1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Thermal conductivity between sheets of two different metals

  1. Jan 26, 2014 #1
    1. The problem statement, all variables and given/known data
    A copper sheet of thickness 2.37mm is bonded to a Aluminum sheet of thickness 1.29mm. The outside surface of the copper sheet is held at a temperature of 100.0°C and the Aluminum sheet at 24.5°C.
    a) Determine the temperature of the copper-aluminum interface.
    b) How much heat is conducted through 1.00m2 of the combined sheets per second?

    LCu = 0.237 cm
    TCu = 373 K
    κCu = 385 (W/m*K)
    LAl = 0.129 cm
    TAl = 297.5 K
    κAl = 205 (W/m*K)

    2. Relevant equations
    PCond=Q/t=Aκ(ΔT/L)

    3. The attempt at a solution

    Part a)
    Now the question doesn't say anything about the sheets being in a steady state so I spent a lot of time trying to solve it without assuming that. After a while I gave up and just assumed it to be in a steady state and I solved for the temperature in between the sheets. My answer was correct

    Cu((TCu-Tx)/LCu) = AκAl((Tx-TAl/LAl)

    After some careful rearranging I solved for Tx

    Tx=335.665 K

    So to assume a steady state is the only way I see to have all the necessary information to find the temperature of the copper-aluminum interface. Can someone explain what in the question tells me that it is in a steady state?

    Part b)
    I would have the calculate the thermal conductivity rate for this dual sheet interface, then use the thermal conductivity equation, set it equal to the thermal conductivity rate I just calculate it with a length of 1.00 m2.

    To do this I need to know the face area of the sheets, but I am only given the thickness. I can't cancel the face area value since I am not setting it equal to anything with that same value on the other side.

    I feel that I am so close to the answer.

    Any help is appreciated. Thanks.
     
  2. jcsd
  3. Jan 26, 2014 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    a) I could turn it around: Do you see anything that would bring in time? Initial temperature or something?
    b) You are close. Look at the dimensions
     
  4. Jan 26, 2014 #3
    For a) Well I know that the thermal conductivity is equal to heat(Q) per unit time(t).

    Q=mCΔT

    but I don't have a mass for either sheet or of the two together. When you say initial temperature, do you mean I can use the initial temperature and final temperature, which was the answer to part a and use it to calculate the heat?
     
  5. Jan 26, 2014 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Making BvU's points slightly differently,
    a) since you are not given an initial temperature or a time, suppose no time has elapsed. The temperature will be the initial temperature, which could be anything at all.
    b)
    Then you've misread the question.
     
  6. Jan 26, 2014 #5
    a) Oh I understand the point that BvU was making. That I shouldn't consider time since nothing was given to indicate that time had elapsed.


    b) Am I missing something in reading the question. The length and width could be any value, the face area would be determined by these values. It doesn't mention anything about it being a square or any thing about the dimensions.
     
  7. Jan 26, 2014 #6

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    a) is clearly asking for a steady state situation. You don't get enough input to do something time dependent.
    b)
    Come on! a length of 1.00 m2 ???
    In your variable list I see k, L and T. What is the meaning of A ?

    Read your 11:46 post just now. There is a lot of symmetry. length and width aren't in your expression, so why worry about them ?

    Look at the dimensions for each factor left and right in your expression. What do you have to do to get the dimension of the item that they are asking for ?
     
  8. Jan 26, 2014 #7

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Oh, and in this jargon L is the thickness of the sheet, not the length or the width!
     
  9. Jan 26, 2014 #8
    Yea that flew right by me, I can't use m2 for the length. I do realize that L is the distance from a side with one temperature to the opposite side with a different temperature.


    and for part b), so the 1.00 m2 given IS the face area.

    Ltotal=LCu+LAl
    Ltotal=0.00366 m

    and the A that you asked about comes from the thermal conductivity equation.
    PCond=Aκ(ΔT/L) where A is the face area of the material.

    So to solve for heat per unit time

    PCond=Aκ(ΔT/L)
    PCond=1.00m2κ(373K - 297.5K)/0.00366 m)

    If this is for the combined sheets, then which value of κ should I use?
     
  10. Jan 26, 2014 #9

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    To answer b) you can use the answer to a) and so avoid worrying about there being two sheets. Pick one. You know the temperature each side of it.
     
  11. Jan 26, 2014 #10

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In your original attempt at solution you have equated the power that flows through one sheet to the power that flows though the other. Correctly. The power can't accumulate anywhere in a steady state.

    You rearranged to solve for T. All you are asked to do now is calulate that power in case A = 1 m^2. If you do it for both sheets, you can check Tx is correct. And you can then tell what the over-all k is ! Preferably in an expression: the number doesn't give any insight.
     
  12. Jan 26, 2014 #11
    Ok, I see progress!

    So the PCond for Aluminum is

    PCond=(1.00 m2)(205 W/(m*K))[(335.665 K - 297.5 K)/(0.00129 m)]
    PCond=6.065*106 W

    For Copper...

    PCond=(1.00 m2)(385 W/(m*K))[(373 K - 335.665 K)/(0.00237 m)]
    PCond=6.065*106 W

    So now that I know the conduction rate is the same, how can I calculate the conduction rate of the combined sheets? I assume I would add the thickness of each to give me a total thickness of 0.00366 m. So this would be my new L value.

    knowing that my power would have to be 6.065*106 W

    PCond=Aκ(ΔT)/Ltotal

    κ=PCondLtotal/AΔT

    κ=6.065*106W(0.00366m)/(1.00m2)(75.5K)

    κ=294.012 W/m*K

    Now I know what the overall κ is. In expression form it is
    κ=PCondLtotal/AΔT

    Now I'm confused. I'm looking for PCond but if I plug in
    κ=294.012 W/m*K
    A=1.00m2
    ΔT=75.5K
    Ltotal=0.00366m

    My answer is the same as it was for the two sheets individually because I literally just used the same values to figure out that κ, so I just did the same equation twice, just in reverse. Am I missing the concept?
     
  13. Jan 26, 2014 #12
    The rate at which heat is conducted through the combined two sheets is the same as the rate at which it conducted through each of them. Think of an electric circuit with two resistors in series. The current flow through each resistor is the same as that through the other resistor, and it is the same as the current through them both considered in combination. It's really pretty much the same thing.
     
  14. Jan 26, 2014 #13

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    To solve part (a) you used the fact that the rate of heat flow through the pair of sheets is the same as the rate through each individually. Essentially, you had two equations involving two unknowns: the rate of heat flow and the temperature at the join. You eliminated the first unknown to find that temperature. Now all you need to do is plug that back in to solve for the other unknown.
    As I said in post #9, you can ignore one sheet and just use the known temperature difference across the other.
     
  15. Jan 26, 2014 #14
    the temperature difference between the outside faces of the two sheets was
    373K-297.5K=75.5K

    Yes that's right, my first unknown was the rate of heat flow and the temperature where the two sheets meet. I figured out the temperature and used that to find out the rate of heat flow. which is 6.065*106 W

    I think what chestermiller was trying to tell me that this is also the same rate of heat flow for the two sheets stuck together as one. I submitted this to my online homework but it told me I was wrong. So I'm not sure if I understood correctly.
     
  16. Jan 26, 2014 #15

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes, that's the rate of heat flow, but the question asks for an amount of heat (in one second). So the answer should be in J or calories or somesuch. Do you get to enter a number only or do you put units in too?
     
  17. Jan 26, 2014 #16
    Yes I can specify the units.

    so 6,065,000 W. or J/S. For one second would come out to 6,065,000 Joules. Right?
     
  18. Jan 26, 2014 #17
    I got the same answer you got. Maybe there is a significant figures issue?

    Chet
     
  19. Jan 26, 2014 #18
    Hmm, that's what I thought but i typed in 6064968.354 J as my answer and it was still incorrect. I don't know why. I do understand what you were telling me so now it doesn't make sense to be a different value other than this.

    I redid all the calculations in part b just to be sure, since the answer I got for part a was marked as correct, I knew everything up to that point was correct. I still arrived at 6064968.354 J

    There's a slight chance this problem may be omitted from the homework assignment. We were told that one of our homework problems was set up incorrectly and would never give us credit. I don't think it's this one though.
     
  20. Jan 26, 2014 #19
    The difficulty is not too few significant figures. It's too many. Try 6.06x106, or 6.1x106.
     
    Last edited: Jan 26, 2014
  21. Jan 26, 2014 #20
    No it still didn't work. I took a screenshot of the problem marking my incorrect answer with my name on the top right to show I was logged in so maybe the professor can take a look at it and determine if I am wrong or right. It's due at midnight tonight so If I was right then he can probably give me credit.

    Thanks for all the help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted