Answer: you would have to remove 80 calories from 1 g of pure water at the freezing point, 0˚ C, to convert it to 1 g of ice at 0˚ C. Explanation:However, 540 calories of energy are required to convert that 1 g of water at 100˚ C to 1 g of water vapor at 100˚ C. This is called the latent heat of vaporization. On the other hand, you would have to remove 80 calories from 1 g of pure water at the freezing point, 0˚ C, to convert it to 1 g of ice at 0˚ C.
