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Thermal Physics: Conductivity

  1. Mar 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Two brass plates, each 0.5 cm thick, have a rubber spacer sheet between them, which is 0.1 cm thick. The outer side of one brass plate is kept at 0 oC, while the outer side of the other is at 100 oC. Find the temperature of the two sides of rubber spacer if the thermal conductivity of brass is approximately 500 times higher than that of rubber.


    2. Relevant equations

    I have Q/t=KA[dT/dx] and Q/t=A(Th-Tc)/sum of (Li/Ki) where Ki=thermal conductivity of each material and Li= thickness of material.



    3. The attempt at a solution
    I have tried setting the Q/t=A(Th-Tc)/sum of (Li/Ki) equal to each other for each brass sheet, like i had split the spacer sheet up, but i dont think this is right. other than that i do not know where to go from the equation. Please help :(
     
  2. jcsd
  3. Mar 23, 2009 #2

    Redbelly98

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    Yes, good.

    What can you say about Q/t/A, for each of the 3 pieces?

    p.s. welcome to PF :smile:
     
  4. Mar 24, 2009 #3
    hmm, for steady state transfer would they be equal? Im really not sure about this :( would i set k[Dt/dx]=k[dT/dx] for the brass and the rubber? with K=500 for brass and K=1 for rubber? and maybe Th = 100 in one case and Tc=0 in the other and get T ? i dont know i am el stupide
     
  5. Mar 24, 2009 #4

    Redbelly98

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    Yes, exactly.

    Close, but there are 3 slabs, so think of it this way:
    k1 [Dt/dx] = k2 [dT/dx] = k1 [dT/dx]​

    Then figure out how Th=100C and Tc=0C go into this.

    There will be two unknown temperatures to solve for.
     
  6. Mar 24, 2009 #5
    what would be the 3rd? would i set th=100 for 1 brass slab, Tc=0 for the other slab; i am unsure what is happening for the 3rd slab :S
     
  7. Mar 24, 2009 #6

    Redbelly98

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    There are two brass slabs, plus a rubber slab.

    Perhaps it will be easier to proceed if you go ahead and replace the dT/dx terms with the appropriate temperatures and slab thicknesses.
     
  8. Mar 25, 2009 #7
    I have: 500(100-Tc)/0.5=500(Th-0)/0.5=(Th-Tc)/0.1

    but i either get 50 or 0 and i somehow don't think this is correct :(
     
  9. Mar 25, 2009 #8

    Redbelly98

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    You're equations are almost correct.

    Where do the temperatures Th and Tc occur? It seems you might have mixed them up, so I'd like to know how you are defining them.

    p.s.
    Don't worry, we are getting close to solving this.
     
  10. Mar 25, 2009 #9
    erm i have said Th=100 at one of the brass plate to the rubber, Th-Tc is the temperature difference across the rubber, and used Tc=0 assuming that the heat from the rubber is being transferred to the brass; i think this is the part that is wrong. ahh its infuriating me now :(
     
  11. Mar 25, 2009 #10

    Redbelly98

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    Okay, let's try this:

    T=100 C on the open face of the hot brass plate.
    T=T1 at the interface between the hot brass plate and the rubber.
    T=T2 at the interface between the cold brass plate and the rubber.
    T=0 C on the open face of the cold brass plate.
     
  12. Mar 26, 2009 #11
    ahhhh
    i give up i can't do it :(
     
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