Question

A 88-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

During a change of state from liquid to solid (like in this problem, from liquid water to ice), the amount of heat that it is liberated is given by

`Q=mL_f`
where
m is the mass of the substance

`L_f` is the latent heat of fusion of the substance.

For ice, the latent heat of fusion is
`L_f=334 kJ/kg`, while the mass of the water in this problem is m=88 kg. If we substitute these data into the equation, we find the amount of heat liberated:

`Q=mL_f =(88 kg)(334 kJ/kg)=29392 kJ=29.4 MJ`