Answer :
Final answer:
The amount of enthalpy required to heat 50.0 g of liquid ethanol from 23.0 °C to ethanol vapor at 78.3 °C is 213,180 J.
Explanation:
To calculate the amount of enthalpy required to heat 50.0 g of liquid ethanol from 23.0 °C to ethanol vapor at 78.3 °C, we need to consider the heat required to increase the temperature and the heat required for vaporization.
Heat required for temperature increase:
The heat required to increase the temperature of a substance can be calculated using the formula Q = m * C * ΔT, where Q is the heat, m is the mass, C is the specific heat, and ΔT is the change in temperature.
Using the given information, we can plug in the values:
Q = (50.0 g) * (2.46 J/g°C) * (78.3 °C - 23.0 °C)
Q = 170,475 J
Heat required for vaporization:
The heat required for vaporization can be calculated using the formula Q = n * ΔHvap, where Q is the heat, n is the number of moles, and ΔHvap is the heat of vaporization.
First, we need to find the number of moles of ethanol using its molar mass.
The molar mass of ethanol (C2H5OH) is 46.07 g/mol.
n = (50.0 g) / (46.07 g/mol) = 1.085 mol
Now we can calculate the heat required for vaporization:
Q = (1.085 mol) * (39.3 kJ/mol) = 42.705 kJ
Total enthalpy required:
To find the total enthalpy required, we add the heat required for temperature increase and the heat required for vaporization:
Total enthalpy = 170,475 J + 42.705 kJ = 170,475 J + 42,705 J = 213,180 J
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