You are asked about the heat released by vapor ethanol so we have to do 3 heat calculations
1. Ethanol vapor going from 98.5 C to the boiling point (78.4)
2. Ethanol going through a phase change (gas to liquid also known as latent heat)
3. Ethanol (liquid) going from 78.4 to -12C
The ethanol turns to solid at -114 C so we do not have solid ethanol here
1. vapor ethanol
∆H = m * Cp * ∆T
∆T = Final temperature – Initial temperature
m is mass
Cp is heat capacity
H is enthalpy so
calculate moles of ethanol, moles = mass / molar mass (molar mass of ethanol is 46)
22 / 46 = 0.478 moles
H = 0.478 moles * 65.7 J / mole * (78.4 - 98.5) = -631.57 J , the negative sign is because heat is being released
2. calculate the heat because of phase change
H = moles * Enthalpy of vaporization
H = 0.478 * 38.56 KJ / mole = 18.441 KJ or 18 441 Joules
3. Calculate the heat for liquid ethanol
∆H = m * Cp * ∆T
H = 0.478 * 113 * ( -12 - 78.4) = -4885.53 Joules
Total heat released = 4885.53 + 18431 + 631.57 = 23 958.84 Joules
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