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Thermodynamics: Water vapor vessel plunged into water ice bath

  1. Jan 11, 2009 #1
    1. The problem statement, all variables and given/known data

    A closed rigid vessel contains 0.1 kg of water vapor (steam) at 1 bar and 200 C. The vessel is plunged into a bath containing water and ice. After the vessel cools and reaches equilibrium with the bath, the bath still contains water and ice.

    a. What is the volume contained by the vessel initially? Finally?

    b. As the vessel cools, at what temperature and pressure does condensation begin?

    c. What is the final temperature and pressure in the vessel?

    d. What is the quality of the water in the final state?


    2. Relevant equations

    I am not sure how to even start. Any push in the right direction would be great.

    Is this an isobaric process?

    3. The attempt at a solution

    Initial V_i = v*m = 2.172*0.1 = 0.2172, where v = specific volume(found in a table)

    If it is an isobaric process, then Q = [change in]U + W = n(c_v + R)[change in]T
     
  2. jcsd
  3. Jan 12, 2009 #2

    Mapes

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    If the vessel is closed and rigid, what variable is absolutely kept constant?
     
  4. Jan 12, 2009 #3
    It would appear that Volume V, mass m, and thus specific volume v are kept constant
     
  5. Jan 12, 2009 #4

    Mapes

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    Agreed.
     
  6. Jan 13, 2009 #5
    So that answers part a, but what about the rest. What is it about condensation (gas to liquid) that helps m solve b?
     
  7. Jan 13, 2009 #6

    Mapes

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    Steam tables.
     
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