. The *apparent* disorder has decreased. The freezer with ice crystals in it looks more organized than when it had liquid water. That's the entire point: apparent disorder as perceived by humans is not the same as thermodynamic entropy.
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Replying to @fchollet @RyanDavidReece
There are some systems where a crystalline state actually has higher entropy than the corresponding liquid state! For instance, consider a packing of chopsticks or other hard rods (a crude model for the nematic liquid crystals in LCDs). Compare the "crystalline" state where ...
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the rods are aligned to the "liquid" state where the rods are pointing in random directions. At sufficiently high fixed density, the *aligned* rods are entropically preferred because they have much more space to move around than the randomly-oriented rods.
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That is, the difference in positional entropy more than outweighs the slight decrease in orientational entropy. This phenomenon is called "order by disorder"; see e.g. this survey by Daan Frenkel for many other examples https://sci-hub.tw/10.1016/S0378-4371(98)00501-9 …
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Replying to @blockspins @fchollet
interesting. as i understand, you are not arguing that this is the case for a snowflake.
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Replying to @RyanDavidReece @fchollet
That's right, the entropy of ice is less than the entropy of liquid water (by about 1220 J/(K * kg) at 0ºC and 1 atm).
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When studying phase transitions, it's usually more convenient to consider closed systems that can exchange energy with their environment (as opposed to isolated systems). Under those conditions, the 2nd law corresponds to the decrease of the ...
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system's Helmholtz free energy F = E - TS, where E is the internal energy, T the temperature and S the entropy. In equilibrium, ∆F=0, so increases of entropy at constant temperature have to be compensated for by corresponding increases of internal energy.
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Since heat is *released* when water freezes, ∆E < 0 and so ∆S < 0 as well. The specific heat of fusion of water is roughly 3.30e5 J/kg – divide this by 273K to get the entropy difference.
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