How much heat is required to change 1.75 L of Ethyl Alcohol (C2H6O) at -50.0°C to a gas at its boiling point?
V = 1.75 L
Ti = -50°C
density = 0.789 g/cm3
boiling point = 78°C
specific heat (c)= 2400 J/kg*C°
Heat of Vaporization (Lv) = 850*103 J/kg
density m = ρV
specific heat Q = mcΔT
latent heat Q = mLv
The Attempt at a Solution
First I determined the mass of the Ethyl Alcohol
m = ρV = (0.789 g/cm3) (1750 cm3) = 1380.75 g = 1.38075 kg
Then I solve for QNET
QNET = (mcΔT)l + mLv
QNET = [1.38 kg (2400 J/kg*C°) (78°C - (-50°C)] + [1.38 kg (850*103 J/kg)] = 1.60 * 106 J
According to the answer sheet, the answer is 4.24*105 J
What am I doing wrong?
Thank you in advance! Any and all help is appreciated!