100 g of water at 25 °C is poured into an insulating cup. 50 g of ice at 0 °C is added to the water. The water is stirred until the temperature of the water has fallen to 0 °C. 18 g of ice remains unmelted. The specific heat capacity of water is 4.2 J / g °C. Which value does this experiment give for the specific latent heat of fusion of ice? B 330 J / g
Q lost H2O = m c t
= 100x4.2x25=10500 J
L = Q gained by ice /m
= 10500 / 50-18
= 328.1J/g
Answers: 3
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