How much heat will be released during condensation and cooling down to 20 ° C of alcohol weighing 850 g?

univerkov | April 19, 2021 | education

Initial data: m (weight of alcohol) = 850 g; tк (final temperature of alcohol) = 20 ºС.

Reference values: tboil (boiling point, ethanol) = 78.4 ºС; C (specific heat of alcohol) = 2500 J / (kg * K); L (specific heat of vaporization of alcohol) = 9 * 10 ^ 5 J / kg.

SI system: m = 0.85 kg.

The released amount of energy during condensation and cooling of alcohol:

Q = Q1 + Q2 = L * m + C * m * (tboil – tk) = 9 * 10 ^ 5 * 0.85 + 2500 * 0.85 * (78.4 – 20) = 889 100 J.

Answer: 889,100 Joules of heat will be released.

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