Answer :
Final answer:
To convert 29.0 g of ethanol at 23.0 ∘C to its vapor at 78.3 ∘C, the total quantity of heat required is 32.22 kJ.
Explanation:
To calculate the quantity of heat required to convert ethanol from a liquid to a vapor, we need to consider two processes: heating the liquid ethanol to its boiling point and vaporizing it at its boiling point. Firstly, we calculate the heat required to raise the temperature of the ethanol:
q1 = mass × specific heat capacity × temperature change
q1 = 29.0 g × 2.46 J/g·C × (78.3 - 23.0) C
q1 = 7455.3 J = 7.5 kJ
Next, we calculate the heat required for vaporization:
q2 = moles × molar heat of vaporization
First, we need to convert grams of ethanol to moles:
moles = mass / molar mass
moles = 29.0 g / 46 g/mol
moles = 0.63 mol
Now, we can calculate q2:
q2 = 0.63 mol × 39.3 kJ/mol
q2 = 24.72 kJ
The total quantity of heat required to convert 29.0 g of ethanol at 23.0°C to a vapor at 78.3°C is the sum of q1 and q2:
q = q1 + q2 = 7.5 kJ + 24.72 kJ = 32.22 kJ
Learn more about quantity of heat here:
https://brainly.com/question/13971985
#SPJ11
