- #1

castrodisastro

- 82

- 0

## 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

**1.00m**of the combined sheets per second?

^{2}**L**= 0.237 cm

_{Cu}**T**= 373 K

_{Cu}**κ**= 385 (W/m*K)

_{Cu}**L**= 0.129 cm

_{Al}**T**= 297.5 K

_{Al}**κ**= 205 (W/m*K)

_{Al}## Homework Equations

**P**

_{Cond}=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

Aκ

_{Cu}((T

_{Cu}-T

_{x})/L

_{Cu}) = Aκ

_{Al}((T

_{x}-T

_{Al}/L

_{Al})

After some careful rearranging I solved for T

_{x}

T

_{x}=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 m

^{2}.

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.