How many calories are required to convert 17g of ice at 0.0 degrees celsius to liquid water at 32.0 degrees celsius. The heat of fusion for water is 80. cal/g
heat required to melt ice,
Q1 = m*Hfus
= 17 g * 80 cal/g
= 1360 cal
heat required to raise temperature of water from 0oC to
32 oC
Q2 = m*C*(Tf - Ti)
= 17 g * 1 cal / g oC * (32-0) oC
=544 cal
Total heat required = 1360 cal + 544 cal = 1904 cal
Answer: 1904 cal
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