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Homework Help: Thermodynamics - Carnot Cycle-esque question?

  1. Oct 11, 2015 #1
    1. The problem statement, all variables and given/known data
    A reversible heat engine produces work from the temperature difference that exists between a mass of m = 9 kg of an ideal gas (cv = 716 J/kgK, R = 287 J/kgK) in a rigid container and a heat reservoir at THR = 285 K. The only heat transfer interaction experienced by the container is with the heat engine. The gas in the container is initially at temperature T1 = 772 K and pressure P1 = 106 Pa. The reversible heat engine operates until the ideal gas and the heat reservoir are in thermal equilibirum (state 2). If the efficiency of the heat engine for process 1-2 can be defined as W/(QH)HE, calculate its value.Give your answer as a percent (without the % sign).

    2. Relevant equations

    • efficiency = (QH - QL)/QH = W/QH
    • QL = MCvTHRln(T2/T1)
    • E2 - E1 = Q1-2 - W1-2
    • S2 - S1 = QL/ THR - change in entropy from or to a heat reservoir
    • QH = MCv(T2 - T1)

    3. The attempt at a solution

    If efficiency = (QH - QL)/QH

    = 1 - (QL/QH)
    = 1 - ((THRln(T2/T1))/(T2-T1))

    I get an answer of 41.683538% but unsure as to what value to use for T2? Would I just use THR (Temp of heat reservoir) seeing as it works at thermal equilibrium? I used the temperature of the reservoir as the T2 value in my solution but not sure if this is correct?
  2. jcsd
  3. Oct 12, 2015 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Somewhat amazingly I got the same answer! (I mean the final expression, not the % number. I don't do numbers!). Vote of confidence, anyway!
    How can T2 be anything but the reservoir temperature THR? The reservoir has infinite heat capacity so at the end (i.e. "state 2") everything has to reach T = THR.
    So far as I can see you got everything right. Congrats!
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