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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 -

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 -
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12 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 -

asked
User MrWater
by
8.9k points

1 Answer

5 votes

Answer:

1200 cal

Step-by-step explanation:

80 cal/g * 15 g = 1200 cal

answered
User Chuma Umenze
by
8.1k points
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