A2294-kg sample of liquid water is cooled from 0°c to it freezes in the process. how much heat is liberated? for water and the specific heat capacity of ice is .

There are some missing data in the text. I've found the complete text on internet:
"A 2294 kg sample of water at 0 degrees C is cooled to -36 degrees Cand freezes in the process. How much heat is liberated? (For water Lf = 334 kJ/kg and Lv = 2257 kJ/kg. The specificheat for ice is 2050 J/kg x K)"

Solution:
We must divide the problem in two steps.

Step 1) Solidification of the water at  into ice. The amount of heat released in this process is given by:

where 
m is the mass of the water
is the latent heat of fusion of ice
Substituting numbers, we find


step 2) the water became ice, now we need to cool it down to . The amount of heat released in this process is

where 
m is the mass of the ice
is the specific heat of the ice
is the variation of temperature
Substituting numbers, we find

where the negative sign simply means the heat is released by the system.

Therefore, the heat released in the whole process is

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