1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted