If you cool water down enough, then a certain amount of energy will go into making the intermolecular forces and it'll freeze into ice. The amount of energy water needs to freeze is its enthalpy of fusion, which is about 6.02 kJ/mol. That's more effective than cooling a fluid that's, let's say...#25^o C# (#298K#). Let's look at a comparison:
#q=msDeltaT#
#m# is mass in #g#, but I'm using one #mol# of water to compare it to #6.02 (kJ)/(mol)#. #s# is specific heat capacity.
If we were to do a sorta realistic example, where water isn't held at a constant temperature (the "cooling fluid", labeled water), and it's in a closed system with more water (the water to be cooled, labeled fluid):
#q = q_(water) + q_(fluid) = m_(water)s_(water)DeltaT_(water) + m_(fluid)s_(fluid)DeltaT_(fluid)#
#0 = (18.015g/(mol))(4.184J/(g*K))(T_(f,water)-273K) + (18.015g/(mol))(4.184J/(g*K))(T_(f,fluid)-298K)#
#-cancel([(18.015g/(mol))(4.184J/(g*K))])(T_(f,water)-273K) = cancel([(18.015g/(mol))(4.184J/(g*K))])(T_(f,fluid)-298K)#
#273K - T_(f,water) = T_(f,fluid) - 298 K#
They'll cool down to the same temperature, so:
#571K = 2T_(f,fluid)#
#T_(f,fluid) = 285.5K = 12.5^oC#
Now how much energy did that take?
#q_(fluid)=18.015cancel(g)/(mol)*4.184J/(cancel(g*K))*(285.5cancel(K)-298cancel(K))=-0.942*10^3 J/(mol) = -0.942 (kJ)/(mol)#
...which is about #16%# of the energy needed to melt ice. So, that means ice is about 6 times as efficient as water in cooling room temperature water down to #0^oC#!
