Question

A 2294-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 .

Answer 1

Answer:935,000k

Explanation: Gradpoint

Answer 2

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
`0^(\circ)C` into ice. The amount of heat released in this process is given by:

`Q_1 = m L_f`
where
m is the mass of the water

`L_f` is the latent heat of fusion of ice
Substituting numbers, we find

`Q_1 = (2294 kg)(334 kJ/kg)=7.66 \cdot 10^5 kJ=7.66 \cdot 10^8 J`

step 2) the water became ice, now we need to cool it down to
`-36^(\circ)C`. The amount of heat released in this process is

`Q_2 = mC_s \Delta T`
where
m is the mass of the ice

`C_s` is the specific heat of the ice

`\Delta T` is the variation of temperature
Substituting numbers, we find

`Q_2 = (2294 kg)(2050 J/kg K)(-36^(\circ)C-0^(\circ)C)=-1.69 \cdot 10^8 J`
where the negative sign simply means the heat is released by the system.

Therefore, the heat released in the whole process is

`Q=Q_1 + Q_2 = 7.66 \cdot 10^8 J + 1.69 \cdot 10^8 J=9.35 \cdot 10^8 J`