ice cube trays holds 325g water. initially water temp of 25 degress C. How many kilojoules of heat must be?

Question

removed to cool and freeze the water at 0 degrees C?

Answer 1

The energy removed is equal to the enerdy released to cool the water to 0C plus the energy released to freeze the water to ice at 0C

The formula for the energy required to change the temperature of a substance is:

q=m*c*(change in temp.), where q=energy, m=mass, c=specific heat, and in the parenthesis is the change in temperature the substance goes through.

The formula for the energy required to change water from liquid to solid is:

q=m*Hf, where Hf=heat of fusion.

Take the specific heat of liquid water to be 4190 J/kg K, the specific heat of ice to be 2100 J/kg K, and the heat of fusion for water to be 334 kJ/kg.

Step 1, calculate the energy released when the water cools:

so, the water goes from 25C to 0C, a temp change of 25C

q=m*c*(change in temp) becomes
q=.325 kg*4190*25, which is q=34043.75 J released

Step 2, figure the energy released to change the water to ice.

The ice then goes from liquid to solid
q=m*Hf, which becomes
q=.325 Kg*334, which is q=108.55 J released

so, total energy (heat) is:

34043.75 +108.55= 34152.3 J

Answer 2

There are two source of heat to remove: cool the water from 25C to 0C and then heat associated with the phase change from liquid to solid. You need the specific heat (Cp) of liquid water and the latent heat of freezing of water. The units on Cp will be joules per gram per degree C. The latent heat will be joules per gram. good luck