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!

Entropy change of a hot, falling object

  1. Aug 14, 2011 #1
    1. The problem statement, all variables and given/known data

    A shipyard worker drops a hot steel rivet (mass 125g, temperature 350 degrees C)
    into a river at temperature 5 degrees C, a distance 30m below. Stating any assumptions
    you make, calculate the entropy change of the universe as a result of this event.
    (specific heat capacity of steel ~0.4 J g-1 K-1).


    2. Relevant equations
    1.jpg
    2.jpg

    3. The attempt at a solution

    3.jpg

    4.jpg

    5.jpg

    6.jpg

    If anybody could tell me if I'm on the right lines at all or what the heck to do with the fact that the rivet's being dropped from 30m I'd greatly appreciate it. Thanks!
     
    Last edited: Aug 14, 2011
  2. jcsd
  3. Aug 14, 2011 #2

    dynamicsolo

    User Avatar
    Homework Helper

    You have traced the change in entropy of the rivet, but I believe there are two more things you need to look at in the "universe" (essentially the rivet-Earth system). How much heat has been added to the river by the change in gravitational potential energy in the rivet-Earth system? Also, the river absorbs the heat of the rivet with virtually no temperature change (it's large enough to serve as a "heat reservoir"), so what is the change in the entropy of the river from that?

    So there is the entropy change of the cooling rivet plus the entropy change from the change in mechanical energy plus the entropy change from the heat transfer from rivet to river.
     
  4. Aug 14, 2011 #3
    Thanks for the reply.

    So, as I see it, so far I've worked out the decrease in the entropy of the rivet.

    The heat flowing into the reservoir from the rivet cooling will be the same as the heat loss from the rivet:

    1.jpg

    That gives me 17250 J.

    Then, if we assume the temperature of the reservoir doesn't change, then, using the temperature of the river in Kelvins for T:

    7.jpg

    That gives me an increase of entropy in the river as 62.1 JK-1

    Finally, the change in gravitational potential energy is:

    8.jpg

    If we assume that this is all converted to heat in the river we can work out a second component for the increase in entropy. I get S = 0.13 JK-1.


    Overall this gives me a net increase in the entropy of the universe as S = 21.93 JK-1.

    Is this the right track or have I completely messed up?
     
  5. Aug 14, 2011 #4

    dynamicsolo

    User Avatar
    Homework Helper

    I agree with those results to the first decimal place (since I rounded off only at the end), so we get a net increase in entropy for the "universe" of 21.9 J/K .

    It is a reasonable safe assumption that just about all of the potential energy change is transferred into heating the river. The "splash" at impact displaces water and ejects some of it upward (briefly), as well as producing (a minute amount of) acoustical energy. But since practically all the water falls back into the river, that mechanical energy winds up heating the river after all. (We can extend "the universe" to include the air over the river, but the tiny amounts of water scattered as microdroplets and vapor and of sound energy probably contribute additions to entropy scarcely worth pursuing...)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Entropy change of a hot, falling object
  1. Change of entropy (Replies: 0)

  2. Change in entropy (Replies: 2)

  3. Entropy changes (Replies: 1)

Loading...