1. Not finding help here? Sign up for a free 30min 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!

Cyclic processes

  1. Aug 2, 2009 #1
    1. The problem statement, all variables and given/known data

    Over several cycles, a refrigerator does 1.51 x 10^4 J of work on the refrigerant. The refrigerant in turn removes 7.55 x 10^4 J as heat from the air inside the refrigerator.

    a. how much energy is transferred as heat to the outside air?
    b. what is the net change in the internal energy of the refrigerant?
    c. what is the amount of work done on the air inside the refrigerator?
    d. what is the net change in the internal energy of the air inside the refrigerator?


    2. Relevant equations

    U=Internal Energy
    Q=Heat
    W=Work

    ΔU = Q-W
    ΔUnet = 0
    Qnet=Wnet


    3. The attempt at a solution

    I'm trying to teach myself physics, so I think I'm getting really confused here. First of all, I get mixed up on when to put negative signs in front of numbers or if I even have to in cyclic processes.

    I got the right answers (given by my online school) but it may have just been luck.

    **I only need help with a and d and somewhat c


    a. -7.55 x 10^4 = (1.51 x 10^4 - x)
    -9.06 x 10^4= -x
    x=9.06 x 10^4 J

    Did i do that right at all?

    b. I know ΔUnet=0 J in cyclic processes

    c. no work done b/c ΔV=0 (how would i know that this is so?)

    d. the answer is ΔU= -7.55 x 10^4 but I'm not sure how to mathematically get it
     
    Last edited: Aug 2, 2009
  2. jcsd
  3. Aug 2, 2009 #2
    a) If you remove heat from the fridge, you have to dump it outside. It should just be 7.55*10^4 J

    d) Internal energy is just a function of temperature. You remove heat, thus its considered a negative change.
     
  4. Aug 3, 2009 #3

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    c) When there is no volume change, there is no displacement of anything due to the force (pressure times area) exerted by the gas. Therefore the gas does no work when its volume is constant.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Cyclic processes
Loading...