A large bottle of water containing 825g of water at 9 degrees celsius. How many KJ's are absorbed to warm the water at 22 celsius room temp?
回答 (1)
2019-03-05 1:08 am
Energy absorbed
= m c ΔT
= (825 g) × (4.184 J/g°C) × (22 - 9)°C
= 44900 J (to 3 sig. fig.)
= 44.9 kJ
= m c ΔT
= (825 g) × (4.184 J/g°C) × (22 - 9)°C
= 44900 J (to 3 sig. fig.)
= 44.9 kJ
收錄日期: 2021-05-01 22:34:09
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