Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Urgent, Easy question with heat

  1. Feb 15, 2010 #1
    1. The problem statement, all variables and given/known data

    A cylindrical copper rod and an iron rod with exactly the same dimensions are welded together end to end. The outside end of the copper rod is held at 130 C, and the outside end of the iron rod is held at 0 C

    What is the temperature at the midpoint where the rods are joined together?

    2. Relevant equations

    Q=MCdT ?
    specific heat of iron =0.46 KJ/ kg k
    specific heat of copper = 0.39 kj/ kg k

    3. The attempt at a solution

    this problem look so easy but I really have no idea how to approah to this problem,, i tried 65 C but its wrong
    Last edited: Feb 15, 2010
  2. jcsd
  3. Feb 15, 2010 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What is the question?
  4. Feb 15, 2010 #3
    Poont probably needs to find the temperature at the copper /iron interface and to assume that the bar is perfectly lagged
  5. Feb 15, 2010 #4
    sorry i forgot to post the question, it is What is the temperature at the midpoint where the rods are joined together?
  6. Feb 15, 2010 #5
    Hello poont.You cannot solve the problem using specific heat capacity.You need to look up thermal conductivity.
  7. Feb 16, 2010 #6
    How do you approch this problem. I have the same one. I thought you could use specific heat also. Guess not.

    You cant use the equation Q/Change T = K(A/L)*change in T

    because since our length is zero, it would cancel everything else. And it doesnt even provide the area.

    Please help
  8. Feb 16, 2010 #7
    Hello Zotelo,Yes you do use that equation.At steady state(when all the temperatures equalise) the heat flow per second (Q/Change using your symbols) is the same all along the bar.In other words K(A/L)* change in T Is the same for both metals.Put the numbers in and things cancel out.You will need to look up k for Cu and Fe.
  9. Feb 16, 2010 #8
    OKay so the k for copper is 400W/mK and for Iron it is 80W/mk
    I am sorry, but I honestly dont understand it. Obviously the units m will cancel out, including the units K. So you are left with W. Thats all I could figure out. I am really lost. I am looking at my book, and theres a couple of examples that I understand. But they provide the Area, and length. Plug and Chug. But this one they only provide the change in temperature. Please help me/us.
  10. Feb 16, 2010 #9
    I got it.... I looked it up on google. The way they solved it...

    T(junction) = T2 + (T1-T2)*kCu/(kCu+kFe)
    = 0 + (100-0)*400/(400+80)
    = 83.33 C

    Its the correct answer... if the copper was at 100 C. (my problem was a different version)

    would you please explain to me how they set up the equation?
  11. Feb 16, 2010 #10
    for copper Q/Change=400(A/l)*T change
    For iron Q/change=80(A/l)* T change
    Q/change is the same for both metals so link the equations and A/l cancels.You need to write an expression for the temperature difference across the copper and iron and pluggitychug.
  12. Feb 16, 2010 #11
    okay thanks. I am still a little bit confused. thanks for the help. :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook