Answer :
Final answer:
To convert 50.0 g of ethanol from 23.0 °C to vapor at 78.3 °C, approximately 49.46 kJ of heat is required. This involves calculating heat for temperature rise and vaporization using specific heat capacity and enthalpy of vaporization respectively.
Explanation:
The heat required to convert 50.0 g of ethanol from a liquid at 23.0 °C to a vapor at 78.3 °C involves two main steps: heating the liquid to its boiling point and then vaporizing it. The heat needed to warm the ethanol up to its boiling point (q1) can be calculated using the specific heat capacity formula:
q1 = m × c × ΔT
Where:
m = mass of ethanol = 50.0 g
c = specific heat capacity of ethanol = 2.46 J/g°C
ΔT = change in temperature = 78.3 °C - 23.0 °C = 55.3 °C
q1 = 50.0 g × 2.46 J/g°C × 55.3 °C = 6789.9 J or 6.7899 kJ
To vaporize the ethanol, you use its enthalpy of vaporization (ΔHvap), but first, you need to determine the number of moles of ethanol involved:
n = m / M = 50.0 g / 46.07 g/mol ≈ 1.085 mol
The heat required to vaporize the ethanol (q2) is:
q2 = n × ΔHvap
ΔHvap = 39.3 kJ/mol
q2 = 1.085 mol × 39.3 kJ/mol = 42.6675 kJ
The total heat needed (Q) is the sum of the heat needed to warm the liquid to its boiling point and the heat required to vaporize it:
Q = q1 + q2
Q = 6.7899 kJ + 42.6675 kJ = 49.4574 kJ
So, approximately 49.46 kJ of heat is required.
