How much heat will be released during the condensation of 2 kg of water vapor at the boiling point?

univerkov | January 4, 2021 | education

Given:
m = 2 kg
water vapor (water)
L = 2.3 * 10 to the 6th degree J / kg
Decision:
Q = Lm, where Q is heat, L is specific heat of vaporization, m is mass
Q = 2.3 * 10 ^ 6 J / kg * 2kg = 4.6 * 10 ^ 6 J
Answer: Q = 4.6 * 10 ^ 6 J

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