How much heat (kJ) must be removed in order to change 4.33 moles of water at 55.0 oC to ice at -23.00oC?

Question

The specific heat capacity of solid water is 2.03 J g-1 oC -1


Use a negative sign in your answer if heat is removed.

Answer 1

You should have provided more info.

c (specific heat capacity) will not remain constant over such a broad temperature range.
Assuming that it will, you have c= 4.184 J g^-1 C^-1 for liquid water

You have mass=mole*MW =4.33*18 g of water.
The freezing point is 0 C, so you need
Q1=mc1ΔT1= 4.33*18*4.184* (0-55.0) = -17.936 kJ

This amount of energy will be removed/lost in order to get liquid water at 0 C. Then you will lose more energy (latent energy) for the transition from liquid to solid at the same temperature.
For water the latent heat is approximately 334 J/g so you have
Q2= -m*334= - 4.33*18*334= -26.032 kJ

Then the specific heat changes, so the amount of energy released for solid water at 0 C to reach -23 C you have

Q3= mc2ΔT2= 4.33*18*2.03*(-23-0) = -3.639 kJ

The total amount of energy released will be the sum of the energies of the three steps

Q= Q1+Q2+Q3 = -47.607 kJ