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Homework Help: Thermodynamics question

  1. Sep 24, 2014 #1
    1. The problem statement, all variables and given/known data
    There are 2 kg of water in an aluminium pot at the temperature of 30C. What is the mass of water vapor (100C) needed to heat the pot with water to 60C? Mass of the pot = 0.5kg

    c(water)=4200 J/kg*C
    c(aluminium)=890 J/kg*C
    T(water vapor)=100*C
    c(water vapor)= 2020 J/kg*C
    m(water vapor)=?

    2. The attempt at a solution


    I supposed that
    But then Q would be less than 0 since the temperature difference is less than 0. What is the right way to write this last equation? And the final equation to find the mass of water vapor?
    Thank you!
  2. jcsd
  3. Sep 24, 2014 #2
    You write Q=c(vapor)*m(vapor)*(T2-T(vapor)), but the formulation says What is the mass of water vapor (100 C), not taking account of the condensation of vapor water. This is, we use m Kg of vapor water to heat up the Al pot, which transfer heat to the water inside it. I'm thinking more of Q=c(vapor)*m(vapor)*(100)

    On the other hand, you are supposing that both Al pot and the water inside it will reach the same temperature (T2). That would be the case according to zeroth law of thermodynamics, in thermal equilibrium (for large t, time). One could put it like: 1) Final T2 in the pot for the water to reach 60c? and 2) mass of vapor water to heat the pot to that T2 calculated in 1. This is my guess...
  4. Sep 24, 2014 #3
    At best, this is a poorly (imprecisely) worded problem. I think what they are asking is how much water vapor has to condense from vapor to liquid water at 100 C to supply the heat required to heat up the pot and its contents to 60C. So, how much heat does it take to heat up the contents?

  5. Sep 25, 2014 #4


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    Maybe, but there are at least two other reasonable interpretations. There is still useful heat available in condensate at 100C, so could take everyting as finishing at 60C. Better still, the initial condensate is useful down to almost 30C, so the minimum vapor required is less again.
  6. Sep 25, 2014 #5
    Yes. I too thought about the possibility of taking the condensate down to 60, and this is certainly also a valid interpretation. As far as taking the condensate all the way down to 30, it doesn't seem like this can be done because of 2nd law constraints. Can you think of a way?

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