# Thermal conductivity between sheets of two different metals

• castrodisastro
In summary, a copper sheet of thickness 2.37mm bonded to an aluminum sheet of thickness 1.29mm has an outside surface held at a temperature of 100.0°C and 24.5°C, respectively. Using the given thermal conductivity values, the temperature of the copper-aluminum interface can be calculated to be 335.665K assuming steady state conditions. To determine the heat conducted through 1.00m2 of the combined sheets per second, the thermal conductivity equation can be used. Using the face area of 1.00m2 and the temperature difference calculated in part a, the heat conducted per second can be found. The thermal conductivity value to be used is the one for either sheet, as
castrodisastro

## Homework Statement

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)

## Homework Equations

PCond=Q/t=Aκ(ΔT/L)

## 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.

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

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?

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)
To do this I need to know the face area of the sheets, but I am only given the thickness.

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.

a) is clearly asking for a steady state situation. You don't get enough input to do something time dependent.
b)
I just calculate it with a length of 1.00 m2.
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 ?

Oh, and in this jargon L is the thickness of the sheet, not the length or the width!

BvU said:
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 ?

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?

castrodisastro said:
If this is for the combined sheets, then which value of κ should I use?
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.

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.

1 person
BvU said:
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.

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?

castrodisastro said:
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?
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.

1 person
castrodisastro said:
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?
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 into 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.

1 person
haruspex said:
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 into 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.

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.

castrodisastro said:
I figured out the temperature and used that to find out the rate of heat flow. which is 6.065*106 W
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?

haruspex said:
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?

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?

I got the same answer you got. Maybe there is a significant figures issue?

Chet

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.

castrodisastro said:
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.
The difficulty is not too few significant figures. It's too many. Try 6.06x106, or 6.1x106.

Last edited:
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.

Well well, you've had three heavyweights (in all modesty) helping you out and all four of us come out at the same point. That should be enough to stand up to the rest of the world. But I, for one, am really curious to hear how this went, so: please let us know and we'll have a virtual toast if we didn't make fools of ourselves!

I have the class on tuesday evenings so I will find out after that if I can get the professor to look at my work.

He is always happy to answer any questions so I should be able to catch him after class.

I'll keep you posted!

## 1. What is thermal conductivity?

Thermal conductivity is the ability of a material to conduct heat. It is a measure of how quickly heat can pass through a material.

## 2. How is thermal conductivity measured?

Thermal conductivity is typically measured in units of watts per meter per Kelvin (W/mK). It is determined by conducting heat through a material of known thickness and measuring the temperature difference across the material.

## 3. How does thermal conductivity differ between two different metals?

The thermal conductivity of a material is dependent on its physical properties, such as density, specific heat, and molecular structure. Therefore, different metals will have different thermal conductivities due to their unique characteristics.

## 4. How does the thermal conductivity between sheets of two different metals affect heat transfer?

The thermal conductivity between two different metals can greatly affect heat transfer. If the two metals have different thermal conductivities, heat will transfer at a different rate between them, resulting in uneven temperature distribution and potentially affecting the overall efficiency of a system.

## 5. Can the thermal conductivity between two different metals be manipulated?

While the thermal conductivity of a material is a physical property that cannot be changed, it is possible to manipulate the thermal conductivity between two different metals by introducing a third material, such as a thermal interface material, to improve heat transfer between them.

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