the specific heat of liquid ethanol,C₂H₅OH(l) is 2.46j/g ⁰C and the heat of vaporization is 39.3 kj/mol the boiling point of ethanol is 78.3⁰C what amount of etnthalpy is required to heat 50 g of liquid ethanol form 23⁰C to ethanol vapor a 78.3⁰C ?

Question

24-06-2025
Chemistry

High School

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the specific heat of liquid ethanol,C₂H₅OH(l) is 2.46j/g ⁰C and the heat of vaporization is 39.3 kj/mol the boiling point of ethanol is 78.3⁰C what amount of etnthalpy is required to heat 50 g of liquid ethanol form 23⁰C to ethanol vapor a 78.3⁰C ?

Answer :

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ethanchris270 ethanchris270 · Answer 1 · 2023-09-24T13:05:20-04:00

Final answer:

The total enthalpy required to heat 50g of liquid ethanol from 23°C to ethanol vapor at 78.3°C is about 49.66 kJ, which includes 6.78 kJ to heat the ethanol to its boiling point, and an additional 42.88 kJ to vaporize it.

Explanation:

To find the enthalpy required to heat 50 g of liquid ethanol from 23°C to ethanol vapor at 78.3°C, we first need to calculate the amount of energy required to heat the ethanol to its boiling point, and then, the energy needed to vaporize it at that point.

First, using the specific heat equation q=mcΔT where q is the energy, m is the mass, c is the specific heat, and ΔT is the change in temperature, we find q=50g * 2.46 J/g°C * (78.3°C - 23°C) = 6781.7 J = 6.78 kJ.

Next, we calculate the energy required to vaporize the ethanol. The molar mass of ethanol, C₂H5OH, is roughly 46.07 g/mol. Therefore, 50g corresponds to 50/46.07 ≈ 1.09 mol. The heat of vaporization is the energy required to vaporize one mole of a substance at its boiling point, so for 1.09 mol of ethanol is it 1.09 mol * 39.3 kJ/mol = 42.88 kJ.

In conclusion, the total enthalpy required is the sum of the energy required to heat the ethanol to its boiling point and the energy required to vaporize it at that point, 6.78 kJ + 42.88 kJ = 49.66 kJ.

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