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Thermo - mixing water in a container.

  1. Nov 15, 2008 #1
    A partition separates the two compartments of a vessel. The vessel is submerged in a constant temperature bath at 200C. Compartment A contains 2.00 kg of water at 200C, 50.00 bar and compartment B contains 0.983 kg of water at 200C, 1.00 bar. The partition is removed and the water in both compartments fills up the entire vessel. Determine the final pressure, and the heat transfer that results from the mixing process, if the final temperature
    is 200C.

    I'm working on a practice test for my thermoclass, and I've been stuck on this problem for a while... What i've tried so far...

    Since this is water, I assumed that you could use density and just say that the volume of the compartments A and B are 2 m^3 and .983 m^3 respectively. I first assumed that it would have something to do with work, but W = PV isn't valid here because this isn't a gas. I'm trying Enthalpy changes, but i'm not given an internal energy for the equation H = U + PV. I am however given several charts, Saturated Steam Tables for temperature and pressure, and a table of Superheated Steam properties.

    So I guess I don't really know how to start finding the Pressure. Any hints to lead me to that?
  2. jcsd
  3. Nov 16, 2008 #2


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    Hi Thyferra,
    Here's an outline of the solution:

    Let's consider the control volume to be the sum of the two volumes. There's a partition in the middle of this control volume. When that partition is removed, there is no work being done by this control volume, so dU = 0.

    Can you determine the internal energy in each of the two sections? m u = U so use the steam tables to determine u and multiply by m to find U. The total internal energy (U1 + U2) for the entire volume after the partition is removed does not change, so now you can find u from the same equation: u = (U1 + U2) / (m1 + m2).

    Now, from the steam tables, you should be able to locate the state of the water given u and density. That state is completely defined by those two variables, so you should be able to look up the temperature and pressure from that state. The temperature is assumed to be something other than 200 C.

    Once you have the state after removal of the partition, you need to determine how much heat is required to raise/lower the temperature to 200 C. Application of the first law gives dU = Q or Uf - Ui = Q. You have Ui so you need Uf to determine Q.

    You can find Uf using your steam tables because you know certain things about the final state, you know temperature and density. From temperature and density, you can find the state and thus find uf. Therefore, you can calulate Uf = uf mf

    Hope that helps.
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