Answer:
You must remove [tex]\text{50.6 kJ}[/tex] .
Explanation:
There are three heat transfers in this process:
Total heat = cool the vapour + condense the vapour + cool the liquid
q = q₁ + q₂ + q₃
q = nC₁ΔT₁ + nΔHcond + nC₂ΔT₂
Let's calculate these heat transfers separately.
Data:
You don't give "the data below", so I will use my best estimates from the NIST Chemistry WebBook. You can later substitute your own values.
C₁ = specific heat capacity of vapour = 90 J·K⁻¹mol⁻¹
C₂ = specific heat capacity of liquid = 115 J·K⁻¹mol⁻¹
ΔHcond = -38.56 kJ·mol⁻¹
Tmax = 300 °C
b.p. = 78.4 °C
Tmin = 25.0 °C
n = 0.782 mol
Calculations:
ΔT₁ = 78.4 - 300 = -221.6 K
q₁ = 0.782 × 90 × (-221.6) = -15 600 J = -15.60 kJ
q₂ = 0.782 × (-38.56) = -30.15 kJ
ΔT = 25.0 - 78.4 = -53.4 K
q₃ = 0.782 × 115 × (-53.4) = -4802 J = 4.802 kJ
q = -15.60 - 53.4 - 4.802 = -50.6 kJ
You must remove [tex]\text{50.6 kJ}[/tex] of heat to convert the vapour to a gas.
