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: Compare final T of reversible/irreversible equations

  1. Oct 5, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose CA = CB = C in equations 2.93 and 2.109 (shown below). Compare the form of expressions for the final temperature.

    2. Relevant equations
    (2.93) [itex]T = \frac{C_{A}T_{A} + C_{B}T_{B}}{C_{A} + C_{B}}[/itex]
    (2.109) [itex]T = T_{A}^{\frac{C_{A}}{(C_{A}+C_{B})}} T_{B}^{\frac{C_{B}}{(C_{A}+C_{B})}}[/itex]

    (2.93) is for an irreversible process
    (2.109) is for a reversible process
    T is the final equilibrium temperature when two bodies are put into thermal contact.
    C is specific heat capacity

    3. The attempt at a solution
    The problem is that I don't know where to start. I'm told in the sentence before equation (2.93) that in that equation, TA > TB. Other than that, I'm not sure exactly how or what I'm supposed to be deducing about the final temperature just from looking at the two equations. I'm told the answer is that the final temperature will be TA but have no idea why. What's my first step? How should I start thinking about it?

    I'd appreciate any guidance. Thanks :)
  2. jcsd
  3. Oct 5, 2011 #2
    I'd say that...
    If we suppose CA = CB = C, as you're told to, start by just replacing both CA and CB with C, and simplify the formulas:

    (2.93) [itex]T = \frac{C T_{A} + CT_{B}}{2C}= \frac{C( T_{A} + T_{B})}{2C}= \frac{ T_{A} + T_{B}}{C}[/itex].

    (2.109) [itex]T = T_{A}^{\frac{C}{(2C)}} T_{B}^{\frac{C}{(2C)}}=T_{A} \ ^{\frac{1}{C}} \cdot T_{B} \ ^{\frac{1}{C}}[/itex]

    Do you have any further ideas now ?

  4. Oct 5, 2011 #3
    If by [itex]\frac{T_{A} + T_{B}}{C}[/itex], you mean [itex]\frac{T_A} + T_{B}}{2}[/itex] for the first one... and replace the C by 2 again in the second one, then yes I realize I can do that. I didn't think to use it here but I did use that in the next question in the homework assignment.. I suppose it would have made more sense to do it here, thanks.

    But no, I don't have any further ideas from that. In the next question on the homework, we're given values for TA and TB and I find that plugging them into 2.93 yields a higher resulting value than plugging the same numbers into 2.109, so that tells me something. But without numbers and just looking at the equations above reduced so the C is gone... am I just supposed to be able to deduce from that that one process is higher than the other simply based on math?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook