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Specific & Latent Heat

  1. Nov 4, 2011 #1
    1. The problem statement, all variables and given/known data

    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

    2. Relevant equations

    density m = ρV

    specific heat Q = mcΔT

    latent heat Q = mLv


    3. 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!
     
  2. jcsd
  3. Nov 4, 2011 #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.
     
  4. Nov 4, 2011 #3
    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?
     
  5. Nov 4, 2011 #4

    SteamKing

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    Only your professor knows for sure.
     
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