Ice and liquid water interacting across a boundary Imagine we have two thermodynamic systems, one a mass of ice and the other an equal mass of liquid water, with both at 273.16K. Each system is isolated, except that they can interact with each other across a boundary that permits the exchange of heat but not matter or work.
What will the two systems look like at equilibrium? Somehow I want to automatically imagine that each system will be identical, a combination of liquid water, ice, and water vapor at 273.16K. But if this is true then the two systems were initially at the same temperature but not in thermodynamic equilibrium, an apparent violation of the zeroth law of thermodynamics. 
 A: You have not specified how the pressure is controlled in the two systems.  If they are each at the triple point pressure of 611.73 Pa there is no reason for heat to exchange and all will stay constant.  If the pressures are different from this (and not on the freezing curve) energy can be released if there is heat flow by transferring heat between the reservoirs.  There is a ratchet effect that will cause a small amount of heat to flow from the water to the ice because if each is 100%  flowing the other way will make a temperature difference.
A: Erik, nothing will happen (as you already know!).  
Water and ice are in equilibrium at the triple point.  No heat can flow at constant temperature.  The water and ice have the same free energy per mole, so no spontaneous change can occur, and the total entropy cannot rise as a consequence of heat transfer.
It is much simpler to just consider putting ice and water into an empty container, removing the air, and sealing it.  Initially the system is out of equilibrium, so some ice will melt, or some water will freeze, and some will evaporate to fill the space.  Once all three phases have reached the triple point, their proportions will not change - they were determined by the initial conditions.
You might ask why the system cannot wander about on the flat surface of constant free energy.  I think one answer is as follows.  To stay at the triple point, the temperature and pressure must be fixed.  Of course the energy is also fixed, so we have three equations, which are (probably) sufficient to determine the proportions of the three phases.  You might ask what happens when there are four phases in equilibrium - but there is a theorem that this cannot happen.
A: The original state is not in thermodynamic equilibrium. The liquid side has a lot more heat than the solid side due to the heat of fusion.
