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Answer :
Approximately 44.02 kJ of heat is required to convert 44.0 g of ethanol from 23.0 °C to a vapor at 78.3 °C.
To calculate the quantity of heat required, we need to consider two steps: heating the ethanol from 23.0 °C to its boiling point and then vaporizing it.
First, let's calculate the heat required to raise the temperature of ethanol from 23.0 °C to its boiling point. We will use the formula:
q = m * C * ΔT
Where:
q is the heat required,
m is the mass of ethanol,
C is the specific heat capacity of ethanol, and
ΔT is the change in temperature.
For this step, the mass of ethanol (m) is 44.0 g, the specific heat capacity (C) is 2.46 J/g°C, and the change in temperature (ΔT) is (78.3 °C - 23.0 °C) = 55.3 °C.
q1 = 44.0 g * 2.46 J/g°C * 55.3 °C = 6,394.23 J = 6.39423 kJ (rounded to 5 decimal places)
Next, let's calculate the heat required for vaporization. The heat of vaporization (ΔHvap) is given as 39.3 kJ/mol. To convert from grams to moles, we need to use the molar mass of ethanol.
Molar mass of ethanol ([tex]C2H5OH[/tex]) = (2 * atomic mass of carbon) + (6 * atomic mass of hydrogen) + atomic mass of oxygen
= (2 * 12.01 g/mol) + (6 * 1.01 g/mol) + 16.00 g/mol
= 46.07 g/mol
The number of moles (n) of ethanol can be calculated using:
n = mass / molar mass
= 44.0 g / 46.07 g/mol
≈ 0.9565 mol (rounded to 4 decimal places)
The heat required for vaporization is given by:
q2 = n * ΔHvap
= 0.9565 mol * 39.3 kJ/mol
≈ 37.63 kJ (rounded to 2 decimal places)
The total heat required is the sum of q1 and q2:
Total heat = q1 + q2
= 6.39423 kJ + 37.63 kJ
≈ 44.02 kJ (rounded to 2 decimal places)
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