A thin brass ring of inner diameter 10cm at 20degrees C is warmed and slipped over an aluminum rod of diameter 10.01cm at 20degrees C. Assuming the average coefficient of linear expansions are constant, to what temperature must this combination be cooled to separate the parts?
Look up the thermal expansion coefficients of alumimum and brass. They should be in units of degC^-1. It is about 18.7*10^-6 for brass and 22.2 for aluminum. You will find that values depend on temperature ramge and alloy. The aluminum must shrink 0.01 cm more than the brass ring to slip it off. Let dT be the change in temperature from 20 C (positive for an increase).
10.01*22.2*10^-6*dT - 10.00*18.7*10^-6*dT = -0.01
dT(22.2*10^-5 - 18.7*10^-5) = -0.01
dT = -0.01/3.5^10^-5 = -285 C
That would put the final temperature at -265C, whch is near absolute zero.
To find the temperature at which the combination must be cooled to separate the parts, we can use the concept of thermal expansion.
The first step is to find the change in diameter of both the brass ring and the aluminum rod.
Given:
Inner diameter of brass ring (D1) = 10 cm
Diameter of aluminum rod (D2) = 10.01 cm
Change in diameter (delta D) = D2 - D1
Change in diameter = 10.01 cm - 10 cm = 0.01 cm
Next, we need to find the average coefficient of linear expansion (alpha) for each material. The coefficient of linear expansion relates the change in size of an object to the change in temperature.
Assuming the average coefficients of linear expansion for brass and aluminum are constant, let's denote them as alpha1 and alpha2, respectively.
To find the temperature at which the combination must be cooled to separate the parts, we can use the formula:
delta D = alpha1 * D1 * delta T1 + alpha2 * D2 * delta T2
where:
delta T1 is the change in temperature for the brass ring
delta T2 is the change in temperature for the aluminum rod
We know that both the brass ring and the aluminum rod were originally at 20 degrees Celsius, so their initial temperatures (T1 and T2) are both 20 degrees Celsius.
The formula can be rearranged to solve for delta T2:
delta T2 = (delta D - alpha1 * D1 * delta T1) / (alpha2 * D2)
Substituting the known values:
delta T2 = (0.01 cm - alpha1 * 10 cm * delta T1) / (alpha2 * 10.01 cm)
Finally, to find the temperature at which the combination must be cooled to separate the parts, we need to solve for delta T1 using the given values of delta D and the coefficients of linear expansion.
Please provide the values for the coefficient of linear expansion (alpha1 and alpha2) for brass and aluminum.
To find the temperature at which the combination of the brass ring and aluminum rod can be separated, we need to use the concept of linear expansion and the coefficient of linear expansion.
The linear expansion of a material can be given as ΔL = α * L * ΔT, where ΔL is the change in length, α is the coefficient of linear expansion, L is the original length of the material, and ΔT is the change in temperature.
In this case, we have a brass ring and an aluminum rod. We can assume that the average coefficient of linear expansion for both materials remains constant over the given temperature range.
Let's start by finding the change in diameter of the brass ring and aluminum rod. The change in diameter can be calculated as ΔD = 2 * α * D * ΔT, where ΔD is the change in diameter, α is the coefficient of linear expansion, D is the original diameter of the material, and ΔT is the change in temperature.
For the brass ring:
ΔD_brass = 2 * α_brass * D_brass * ΔT
For the aluminum rod:
ΔD_aluminum = 2 * α_aluminum * D_aluminum * ΔT
Given that the inner diameter of the brass ring is 10 cm and the outer diameter of the aluminum rod is 10.01 cm, we can calculate the initial difference in diameter (ΔD_initial) between the brass ring and aluminum rod:
ΔD_initial = D_aluminum - D_brass
Now, we want to find the temperature at which the difference in diameter becomes zero (ΔD_final = 0). At this temperature, the brass ring and aluminum rod can be separated.
To find this temperature, we can set up the following equation:
ΔD_brass + ΔD_initial + ΔD_aluminum = 0
Simplifying the equation:
2 * α_brass * D_brass * ΔT + ΔD_initial + 2 * α_aluminum * D_aluminum * ΔT = 0
Now we can plug in the given values:
2 * α_brass * 10 cm * ΔT + (10.01 cm - 10 cm) + 2 * α_aluminum * 10.01 cm * ΔT = 0
Simplifying further:
20 * α_brass * ΔT + 0.01 cm + 20.02 * α_aluminum * ΔT = 0
Rearranging the equation to solve for ΔT:
20 * α_brass * ΔT + 20.02 * α_aluminum * ΔT = -0.01 cm
Combining the terms with ΔT:
(20 * α_brass + 20.02 * α_aluminum) * ΔT = -0.01 cm
Solving for ΔT:
ΔT = (-0.01 cm) / (20 * α_brass + 20.02 * α_aluminum)
To find the temperature at which the combination can be cooled to separate the parts, you would need to know the values of the coefficient of linear expansion for brass (α_brass) and aluminum (α_aluminum). Plug in the values and calculate ΔT using the equation above.