# Does liquid water remain at thermal equilibrium with ice at the melting point of ice?

Suppose, I have a glass of water. Now, I begin cooling it until it reaches $$0^{\circ}\rm C$$. Just after it reaches $$0^{\circ}\rm C$$, I stop cooling it.

Again suppose I have an ice cube. Now I begin heating it until it reaches $$0^{\circ}\rm C$$. Just after it reaches $$0^{\circ}\rm C$$, I stop heating it.

Now, I take the ice cube and the glass of water in a controlled room where the temperature is $$0^{\circ}\rm C$$. Now, if I drop the ice cube in the glass of water, what will happen? Will the ice melt or will the water freeze? I think they will both stay the same.$$\tag{1}$$

My book defined the melting point of water like this:

At standard pressure, the temperature at which pure ice remains in equilibrium with water, that is the temperature at which pure ice begins to melt is known as the lower fixed point, ice point or the melting point.

Now, my book forgot to specify what type of equilibrium the ice and water will be in: will it be thermal equilibrium?$$\tag{2}$$

What are the answers to $$(1)$$ & $$(2)$$?

In an ideal scenario, the water and ice will stay in thermodynamic equilibrium with each other as there is no temperature gradient between them for heat transfer to occur. However, in a practical experiment, water at $$0^\circ$$C will immediately turn to ice when disturbed. This is because the enthalpy of fusion of ice is quite low compared to its specific heat $$(\Delta H_{\rm fus} \approx \frac1{10}C_{\rm water} )$$. The dropped ice cube will serve as a seeding point for new ice crystals to grow, causing the rest of the water to also turn to ice.