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Entropy change of universe

  1. Jan 4, 2012 #1
    1. The problem statement, all variables and given/known data
    A system absorbs 200 J of heat from a reservoir at 300 K and rejects 100 J to a reservoir at 200 K as it moves from state A to B. During this "NON-QUASI-STATIC" process 50 J of work is done by the system. What is Entropy change of universe ?

    3. The attempt at a solution

    My intuition says, it must be some value greater than zero.
     
  2. jcsd
  3. Jan 4, 2012 #2

    Andrew Mason

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    Start with the definition of ΔS: ΔS = ∫dQrev/T

    You have to break the entropy change into four parts:

    1. the heat flow out of the hot reservoir to the system changes the entropy of the hot reservoir.

    2. the heat flow into the system changes the entropy of the system.

    3. the heat flow out of the system to the cold reservoir changes the entropy of the system

    4. the heat flow into the cold reservoir changes the entropy of the cold reservoir.

    For each of the 4 parts you have to determine the ΔS (be careful about the sign). Then add them up to find the total entropy change.

    Hint: 1 and 4 are easy. But in order to find the ΔS of the system you have to know the heat flow Q and the temperature at which the heat flow occurs. Does the problem provide you with more information?

    AM
     
  4. Jan 5, 2012 #3
    You are right. I also thought the same way but I have some slight confusions:
    1) The question was asked for "NON-QUASI-STATIC" and "QUASI-STATIC". The QUASI-STATIC part I was able to answer as in that case 2 , 4 will be zero. So how does "NON-QUASI-STATIC" entropy calculation differ from "QUASI-STATIC" one.

    2) For getting the value of 2, 4 I will need the temperature at which the system(2,4) is working which is not given in problem.
     
  5. Jan 5, 2012 #4

    Andrew Mason

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    I don't think 2 and 4 will be 0 in any case. If processes are quasi-static, the flow of heat into the system will be at a temperature arbitrarily close to the reservoirs. But since there is heat flow in both stages, there is an entropy change of the system for each stage. The two could sum to zero if there was a complete cycle involved. But in this case you can see that the system experiences a net increase in internal energy, which means that we are dealing with only part of the cycle.

    That was my thought as well. You don't need to be given the temperature if you can work it out but there is not sufficient information here to do that.

    Maybe the answer is just: ΔS > 0.

    AM
     
  6. Jan 5, 2012 #5
    yes the answer given is just ΔS > 0. I just wanted to know the figures.
     
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