You have a brass ring of inside diameter 10.00 cm that is at 20.0oC. You want to slip this ring over an aluminum rod that is at 20.0oC and has a diameter of 10.01 cm.

a. To what temperature do you have to heat the ring to slip it over the rod?
b. Once you have the ring over the rod they are allowed to cool back to room temperature (20.0oC). To what temperature must the combination be cooled to separate the two metals?
c. If the diameter of the rod was 10.02 cm would you still be able to get the brass ring off?

I know that the diameter increases in proportion to alpha and delta T. We want the brass diameter to expand 0.01 cm more than the aluminum though I don't know what to do after that. Any help would be much appreciated!

To answer these questions, we can use the principle of thermal expansion and the equation that relates the change in length or diameter due to thermal expansion, called the thermal expansion equation.

a. To find the temperature at which the brass ring will slip over the aluminum rod, we need to calculate the change in diameter for both the ring and the rod. We can use the thermal expansion equation:

ΔL = α * L0 * ΔT

where ΔL is the change in length or diameter, α is the coefficient of linear expansion, L0 is the initial length or diameter, and ΔT is the change in temperature.

Let's calculate:

For the brass ring:
ΔL(brass ring) = α(brass) * D(brass ring) * ΔT(brass ring)

For the aluminum rod:
ΔL(aluminum rod) = α(aluminum) * D(aluminum rod) * ΔT(aluminum rod)

Since we want the brass diameter to expand 0.01 cm more than the aluminum, we can set up the equation:

ΔL(brass ring) = ΔL(aluminum rod) + 0.01 cm

Now, we can set up the equation using the given values:

α(brass) * D(brass ring) * ΔT(brass ring) = α(aluminum) * D(aluminum rod) * ΔT(aluminum rod) + 0.01 cm

Substituting the given values:
α(brass) * 10.00 cm * ΔT(brass ring) = α(aluminum) * 10.01 cm * ΔT(aluminum rod) + 0.01 cm

We can solve this equation to find ΔT(brass ring), the temperature change for the brass ring. Rearranging the equation:

ΔT(brass ring) = [α(aluminum) * D(aluminum rod) * ΔT(aluminum rod) + 0.01 cm] / (α(brass) * D(brass ring))

Substituting the respective values for α and D, we can solve for ΔT(brass ring).

b. To find the temperature at which the combination must be cooled to separate the brass ring and aluminum rod, we need to consider the contraction due to cooling. Using the same thermal expansion equation, the change in length or diameter due to cooling is given by:

ΔL = α * L * ΔT

Setting up the equation, we have:

ΔL(brass ring) + ΔL(aluminum rod) = 0

We can solve this equation for ΔT, the change in temperature:

α(brass) * D(brass ring) * ΔT(brass ring) + α(aluminum) * D(aluminum rod) * ΔT(aluminum rod) = 0

Rearranging the equation:

ΔT(brass ring) = -[α(aluminum) * D(aluminum rod) * ΔT(aluminum rod)] / [α(brass) * D(brass ring)]

Again, substituting the respective values for α and D, we can solve for ΔT(brass ring).

c. If the diameter of the rod was 10.02 cm, we can repeat the calculations above using the new diameter of the rod in the equations. We can find the temperature at which the brass ring will slip over the aluminum rod and determine whether it is possible to get the brass ring off. If the resulting ΔT(brass ring) is positive, it means that the ring would expand enough to slip off the rod. If the resulting ΔT(brass ring) is negative, it means that the ring would not be able to expand enough to slip off the rod.