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.