Question

Distillation is a process of vaporization a substance and chilling the vapor to collect it back the liquid form. How much heat is removed from 74.2 grams of ethanol vapor at 83 °C (Tb = 78.37 °C) if the collected liquid ethanol has a temperature of 26ºC?

Answer 1

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)