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Homework Help: Carnot's Engines

  1. Dec 4, 2011 #1
    A lawnmower engine with an efficiency of 0.22 rejects 9900 J of heat every second. What is the magnitude of the work that the engine does in one second?

    e = 0.22
    Qc = 9900 J/s

    e = W/Qh
    Qh = Qc + W

    e = W / Qc + W
    0.22 = W / 9900 + W
    0.22(9900 + W) = W
    2178 + 0.22W = W
    W = 2792.31 J/s?

    I checked my answer on cramster.com but they have something different, but I believe my answer to be right. Can anybody verify if I am doing this correctly or not? Thank you.
  2. jcsd
  3. Dec 4, 2011 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    Your answer should be expressed in the correct number of significant figures. You may want to simplify your answer algebraically before plugging in the numbers:

    Qh = Qc/(1-e)

    W = eQh = e(Qc/(1-e))

    Also, you have to put your answer for W in the correct units. J/s is a unit of power.

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