The heat needed to freeze the water is 4400 calories
From the question given above, the following data were obtained:
Mass of water (m) = 55 g
Heat (Q) =?
NOTE:
Enthalpy of fusion of water (ΔHբ) = 79.7 cal/g
The heat needed to freeze the water can be obtained as illustrated below:
Q = m•ΔHբ
Q = 55 × 79.7
Q = 4383.5 calories
Q = 4400 calories ( 2 sf)
Therefore, the heat needed to freeze the water is 4400 calories
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