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Homework Help: Need help with a question involving heat engines, the carnot cycle and entropy.

  1. Jun 21, 2010 #1
    question.jpg



    3. The attempt at a solution
    For
    a) I imagine it is easier to control the temperature of the hot reservoir because you could use a constant heat source e.g. bunsen burner.

    c) melting/freezing water.

    d) Entropy of the environment increases and the entropy of the universe always increases.

    Any help with b) or e) would be massively appreciated
     
  2. jcsd
  3. Jun 21, 2010 #2
    b)

    [tex]\eta=\frac{W}{Q_{in}}=1-\frac{|Q_{out}|}{Q_{in}}=1-\frac{T_{min}}{T_{max}}[/tex]

    If you raise the temp of hot reservoir by [tex]\Delta T[/tex] you'll have: [tex]T_{max}'=T_{max}+\Delta T[/tex], and if you lower the cold reservoir by the same amount you'll have:
    [tex]T_{min}'=T_{min}-\Delta T[/tex].
    So the new efficiencies will be:
    [tex]\eta_1=1-\frac{T_{min}}{T_{max}'}=\frac{T_{max}-T_{min}+\Delta T}{T_{max}+\Delta T}[/tex]
    and
    [tex]\eta_2=1-\frac{T_{min}'}{T_{max}}=\frac{T_{max}-T_{min}+\Delta T}{T_{max}}[/tex]
    If you rearrange those two:
    [tex]\eta_1(T_{max}+\Delta T)=\eta_2T_{max}[/tex]

    So it is obvious that [tex]\eta_1>\eta_2[/tex] the efficiency of Carnot cycle is greater when you take away the temp from cold reservoir. :)
     
  4. Jun 25, 2010 #3
    thats great, thanks

    anybody have any idea for e), even just a prod in the right direction would be excellent.
     
  5. Jun 25, 2010 #4
    For e), the condition for minimum work to be achieved is the coefficient of performance [tex]\eta = \frac{dQ_{cool}}{dA} = \frac{T_{cool}}{T_{hot}-T_{cool}}[/tex]
    Where:
    [tex]dA[/tex] is the work needed to lower the temperature by [tex]dT[/tex]
    [tex]dQ_{cool}[/tex] is the heat taken from the water/ice.
    [tex]T_{hot}[/tex] is the temperature of the sink, which is the air in this problem.
    [tex]T_{cool}[/tex] is the temperature of the source, which is the water/ice. Notice that [tex]T_{cool}[/tex] varies during the process.

    There is another equation relating [tex]dQ_{cool}[/tex] with [tex]dT[/tex] (actually 2 equations, each corresponds to one process). Compute the integral, and you will obtain the total work needed.
     
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