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Work of a heat engine

  1. Dec 6, 2006 #1
    1. The problem statement, all variables and given/known data

    An engine works at 26% efficiency. The engine raises a 6-kg crate from rest to a vertical height of 11 m, at which point the crate has a speed of 5 m/s. How much heat input is required for this engine?

    2. Relevant equations

    e=work output / heat input = W/Qin

    3. The attempt at a solution

    I assume the velocity is insignificant since W=Fd; W=(6*.981)*(11)= 647.46J

    Qin=.0004016 J

    the value for calculated heat is wayy too low; where did I screw up? Thanks
  2. jcsd
  3. Dec 7, 2006 #2

    Andrew Mason

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    For starters, you have calculated 1/Qin not Qin.

    You are also ignoring the work that is required to give the load the kinetic energy. Add that to your calculation for W and then do the algebra properly and you will be fine.

  4. Dec 7, 2006 #3
    for KE do I just say that delta E = 1/2mv^2 + mgh = W = eQ ?

    .5(6kg)(5^2)= 75J

    mgh = 647.46J
    What do i do with these two values?
  5. Dec 7, 2006 #4

    Andrew Mason

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    Looks good to me.

    Why not put them into your equation and find Q?

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