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Thermal Conduction

  1. Apr 27, 2013 #1
    1. The problem statement, all variables and given/known data
    A rod, with sides insulated to prevent heat loss, has one end immersed in boiling water at 100C and the other end immersed in a water/ice mixture at 0C. The rod has uniform cross-sectional area of 4.04 cm^2 and length 91cm. Under steady state conditions, heat conducted by the rod melts the ice at a rate of 1.0g every 34 seconds. What is the thermal conductivity of the rod?


    2. Relevant equations
    H=dQ/dt=k*Area(TempChange)(1/length)
    Heat of Fusion of water is 3.34*10^5 J/kg


    3. The attempt at a solution
    (3.34*10^5*91)/(34seconds*100C*4.04cm^2) = 2200

    My answer key says 220 W/m*K, i've tried converting 91cm to .91m and 4.04cm^2 to .000404m^2 and I get the same 2200 answer. I think I'm making a conversion error but I'm not sure, please advice, thank you.
     
  2. jcsd
  3. Apr 27, 2013 #2
    have you realised that it is 1x10^-3 kg melted in 34secs
     
  4. Apr 27, 2013 #3

    cepheid

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    Your work is off by a factor of 1000 because you forgot to multiply the heat of fusion of water (J/kg) by the rate at which mass is melting (0.001 kg). Only 334 J of heat is being transferred in 34 sec, NOT 334,000 J.
     
  5. Apr 27, 2013 #4
    hmm I believe I follow what you are saying. I recalculated with 334J and my answer comes to 2.20 but that's still not what the answer key says. Am I just dense or is maybe the key incorrect?
     
  6. Apr 27, 2013 #5
    wait, maybe the 2.20 is in W/cm*K and the 220 on the key is W/m*K? ....nevermind, that would be backwards
     
  7. Apr 27, 2013 #6

    cepheid

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    I don't know what to tell you, because it's just arithmetic at this point. You are messing it up somewhere and just need to be meticulous and get it right.
     
  8. Apr 27, 2013 #7
    (334*0.91)/(34seconds*100C*.000404) = 221
     
  9. Apr 27, 2013 #8
    Thank you Chester, I'd been messing with this over and over and apparently it's like trying to grammar check your own novel, I never tried both converting to meters, and fixing my gram and kilogram mistake. Thank you so much :) I don't know how I missed it now.
     
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