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Maximum Mass You Could Boil with 1000 J of Heat (Thermodynamics)

  1. Nov 30, 2013 #1
    1. The problem statement, all variables and given/known data
    What is the maximum mass of ethyl alcohol you could boil with 1000J of heat, starting from 18∘C?


    2. Relevant equations
    Q=mcΔT

    c ethyl alcohol = 2400 J/(kg*K)
    T boil ethyl alcohol = 78 C = 351 K
    T starting = 18 C = 291 K

    3. The attempt at a solution

    m=Q/(c*ΔT)

    m= (2400 J)/[(2400 J/kg*K)(351 K -291 K)] = .00694 kg (or 6.94E-3 kg)

    It tells me this is wrong. I am probably overlooking something quite obvious. Thanks :)
     
  2. jcsd
  3. Nov 30, 2013 #2

    Borek

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    So far you calculates mass than can be heated to the boiling point. It has not even started to boil, not to mention boiling away.
     
  4. Nov 30, 2013 #3

    haruspex

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    Well, the OP doesn't say 'boil away'. When I boil water, I bring it to the boil - I don't keep boiling until there's none left. Maybe it's not been quoted correctly.
     
  5. Nov 30, 2013 #4

    Borek

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    My first idea was it means "boil away", as opposed to "bring to boil". But after your comment I see the wording as ambiguous.

    Can be my English fails me and I am completely off.
     
  6. Nov 30, 2013 #5

    SteamKing

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    Just a nit pick: a temperature difference of 1 degree centigrade is the same temperature difference of 1 degree Kelvin. (You don't need to convert Celsius temps. to Kelvin temps. if you are interested only in temperature differences.)
     
  7. Nov 30, 2013 #6
    Borek, you were right (and your English is just fine! :) ). Here is how you get the right answer:

    L vaporization ethyl = 8.79 x 10^5 J/kg

    Qtotal = Q1 + Q2 = (Energy to heat mass up to boiling point) + (Energy used to boil all of mass)

    1000 J = mcΔT + mLv

    1000 J = m(cΔT + Lv)

    m = 1000 J/(cΔT + Lv)

    m = 1000 J/[(2400 J/kg*K)(351 K - 291 K) + 8.79 x 10^5 J/kg)]

    m = 9.78 x 10^-4 kg = .978 g
     
    Last edited: Nov 30, 2013
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