How Much Heat is Needed to Vaporize Ethyl Alcohol from -50°C to Boiling Point?

[format1998]

Homework Statement

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?

Ethyl Alcohol
V = 1.75 L
T_i = -50°C

46 g/mol
density = 0.789 g/cm³
boiling point = 78°C
specific heat (c)= 2400 J/kg*C°
Heat of Vaporization (L_v) = 850*10³ J/kg

Homework Equations

density m = ρV

specific heat Q = mcΔT

latent heat Q = mL_v

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 Q_NET

Q_NET = (mcΔT)_l + mL_v

Q_NET = [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!

 

[SteamKing]

The answer sheet apparently has only accounted for the heat required to raise the temperature of the liquid ethanol from -50C to 78C.

 

[format1998]

So, did I do it correctly? It did say that "to a gas" . From my understanding of the problem as worded, I have to account for the amount of heat required to bring it to its boiling point, as well as the amount of heat required to turn it into a gas, right? Or did I do it incorrectly and I'm only suppose to calculate the amount of heat needed to bring it to the boiling point?

 

[SteamKing]

Only your professor knows for sure.