How much heat must be removed from 456 g of water at 25.0°c to change it into ice at?

This question can be simply solved by using heat formula,
    Q = mCΔT

Q = heat energy (J)
m = Mass (kg)
C = Specific heat capacity (J / kg K)
ΔT = Temperature change (K)

when water freezes, it produces ice at 0°C (273 K)
hence the temperature change is 25 K (298 K - 273 K)
C for water is 4186 J / kg K or 4.186 J / g K
By applying the equation,
 Q = 456 g x 4.186 J / g K x 25 K
     = 47720.4 J
     = 47.72 kJ

hence 47.72 kJ of heat energy should be removed.