更新1:
A 45 g piece of ice at 0.0 ∘C is added to a sample of water at 8.0 ∘C. All of the ice melts and the temperature of the water decreases to 0.0 ∘C. How many grams of water were in the sample? Can't figure this out. Thanks!
Chemistry?
回答 (2)
2017-10-18 12:19 pm
Latent heat of fusion of ice at 0.0°C = 334 J/g
Heat capacity of water = 4.18 J/g°C
Heat released by water = (m g) × (4.18 J/g°C) × [(8.0 - 0.0)°C] = 33.44m J
Heat absorbed by ice = (45 g) × (334 J/g) = 15030 J
Heat released by water = Heat absorbed by ice
33.44m = 15030
m = 449.5 ≈ 450 (to 2 sig. fig.)
Mass of water = 450 g
Heat capacity of water = 4.18 J/g°C
Heat released by water = (m g) × (4.18 J/g°C) × [(8.0 - 0.0)°C] = 33.44m J
Heat absorbed by ice = (45 g) × (334 J/g) = 15030 J
Heat released by water = Heat absorbed by ice
33.44m = 15030
m = 449.5 ≈ 450 (to 2 sig. fig.)
Mass of water = 450 g
2017-10-18 12:27 pm
Thanks! You're a lifesaver!
收錄日期: 2021-04-18 17:55:20
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https://hk.answers.yahoo.com/question/index?qid=20171018040621AAb8zj3