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Chem Engr thermo prob

  1. Nov 2, 2012 #1
    1. The problem statement, all variables and given/known data
    A young engineer notices in her plant that 1-kg blocks of brick are routinely removed from a 800 ıC oven and are let to cool in air at 25 ıC. Conscious about cost-cutting and efficiency, she wonders whether some work could be recovered from this process. Calculate the maximum amount of work that could be obtained. The CP of brick is 0.9 kJ/kg K. Can you come up with devices that could extract this work?


    2. Relevant equations
    Q=ΔH, ΔU=W+Q, η=1-Tl/Th,


    3. The attempt at a solution
    I know that a reversible process is the best option do so. This is most likely going to be done with a carnot cycle to extract the work. I'm not sure if the T of the air given is going to be any help. The bricks being 1kg can be relevant to how much thermal energy it has. Don't do this with Q=mCpΔT because we used in basic chemistry thermo so he doesn't want us to use that in this class.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 2, 2012 #2
    It isn't clear to me why W=m*Cp*ΔT isn't appropriate. Assuming the air is a large heat sink of constant temperature the brick will cool to air temperature. That equation should be appropriate.
     
  4. Nov 2, 2012 #3
    That equation would work for the brick but not air or water since Cp is actually calculated from tabulated values which depend on temperature and pressure.
     
  5. Nov 6, 2012 #4

    rude man

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    Homework Helper
    Gold Member

    Carnot is out since that cycle operates between two constant temperatures. Your T2 is constant at 25C but your T1 is not.

    Reversible cycle, yes. Be guided by the 1st and 2nd laws. While the entropy increase S2 of the cooling medium is just Q2/T, T = 25 + 273, S1 will be an integral over the temp. drop of the brick and will of course be negative.

    So Q1 = Q2 + W but S2 + S1 > 0.
     
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