How much heat is required to evaporate 20 g of water taken at the boiling point (at normal atmospheric pressure)?
| June 30, 2021 | education
Data: m (mass of evaporated water) = 20 g = 20 * 10 ^ -3 kg; L (beats heat of vaporization of water under normal conditions) = 2.3 * 106 J / kg.
Required heat: Q = m * L = 20 * 10 ^ -3 * 2.3 * 10 ^ 6 = 46 * 10 ^ 3 J.
Answer: For vaporization of 20 g of water, 46 kJ of heat is required.
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