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Homework Help: Carnot process question

  1. Jan 28, 2010 #1
    1. The problem statement, all variables and given/known data
    Four identical blocks of copper, of mass 20Kg each, are 'perfectly' isolated from the environment. One is kept at a temperature of 573K and the other three at 288K. The heat capacity of copper is 375 J/K/Kg. Assume a carnot process is being used to exploit the temperature differences between the se 4 blocks of copper. How much work can maximally be extracted?
    (Hint: the temperature of the copper blocks change continuously from initial to final temperature. Remember that a carnot process is reversible, the total entropy change is zero)


    2. Relevant equations
    1. [tex]T_{f}=T_{1}^{\frac{m_{1}}{m_{1}+m_{2}}}*T_{2}^{\frac{m_{2}}{m_{1}+m_{2}}}[/tex]


    3. The attempt at a solution
    "Treating it as 1 block of 573K at 20Kg and 1 block of 288K at 60Kg and using T1 = 573, T2 = 288 and m1 = 20, m2 = 60 i get;

    [tex]T_{mix}=359.25K[/tex]

    [tex]T_{f}=573^{\frac{20}{80}}*288^{\frac{60}{80}}[/tex]

    [tex]T_{f}=342.04K[/tex]

    now using [tex]\Delta{E}=mc\Delta{T}[/tex]

    [tex]\Delta{E}=80*375*(359.25-342.04)=516300J[/tex]

    [tex]\Delta{S}=\frac{\Delta{E}}{T}-\frac{\Delta{W}}{T}[/tex]

    As theres no entropy change the delta S terms is zero which leaves,
    [tex]\frac{\Delta{E}}{T}=\frac{\Delta{W}}{T}[/tex]

    the T's cancel and your left with extractable work = 516300J"


    Does this seem correct? im not sure about the 'Tmix' part and the last part where delta E is converted to delta W.
     
  2. jcsd
  3. Jan 28, 2010 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    I am having difficulty following your reasoning. I think you have to apply the equation for efficiency to determine the work as a function of temperature. Initially, the extractable work is:

    [tex]dW = \eta dQ_h = (1-\frac{T_c}{T_h}) cmdT [/tex]

    Write out the relationship between Tc and Th and do the integration.

    AM
     
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