Answer:
72 kJ of heat is removed.
Explanation:
First, the ethanol vapor will reduce its temperature until the temperature of the boiling point, then it will occur a phase change from vapor to liquid, and then the temperature of the liquid will decrease. The total heat will be:
Q = Q1 + Q2 + Q3
Q1 = n*cv*ΔT1, Q2 = m*Hl, and Q3 = n*cl*ΔT2
Where n is the number of moles, cv is the specific heat of the vapor (65.44 J/K.mol, cl is the specific heat of the liquid (111.46 J/K.mol), Hl is the heat of liquefaction (-836.8 J/g), m is the mass, and ΔT is the temperature variation (final - initial).
Q = n*cv*ΔT1 + m*Hl + n*cl*ΔT2
The molar mass of ethanol is 46 g/mol, and the number of moles is the mass divided by the molar mass:
n = 74.2/46 = 1.613 moles
Q = 1.613*65.44*(78.37 - 83) + 74.2*(-836.8) + 1.613*111.46*(26 - 78.37)
Q = -72000 J
Q = -72 kJ (because it is negative, it is removed)
