What fraction of water supercooled down to the temperature `t = -20 ^@C` under standard pressure turns into ice when the system passes into the equilibrium state ? At what temperature of the supercooled water does it turn into ice completely ?

Explanation

From step 1, we know that 15.98 kJ is neede to raise the temperature of the ice to the melting point. The remaining heat `Q_("rem")` is then 210 kJ -15.98kJ, or abou 194kJ . From step 2, we can see that this amount of heat is insufficient to melt all the ice. Because the melting of the ice is incomplete, we must end up with a mixture of ice and liquid, the temperature of the mixture must be freezing point, `0^(@)C`.
Calculations `:` We can find the mass m of ice that is melted by the available energy `Q_("rem")` by using Equation with `L_(F)`.
`m = ( Q_("rem"))/( L_(F) ) = ( 194kJ)/( 333kJ//kg ) = 0.583 kg ~~ 580 g `.
Thus, the mass of the ice that remains is 720g - 580 g, or 140 g and we have
580 g water and 140 g ice, at `0^(@)C`.