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Water vapor Thermodynamics

  1. Mar 7, 2014 #1
    1. The problem statement, all variables and given/known data

    How much water vapor (100 ° C) should be added to 120 g of ice (0 ° C) so that the final system has a final temperature of 35° C.


    2. Relevant equations
    Qvapor = lvapor * mvapor during condensation, where lvapor =2260 kJ/kg
    Q1 = cwater*mvapor*ΔT during cooling, and cwater = 4,19 kJ/kg

    Qmelting =lmelting * mice during melting, and lmelting =333 kJ/Kg
    Q2 = cwater*mice*ΔT during heating


    3. The attempt at a solution

    Energy which is emitted by the vapor is given by: Qvapor + Q1 = 2260m + 4,19*m*100

    Energy taken up/used by the ice is given by: Qmelting + Q2 =
    333*0,120 + 4,19*0,120*35

    I then set energy emitted = energy taken up by the ice and solved for m

    2260m + 4,19m*100 = 333*0,120 + 4,19*0,120*35
    m= 0,0215 kg or 21,5 g.

    Do you think this is the correct way to solve this problem?
     
  2. jcsd
  3. Mar 7, 2014 #2

    DrClaude

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    Your approach is correct, but your numbers are not:
     
  4. Mar 7, 2014 #3
    Ok. I don´t see which number could possibly be wrong? The temperature change is + 100 degrees, 4,19 kJ/kg is cwater, and 2260 is lvapor (according to my somewhat dated textbook).
     
  5. Mar 7, 2014 #4

    DrClaude

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    That's the number that's incorrect.
     
  6. Mar 7, 2014 #5
    I tried calculating m using both -100 degrees and +135 degrees, and still got the wrong answer.
     
  7. Mar 7, 2014 #6

    DrClaude

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    Don't plug in numbers randomly.

    What does ##\Delta T## stand for in the equation ##Q = c m \Delta T##?
     
  8. Mar 7, 2014 #7
    ΔT stands for Tfinal - Tinitial, but using Tfinal = 35 and Tinitial= 100 gives a ΔT = -65, and a m = 0,0289 g, which is also the wrong answer.
     
  9. Mar 7, 2014 #8

    BvU

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    Could it be you forgot the 4,19*0,120*35 ? Or the 4,19*m*65 ? My guess: the latter!
     
  10. Mar 7, 2014 #9

    DrClaude

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    Careful with the sign. You have defined ##Q_1## as the amount of energy given by the condensed steam to the final mixture.
     
  11. Mar 7, 2014 #10
    I still don´t get it. Did you mean the negative sign? I tried using + 65, but the answer still turns out wrong (m = .00695). The answer is supposed to be 22,7 g.
     
  12. Mar 7, 2014 #11

    DrClaude

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    Solving 2260m + 4,19m*65 = 333*0,120 + 4,19*0,120*35 for m, you should indeed get m=0.0227 kg.
     
  13. Mar 7, 2014 #12
    I must´ve pressed something wrong on my calculator. Was a bit hasty...as usual. Now I got the right answer though. Thanks!
     
  14. Mar 7, 2014 #13

    BvU

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    Fr0m the 28.9 g (the 0.0289 was kg) it follows you forgot the 4,19*m*65. Elementary, my dear Watson.
     
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