The human body obtains 1164 kJ from a candy bar Of this energy were used to vaporize water at 100 ∘C,?
2018-07-18 1:13 pm
how much water in liters could be vaporized? (Assume that the density of water is 1.0 g/mL.) The heat of vaporization of water at 100 ∘C is 40.7 kJ/mol.
回答 (1)
2018-07-18 1:31 pm
Method 1 :
Number of moles of H₂O (water) evaporated = (1164 kJ) / (40.7 kJ/mol) = 28.6 mol
Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol
Mass of H₂O evaporated = (28.6 mol) × (18.0 g/mol) = 515 g
Volume of H₂O evaporated = (515 g) / (1.0 g/mL) = 515 mL = 0.515 L
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Method 2 :
Water evaporated
= (1164 kJ) × (1 mol / 40.7 kJ) × (18.0 g / 1 mol) × (1.0 mL / 1.0 g) × (1 L / 1000 mL)
= 0.515 L
Number of moles of H₂O (water) evaporated = (1164 kJ) / (40.7 kJ/mol) = 28.6 mol
Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol
Mass of H₂O evaporated = (28.6 mol) × (18.0 g/mol) = 515 g
Volume of H₂O evaporated = (515 g) / (1.0 g/mL) = 515 mL = 0.515 L
====
Method 2 :
Water evaporated
= (1164 kJ) × (1 mol / 40.7 kJ) × (18.0 g / 1 mol) × (1.0 mL / 1.0 g) × (1 L / 1000 mL)
= 0.515 L
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https://hk.answers.yahoo.com/question/index?qid=20180718051323AAUVKty