Water is at the triple point, and the working fluid includes water, ice, and water vapor. The system volume remains unchanged, maintaining energy exchange with the large heat source while keeping the temperature constant. The equilibrium state has the following equation:
mg+ml+ms=m0
mg/ρg+ml/ρl+ms/ρs=V
There are two equations, three unknown variables,, and an infinite number of equilibrium solutions in a system. The three-phase distribution of matter cannot be determined, and thermodynamics requires that the equilibrium state of a system be unique, which is inconsistent with physical facts. The thermodynamic direction advocated by the second law of thermodynamics will also be lost, such as volume expansion, which cannot determine whether it is the evaporation of liquid water or the transformation of ice into water vapor.
When you specify the heat exchange you talk about, you'll get the third equation.
You can show that if you inset heat H into your system, you must vaporize H*(vl - vs)/(he*(vl - vs) - hf*(vg - vl)) moles of liquid, and melt -H*(vg - vl)/(he*(vl - vs) - hf*(vg - hf)) moles of ice.
(hf and he - fussion and evaporation enthalpies; the v's - molar volumes)
Under isothermal conditions (large heat source), there is reversible thermal equilibrium between water, ice, and water vapor at the triple point. The total energy of water, ice, and water vapor can vary. There are only two definite equations, volume conservation and mass conservation.
...the very words you use ('large heat source') contradict what you say. There's a third very definite equation that stands behind it.
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