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.
