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Thermodynamics Change in Internal Energy?

  1. Feb 25, 2017 #1
    1. The problem statement, all variables and given/known data
    A closed, rigid tank contains 2 kg of water, initially a two phase liquid–vapor mixture at T1 = 70°C. Heat transfer occurs until the tank contains only saturated vapor at T2 = 120C.
    Determine the heat transfer for the process, in kJ.
    answer choices:
    3701kJ
    119.4kJ
    4835kJ
    1558kJ
    2. Relevant equations
    Steam Tables
    ΔU=Q-W(W=0)

    3. The attempt at a solution
    Since we know T1, we can use the tables to identify the specific internal energy of saturated liquid and gas and add them together to find u1. We can also do the same with T2 but neglecting the specific internal energy of fluid since it is only a saturated vapor at this point. Subtract u2 from u1 then multiply by the mass. Since it is a rigid tank we can also neglect work done. Here is my work:
    Q=m(u2-u1)
    Q=2kg(2529.3kJ/kg - (292.95kJ/kg+2469.6kJ/kg))
    Q=-466.5kJ (this answer is not given as one of the choices so i am assuming that it is wrong)
     
  2. jcsd
  3. Feb 25, 2017 #2
    Nevermind i figured it out. I did this completely wrong
     
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