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Thermodynamics and entropy of reservoirs

  1. Mar 1, 2010 #1
    1. Two identical bodies at constant heat capacity and temperatures T1 and T2 are used as reservoirs for a heat engine. If the bodies remain at constant pressure and undergo no change of phase, show that the amount of work obtainable is: W = cp(T1+T2-2Tf) where Tf is the final temperature attained by both bodies and cp is the heat capacity at constant pressure.



    2. cp = heat capacity at constant pressure
    ΔS = -Q/Tf = -cp(Tf - Ti)/Tf (entropy change in reservoir)




    3. I'm having a bit of trouble with this problem. I figured I'd start by making a T-S diagram using the T1 and T2 temperatures where T2 > T1. Then the net heat (Q) would be ΔS(T2-T1). Since ΔU = 0 for a cycle the net heat would have to equal the net work. So, W = ΔS(T2-T1). To find ΔS I used the equation for a heat flow into a reservoir which is ΔS = -cp(Tf - Ti)/Tf. So for both reservoirs I have ΔS1 = -cp(Tf - T1)/Tf and ΔS2 = -cp(Tf - T2)/Tf. Then I said: ΔS = ΔS1 + ΔS2. I find this to be: ΔS = (cp/Tf)(T1 + T2 -2Tf). Which looks similar to what I'm supposed to be getting. But it can't be right because I still have to multiply that by (T2 - T1) to get the work obtainable from the above equation. Can someone please help me out with this? What am I doing wrong? Thanks!
     
  2. jcsd
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