Answer :
Final answer:
In order to convert 25.0 g of ethanol to vapor at its boiling point, we need to calculate the quantity of heat required. Using the specific heat capacity and enthalpy of vaporization of ethanol, we can calculate the heat lost during vaporization and the heat gained during heating. Adding these two quantities together, the total heat required is 29.09 kJ.
Explanation:
To calculate the quantity of heat required to convert the ethanol to vapor, we will need to use two equations: the heat gained equation and the heat lost equation.
The heat gained equation is q = m * c * ∆T, where q is the heat gained, m is the mass of the substance, c is the specific heat capacity, and ∆T is the change in temperature.
The heat lost equation is q = n * ∆Hvap, where q is the heat lost, n is the moles of the substance, and ∆Hvap is the enthalpy of vaporization.
Using the given information, we can calculate the quantity of heat required:
- Calculate the moles of ethanol: moles = mass / molar mass = 25.0 g / 46.07 g/mol = 0.543 mol
- Calculate the heat lost during vaporization: q = n * ∆Hvap = 0.543 mol * 39.3 kJ/mol = 21.37 kJ
- Calculate the heat gained during heating: q = m * c * ∆T = 25.0 g * 2.46 J/g •C * (78.3 – 23.0) °C = 7.72 kJ
- Add the two quantities of heat together: 21.37 kJ + 7.72 kJ = 29.09 kJ
Therefore, 29.09 kJ of heat is required to convert 25.0 g of ethanol to a vapor at 78.3 °C.
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