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!

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

    User Avatar
    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.

    AM
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Carnot's Engines
  1. Carnot engine (Replies: 7)

  2. Carnot Engine (Replies: 2)

  3. Carnot Engine (Replies: 4)

Loading...