my teacher said that a broken glass can't fix itself spontaneous because of the second law of thermodynamics, entropy won't decrease in a isolated system. But when the temperature is at 0k, then entropy becomes zero. So if the broken glass can return to its original state, then this doesn't violate any laws, right? Does it means this is possible?
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1$\begingroup$ Your teacher is referring to entropy of kinetic energy at 0 K. If you pick up a tea cup at 0 K and smash it to the ground, it will most definitely break. $\endgroup$– NeilCommented Mar 22, 2016 at 8:15
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$\begingroup$ Is it possible? No, it isn't. The glass is not a closed system, not even at low temperatures. 0K is not a physical temperature, to begin with. $\endgroup$– CuriousOneCommented Mar 22, 2016 at 8:24
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$\begingroup$ @CuriousOne: I can put the glass in a box, which is a closed system. $\endgroup$– alstCommented Mar 22, 2016 at 8:28
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$\begingroup$ @Neil: what's the meaning of entropy of kinetic energy? Do you mean that the total entropy is not zero at 0k? $\endgroup$– alstCommented Mar 22, 2016 at 8:29
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$\begingroup$ @alst Kinetic energy refers to movement of an object in general, but more specifically in this case I'm referring to the internal jittering of the atoms that is directly associated with temperature. At 0 K, the atoms don't jitter. This creates some strange properties at that temperature, but you'd certainly never see entropy reversed on that object. $\endgroup$– NeilCommented Mar 22, 2016 at 8:29
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1 Answer
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Zero Kelvin is not attainable by a physical system.
Why is absolute zero (0 kelvin or −273.15°C) an impossible goal?
Practically, the work needed to remove heat from a gas increases the colder you get, and an infinite amount of work would be needed to cool something to absolute zero. In quantum terms, you can blame Heisenberg’s uncertainty principle, which says the more precisely we know a particle’s speed, the less we know about its position, and vice versa. If you know your atoms are inside your experiment, there must be some uncertainty in their momentum keeping them above absolute zero – unless your experiment is the size of the whole universe.
Let us not quibble about this, let us suppose a crystal very close to absolute zero in a thermally isolated box with. It has its lowest possible entropy.
To break the lattice energy must be supplied. Suppose an equally cold stone detaches itself from the roof (gravity supplying the energy) and falls on the crystal. The system is now broken crystal, fallen stone and the kinetic energy has increased or statistically the number of microstates is orders of magnitude than it was at t=0, i.e. the entropy has been increased.
There is no way the crystal pieces can be connected again without energy being supplied, which will introduce more microstates. To get back to the original entropy of the crystal energy has to be supplied outside the system, to cool it.
So the broken pieces can be back together again in an open system, where one has to count the total and entropy will be increasing.
