What Temperature Allows a Brass Plug to Fit into an Iron Ring?

[davidkis]

Homework Statement

A brass plug is to be placed in a ring made of iron. At room temperature (20°C), the diameter of the plug is 8.764 cm and that of the inside of the ring is 8.754 cm. They must both be brought to what common temperature in order to fit?

Homework Equations

Change in V=B*V0*Change in T
B for brass = 56x10^-6
B for Iron = 35 x 10^-6

The Attempt at a Solution

I had no idea how to start i notice that the change in the coefficent of expansion is going to change the items dimensions but i honestly don't know where to begin, do i combine the 2 separate formulas? how would i solve for the temperture?

 

[rl.bhat]

Here you need not consider the volume expansion. Brass plug will pass through iron ring if their radii are the same.
Rb = Rb0( 1 + αb*Δt)
Ri = Ri0( 1 + αi*Δt)
where α is the coefficient of linear expansion which is equal to B/3.
Equate Rb and Ri and solve for Δt

 

[kerimek]

I would approach this problem by considering the principles of temperature and kinetic theory. According to the kinetic theory of matter, the temperature of an object is related to the average kinetic energy of its particles. This means that as the temperature increases, the particles within the object will vibrate faster and take up more space.

In this problem, we are dealing with two materials, brass and iron, which have different coefficients of thermal expansion (B). This means that they will expand or contract at different rates when subjected to a change in temperature. In order for the brass plug and iron ring to fit together, they must have the same dimensions, which means they must be at the same temperature.

To solve for the common temperature at which they will fit, we can use the formula for change in volume (ΔV) as follows:

ΔV = B * V0 * ΔT

Where B is the coefficient of thermal expansion, V0 is the original volume, and ΔT is the change in temperature.

Since we want the final volume of the brass plug and iron ring to be the same, we can set their change in volumes equal to each other:

Bbrass * V0brass * ΔT = Biron * V0iron * ΔT

We can then rearrange the equation to solve for ΔT:

ΔT = (Biron * V0iron) / (Bbrass * V0brass)

Now, we know the values for B and V0 for both brass and iron, so we can plug them into the equation to solve for ΔT:

ΔT = (35 x 10^-6 * 8.754 cm) / (56 x 10^-6 * 8.764 cm)

ΔT = 5.39°C

Therefore, the brass plug and iron ring must both be brought to a temperature of 5.39°C in order to fit together. This is the common temperature that will result in the same change in volume for both materials.