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3 votes
3 votes
If the heat of fusion of water is 80 cal/g, the amount of heat energy required to change 15.0 grams of ice at 0°C to

15.0 grams of water at 0°C is -

User Chinthakad
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1 Answer

23 votes
23 votes

Answer:

1200 cal

Step-by-step explanation:

80 cal/g * 15 g = 1200 cal

User Mzedeler
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