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Specific & Latent Heat

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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
Ti = -50°C

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

Homework Equations



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!
 

Answers and Replies

  • #2
SteamKing
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The answer sheet apparently has only accounted for the heat required to raise the temperature of the liquid ethanol from -50C to 78C.
 
  • #3
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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?
 
  • #4
SteamKing
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Only your professor knows for sure.
 

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