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 [tex]0^{\circ}C[/tex] into ice. The amount of heat released in this process is given by:
[tex]Q_1 = m L_f[/tex]
where
m is the mass of the water
[tex]L_f[/tex] is the latent heat of fusion of ice
Substituting numbers, we find
[tex]Q_1 = (2294 kg)(334 kJ/kg)=7.66 \cdot 10^5 kJ=7.66 \cdot 10^8 J[/tex]
step 2) the water became ice, now we need to cool it down to [tex]-36^{\circ}C[/tex]. The amount of heat released in this process is
[tex]Q_2 = mC_s \Delta T[/tex]
where
m is the mass of the ice
[tex]C_s[/tex] is the specific heat of the ice
[tex]\Delta T[/tex] is the variation of temperature
Substituting numbers, we find
[tex]Q_2 = (2294 kg)(2050 J/kg K)(-36^{\circ}C-0^{\circ}C)=-1.69 \cdot 10^8 J [/tex]
where the negative sign simply means the heat is released by the system.
Therefore, the heat released in the whole process is
[tex]Q=Q_1 + Q_2 = 7.66 \cdot 10^8 J + 1.69 \cdot 10^8 J=9.35 \cdot 10^8 J[/tex]
