Answer:Heat added or subtracted for a phase change = Latent heat x Mass

Q = L heat M

Where,

Q = heat (calories or joules)

L heat = latent heat (calories/GRAM or joules/gram)

M = mass (grams)

If liquid water at 100°C is changed into steam, the heat added (the latent heat of vaporization) is 540 calories for every gram of water. If steam at 100°C is changed into the water at 100° C, 540 calories for every gram of steam MUST be subtracted. If ice at 0°C is changed into liquid water at 0°C, the heat added (the latent heat of melting) is 80 calories for every gram of ice. If liquid water at 0°C changes into ice at 0°C, 80 calories for every gram of liquid water must be subtracted.

Latent heats can be very large. For example, the latent heat of vaporization of water is 540 cal/g and the latent heat of freezing of water is 80 cal/g. Therefore, changing a given quantity of water to steam REQUIRES 5.4 times as much heat as WARMING it from 0°C (+32°F) to 100°C (212°F), and melting ice requires as much heat as warming water from 20°C (68°F) to 100°C.

Step-by-step explanation: