# How much heat must be removed to freeze a tray of ice cubes at 0C? The mass of the water is 225 g?

$18 \text{kcal}$.
The latent heat of water freezing is $80 \text{cal/g}$ -- that amount of heat has to be removed to freeze water at a constant temperature equal to its melting/freezing point 0°"C".
For $225 \text{g}$ of water, the total latent heat is
(80"cal/g")×(225"g")=18,000"cal"=18"kcal"