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Entropy of Macroscopic Collision

  1. Apr 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Two cars each of mass 1410 kg traveling at 91 km/h in opposite directions collide head-on and come to a disastrous halt; see Figure P.38. Assume that the surrounding air and ground temperature remains fixed at 18°C. Calculate the total entropy change of the cars and environment system that results from the collision.



    2. Relevant equations



    3. The attempt at a solution
    I understand how to do most problems involving gases or mixing liquids or engines etc. which involve usually microscopic untangible processes but I'm not sure what to make of this. As soon I see collision I think momentum but I don't know where to begin to relate collisions to the formulas and equations in my textbook which deal with volumes, pressures, and mols to this. any hint as a jumpstart would be appreciated! thank you
     
  2. jcsd
  3. Apr 22, 2009 #2
    S(total)=S(car) + S(car 2) + S(environment) where the values are changes in entropy

    delta-S=delta-q/T, where delta-q=heat and T is a constant temperature?

    how do I relate velocities and masses to heat?
     
  4. Apr 22, 2009 #3
    I am really just guessing here, but perhaps all of that kinetic energy is assumed to be converted to heat energy ?
     
  5. Apr 22, 2009 #4
    wait since
    Q + W'=delta-U, where W'=work done by nonconservative/other forces i.e. the work done by the surroundings on the system

    W'=(delta-KE) + (delta-PE)...in this case PEi=0 and PEf=0
    thus W'=1/2mvf^2- 1/2mvi^2 for both cars

    I know just intuivitely that when a car crashes the KE of both cars is translated into heat energy and some into sound and even light (which revert back to heat)...
     
  6. Apr 22, 2009 #5
    oh haha ok maybe that's the assumption to make because otherwise calculating the delta-U would be difficult or is the delta-U of the entire system=0 i.e. unchanged?
     
  7. Apr 22, 2009 #6
    well wouldn't it be? it's an isothermal process?
     
  8. Apr 22, 2009 #7
    uhh nevermind I got it
    took a little reasoning but not too difficult
    isothermal so delta-T=0 i.e. delta-U=0
    Q + W'=U-->Q=-W'
    W'=(delta-KE)
    delta-S= Q/T

    thanks
     
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