Solving Thermal Expansion for Separation of Metals

[Lorinnn]

Homework Statement

A brass ring of diameter 10.00cm at 20.0*C is heated and slipped over an aluminum rod of diameter 10.01cm at 20.0*C. Assuming the average coefficients of linerar expansion are constant, to what temperature must the combination be cooled to separate the two metals?

Homework Equations

$$\alpha$$ of brass = 19e-6
$$\Delta$$A = $$\gamma$$ A_o$$\Delta$$T

The Attempt at a Solution

That's the equation i believe i should be using, but somehow i don't really even know where to begin except for plug and chug. I'm supposed to get the answer of -179*C. =[ please help!

 

[anathema]

Hello, thank you for your question. To solve this problem, we will use the equation for thermal expansion:

\DeltaA = \gamma A_o\DeltaT

where \DeltaA is the change in area, \gamma is the coefficient of linear expansion, A_o is the original area, and \DeltaT is the change in temperature.

First, we need to find the change in area of the brass ring and aluminum rod when they are heated and expanded. We know that the original diameter of the brass ring is 10.00cm and the original diameter of the aluminum rod is 10.01cm. We also know that the average coefficient of linear expansion for brass is 19e-6. Therefore, we can calculate the change in area for the brass ring as:

\DeltaA_{brass} = (19e-6)(\pi)(10.00cm)^2\DeltaT

Similarly, we can calculate the change in area for the aluminum rod as:

\DeltaA_{aluminum} = (23e-6)(\pi)(10.01cm)^2\DeltaT

Now, we need to find the temperature at which the areas of the brass ring and aluminum rod are equal. This will be the temperature at which the combination can be cooled to separate the two metals. So, we set \DeltaA_{brass} = \DeltaA_{aluminum} and solve for \DeltaT:

(19e-6)(\pi)(10.00cm)^2\DeltaT = (23e-6)(\pi)(10.01cm)^2\DeltaT

Simplifying and solving for \DeltaT, we get:

\DeltaT = \frac{(23e-6)(10.01cm)^2}{(19e-6)(10.00cm)^2} = 1.2153

Therefore, the combination must be cooled by 1.2153 degrees Celsius to separate the two metals. Since the original temperature was 20.0 degrees Celsius, the final temperature at which the combination must be cooled is:

20.0 - 1.2153 = 18.7847 degrees Celsius

This is approximately -179 degrees Celsius, as mentioned in the forum post. I hope this helps you understand how to approach this problem. Let me know if you have any further questions. Good luck!