Answer :
The enthalpy required to heat 50 g of liquid ethanol from 23 °C to ethanol vapor at 78.3 °C is 50. kJ.
We want to heat 50 g of liquid ethanol from 23 °C to ethanol vapor at 78.3 °C. We can divide this process in 2 parts:
Heating of liquid ethanol from 23 °C to 78.3 °C (boiling point).
Vaporization of ethanol at 78.3 °C.
1. Heating of liquid ethanol from 23 °C to 78.3 °C
The amount of heat required for this part (Q₁),
Q₁ = c × m × ΔT
Q₁ = (2.46 J/g °C) × 50 g × (78.3 °C - 23 °C) × (1 kJ/1000 J) = 6.8 kJ
where,
c = specific heat of liquid ethanol.
m = mass of ethanol.
ΔT = change in the temperature.
2. Vaporization of ethanol at 78.3 °C.
The amount of heat required for this part (Q₂),
Q₂ = (m/M) × ΔH°vap
Q₂ = [50 g/(46.07g/mol] × (39.3 kJ/mol) = 43 kJ
where,
m = mass of ethanol.
M = molar mass of ethanol.
ΔH°vap = enthalpy of vaporization of ethanol.
The total amount of heat required (Q) = the sum of the heat required in each step.
Q = Q₁ + Q₂ = 6.8 kJ + 43 kJ = 50. kJ
Let the process is carried out at constant pressure and the enthalpy required for the process = 50. kJ.
The enthalpy required to heat 50 g of liquid ethanol from 23 °C to ethanol vapor at 78.3 °C is 50. kJ.
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