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: Heat transfer in an Isothermal process?

  1. Oct 6, 2012 #1
    1. The problem statement, all variables and given/known data
    I have a general concept inquiry. I was doing some thermal homework, and calculating values of Q in a given cycle. One process was isothermal, however it turned out that Q was not zero, causing me to wonder: How can there be a heat transfer in an isothermal process?

    2. Relevant equations

    3. The attempt at a solution
    In an isothermal process the temperature does not change. Looking at the formula, Q=mcΔT, if there is no change in temperature, one would therefore find that Q=O. However, if you look at it from conservation of energy, (ΔU=Q-W), the ΔU is zero for an isothermal process, making Q=W, and thereby concluding that there is a heat transfer. Thus I am confused. Could someone please provide some insight into this?
  2. jcsd
  3. Oct 7, 2012 #2
    The energy of a given amount of gas is determined solely by its temperature. So if you add heat to it and want to keep its temperature constant, you will have to let the gas do the work equal to the amount of the added heat. That's what Q = W means.
  4. Oct 7, 2012 #3


    User Avatar
    Homework Helper

    Q=mcΔT is valid with different c-s -specific heat capacity values- in different processes. Remember, c=(1/m)Q/ΔT while an other state-parameter stays constant. For a constant-volume process, it is cv. In case of an isobaric process, it is cp , and it is greater than cv as part of the heat covers the work done by the gas during expansion, according to ΔU=Q-W. . The more isotherm is a process, the more is the part of work. c is not defined for the case ΔT=0. Or you can imagine that it is infinite.

    Last edited: Oct 7, 2012
  5. Oct 7, 2012 #4
    I see what you are saying; that makes sense to me now. Thank you both for your explanations :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook