Solve Heat Transfer Problem: Copper & Iron Rods Joined Together

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A cylindrical copper rod and an iron rod are welded together, with the copper end at 130°C and the iron end at 0°C. To find the temperature at the junction, thermal conductivity rather than specific heat capacity must be used. The correct formula involves the thermal conductivities of both materials, leading to the equation T(junction) = T2 + (T1-T2)*kCu/(kCu+kFe). Using the thermal conductivities of copper (400 W/mK) and iron (80 W/mK), the temperature at the midpoint is calculated to be approximately 83.33°C. Understanding the relationship between heat flow and temperature difference is crucial for solving this type of heat transfer problem.
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Homework Statement



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?

Homework Equations



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

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

You can't 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 doesn't even provide the area.

Please help
 
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.
 
Dadface
OKay so the k for copper is 400W/mK and for Iron it is 80W/mk
I am sorry, but I honestly don't 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 there's 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.
 
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?
 
  • #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.
 
  • #11
okay thanks. I am still a little bit confused. thanks for the help. :)
 
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