Phy (water)

Assuming the heat loss per unit time to the surroundings Q is constant when the water cools and when it freezes to become ice.
Suppose it takes time t for a mass of m kg of water to cool from room temperrature (say, at 25 deg.C) to its freezing point.
Thus, Qt = mc.(25)
where c is the specific heat capacity of water (=4200 J/kg.C)
When the m kg of water freezes, suppose it takes time T for all the water to form solid ice completely,
then, QT = mL
where L is the latent heat of fusion of water (=334000 J/kg)
hence, QT/Qt = mL/m.c(25)
T/t = L/25.c = 334000/(25x4200) = 3.18
therefore, the time needed for freezing is about 3 times the needed for cooling of the water.

because more energy is required to melt the ice to water. it involves latent heat & haat due to specific heat.