How much heat will be released during the condensation of 300 g of alcohol?

Initial data: m (mass of alcohol) = 300 g.

Reference data: L (specific heat of vaporization of alcohol) = 0.9 * 10 ^ 6 J / kg.

SI system: m = 300 g = 0.3 kg.

The amount of heat that will be released during the condensation of alcohol is calculated by the formula: Q = L * m.

Let’s calculate: Q = 0.9 * 10 ^ 6 * 0.3 = 0.27 * 10 ^ 6 J = 270 kJ.

Answer: Upon condensation of 300 grams of alcohol, 270 kJ of heat will be released.




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