How much heat must be removed to freeze a tray of ice cubes at 0 degrees celcius if the water has a mass of 225g?

To be able to answer this question, we just have to determine the heat of fusion of the water. From reliable sources, the heat of fusion of water is equal to 80 cal/g. Multiplying this value to the mass given,
                   Heat = (225 g)(80 cal/g) 
                            = 18000 cal
Therefore, the amount of heat that must be removed is equal to 18000 cal or 18 kcal.