The Cp of liquid water is 4.184 J/g°C. How much energy is needed to heat a 52.0 gram sample of water from 0.0°C to 100.0°C?
回答 (1)
2021-03-02 11:15 pm
✔ 最佳答案
Energy needed= m Cp ΔT
= (52.0 g) × (4.184 J/g°C) × [(100.0 - 0.0)°C]
= 21800 J
= 21.8 kJ
收錄日期: 2021-04-25 13:53:56
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https://hk.answers.yahoo.com/question/index?qid=20210302150114AARxu56