joules needed to melt 20.0 g of ice at 0 ∘C and to warm the liquid to 40.0 ∘C?
2016-11-29 3:51 am
kilojoules needed to melt 27.0 g of ice at 0 ∘C, warm the liquid to 100 ∘C and change it to steam at 100 ∘C
回答 (1)
2016-11-29 4:38 am
(333.6 J/g) x (20.0 g) = 6672 J to melt the ice
(4.184 J/g·°C) x (20.0 g) x (40.0 - 0.0)°C = 3347.2 J to warm the melted ice to 40.0°C
6672 J + 3347.2 J = 10019.2 = 10000 J total
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(333.6 J/g) x (27.0 g) = 9007.2 J to melt the ice
(4.184 J/g·°C) x (27.0 g) x (100.0 - 0.0)°C = 11296.8 J to warm the melted ice to its boiling point
(2257 J/g) x (27.0 g) = 60939 J to vaporize the water
9007.2 J + 11296.8 J + 60939 J = 81243 J = 81.2 kJ total
(4.184 J/g·°C) x (20.0 g) x (40.0 - 0.0)°C = 3347.2 J to warm the melted ice to 40.0°C
6672 J + 3347.2 J = 10019.2 = 10000 J total
-------------------
(333.6 J/g) x (27.0 g) = 9007.2 J to melt the ice
(4.184 J/g·°C) x (27.0 g) x (100.0 - 0.0)°C = 11296.8 J to warm the melted ice to its boiling point
(2257 J/g) x (27.0 g) = 60939 J to vaporize the water
9007.2 J + 11296.8 J + 60939 J = 81243 J = 81.2 kJ total
收錄日期: 2021-04-27 23:31:15
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20161128195101AA7sxXr