How much energy in kilojoules is released when 25.0 g of ethanol (C2H5OH) vapor at 93.0 C is cooled to -10 C?

Ethanol has mp = -114.1 oC, bp = 78.3 oC, ΔHvap = 38.56 kJ/mol, and ΔHfus = 4.93 kJ/mol. The molar heat capacity is 112.3 J/(K.mol) for the liquid and 65.6 J/(K.mol) for the vapor

I answered the questions:

25.0g[(1.7)(93-78.3)+841+2.44(78-(-10))] and got 27.030 kJ but that isn't correct!!

2 answers

q1 = energy released on vapor cooling from 93 C to 78.3 C
q1 = mass ethanol x specific heat ethanol vapor x (Tfinal - Tinitial) note: Tf is 78.3 and Ti is 93.0.

q2 = energy released on changing from vapor at 78.3 to liquid at 78.3
q2 = mass ethanol x heat vaporization

q3 = energy released on lidquid cooling from 78.3 to -10. Use same formula as in q1 but with correct specific heat.

Total energy released = q1 + q2 + q3
17.83kJ