To Flow or Not to Flow
Some ice cubes are floating in a glass and the temperature of the ice and water are both equal to 0C.
Consider the system of particles in the ice cube and the system of particles in the water and call these systems A and B respectively. Now consider the abstract systems "particles in ice state" and "particles in water state" and call these A' and B' respectively.
To be clear, at the beginning A and A' coincide and so B and B'. However, if an ice particle becomes a water particle then they no longer coincide- A now contains both ice and water particles while A' consists of only particles in the ice state.
Will heat flow from B to A? How about from B' to A'?
Assumption: the environment is at 0C, conditions are such that water's freezing point is at 0C.
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1 solution
This is a rather subtle question so I will provide a detailed discussion for it: On the face of it we have the Zeroth Law of Thermodynamics which tells us that heat will flow between two systems brought into thermal contact until both systems are in thermal equilibrium. Notice the proviso- both systems must be in thermal contact only i.e. there cannot be any particle exchange between the two.
In this instance we have both the ice and the water at precisely 0C. If the ice were colder when the water reaches zero, then some water will freeze. If the water is warmer when the ice reaches zero, then some ice will melt. If they both reach zero at the same time then neither one is allowed to supply the extra energy needed to change the state of the other.
Heat exchange, just like temperature, is a macroscopic phenomenon. It is rather ill defined on the microscopic scale. To make this more concrete consider the following:
Suppose a bit of water gives enough heat to melt a bit of ice and becomes frozen itself. Has there been a heat exchange between A and B? Yes! We had some particles in B that became ice particles (still in B) and some particles in A that became water particles (still in A).
(This can never happen in this way- If there is such a heat exchange then some bit of water must be warmer on account of local temperature fluctuations- and cannot therefore lose enough heat to become ice-and there will correspondingly be some water elsewhere which will be slightly cooler and will in fact become ice to compensate for the ice that was melted.)
However has there been a heat exchange between A' and B'? No: there was no (net) thermal heat exchange and no (net) particle exchange: the bit of water that lost heat joined the ice system and the heat that went to the ice system melted some ice thereby restoring that heat to the water system.
There is an equilibrium in exchange of particles between A'and B' but not between A and B- that is where the subtlety lies. Equilibrium does not mean there is no local heat or particle exchange, it means there is no global heat or particle exchange.
This is by no means a basic problem, it brings to light many subtle considerations of thermodynamics. More to the point, it forces you to think more carefully about what happens, and how things are defined.
But what about the Zeroth law? We see that ice cubes do not remain as cubes, and in fact we end up with a slushy mixture of ice and water, because that state is favorable on account of it having higher entropy than the far less probable arrangement of ice cubes (no detailed calculation necessary- think how likely is it for a cup of water in the freezer to spontaneously produce ice cubes). We know that there must have been local heat exchanges since we end up with that slushy mess so that the zeroth law is not violated either.
But then why can't we just appeal to straight away to the zeroth law? Because in the case of A and B, there is more than just thermal contact- there is particle exchange whose equilibrium we cannot speak for: some or even all of the particles in the A may remain in the ice state or join the water state.
But in the case of A' and B', we know that we must have an equilibrium in particle exchange on account of the state of thermal equilibrium between the two.