Calculating DeltaT for Brass and Iron Plugs

[KU_Mustang]

Ok, so i understand that DeltaT = Delta L /coff Lo

But I am not sure how to figure it with two different substances, Given:
A brass plug is to be placed in a ring made of iron. At room temperature (20oC), the diameter of the plug is 8.759 cm and that of the inside of the ring is 8.741 cm. They must be brought to what common temperature (in Co) in order to fit? The coefficient of linear expansion for brass is 10x10-6(Co)-1 and that for iron is 12x10-6(Co)-1.

Can someone please explain to me what i should be doing since there are two different materials?

 

[OlderDan]

Linear expansion will change the linear dimensions of both objects. Circumference and diameter are linear dimensions. Since they have different coefficients, the iron will expand more than the brass. Heat them up until the diameters are the same. Then they will fit together.

 

[thepatientmental]

To calculate the common temperature (DeltaT) for the brass and iron plugs, you will need to use the coefficient of linear expansion (coff Lo) for both materials. As you mentioned, DeltaT = Delta L / coff Lo. This means that you will need to calculate the change in length (Delta L) for both the brass plug and the iron ring at the given temperatures.

To do this, you will need to use the formula Delta L = Lo * coff * DeltaT, where Lo is the original length, coff is the coefficient of linear expansion, and DeltaT is the change in temperature.

For the brass plug, the original length (Lo) is equal to the diameter (8.759 cm), and the coefficient of linear expansion (coff) is 10x10-6(Co)-1. The change in temperature (DeltaT) is the final temperature (Tf) minus the initial temperature (Ti). In this case, the final temperature would be the common temperature that the brass and iron plugs need to be brought to, and the initial temperature would be the room temperature (20oC). So the formula for the brass plug would be: Delta L = 8.759 cm * 10x10-6(Co)-1 * (Tf - 20oC).

For the iron ring, the original length (Lo) is equal to the diameter (8.741 cm), and the coefficient of linear expansion (coff) is 12x10-6(Co)-1. The change in temperature (DeltaT) is also the final temperature (Tf) minus the initial temperature (Ti). So the formula for the iron ring would be: Delta L = 8.741 cm * 12x10-6(Co)-1 * (Tf - 20oC).

Since the brass plug and iron ring must have the same change in length in order to fit, you can set the two formulas equal to each other and solve for Tf. This will give you the common temperature that they need to be brought to.

So the final equation would be: 8.759 cm * 10x10-6(Co)-1 * (Tf - 20oC) = 8.741 cm * 12x10-6(Co)-1 * (Tf - 20oC). You can then solve for Tf, which would be the