Calculate the amount of heat that is required to convert 200g alcohol into steam. located at a temperature of 28C?

Problem data: m (mass of alcohol) = 200 g (in SI m = 0.2 kg); t0 (temperature at which the alcohol was located) = 28 ºС.

Reference values: Cc (specific heat of alcohol) = 2500 J / (kg * K); tpap (alcohol vaporization temperature) = 78 ºС; L (specific heat of vaporization of alcohol) = 0.9 * 10 ^ 6 J / kg.

The amount of heat for vaporization of the alcohol taken: Q = Cc * m * (tp – t) + L * m.

Calculation: Q = 2500 * 0.2 * (78 – 28) + 0.9 * 10 ^ 6 * 0.2 = 205,000 J (205 kJ).

Answer: You will need 205 kJ of heat.