a brass ring of diameter 10.00cm at 20.0 degrees Celsius is heated and slipped over an aluminum rod of diameter 10.01cm at 20.0 degrees Celsius. Assuming the average coefficients of linear expansion are constant, (a) to what temperature must this combination be cooled to separate them? is this attainable?
(b) What if? What if the aluminum rod were 10.02cm in diameter?
thanks.
diameter increases in proportion to alpha and delta T. You want the brass diameter to expand 0.01 cm more than the aluminum
(T-20)[10.00*alphab - 10.01*alphaa] = 0.01
Look up the alpha values and solve for T
It can be done if a melting point is not exceeded.
The solution to this problem
To determine the temperature at which the brass ring and aluminum rod can be separated, we can use the principle of linear expansion.
(a) Let's first calculate the change in diameter for both the brass ring and the aluminum rod:
Change in diameter of brass ring: Δd_brass = (diameter of brass ring at final temperature) - (initial diameter of brass ring)
Change in diameter of aluminum rod: Δd_aluminum = (diameter of aluminum rod at final temperature) - (initial diameter of aluminum rod)
Since the diameters are given at the same initial temperature, we can express the change in diameter as a function of the length change:
Δd = α * d * ΔT
where ΔT is the change in temperature, α is the coefficient of linear expansion, and d is the initial length.
Assuming that the average coefficients of linear expansion are constant, let's calculate the temperature required to cool the combination so that the rings can be separated.
Given:
Initial diameter of brass ring (d_brass_initial) = 10.00 cm
Initial diameter of aluminum rod (d_aluminum_initial) = 10.01 cm
Coefficient of linear expansion for brass (α_brass) = α_aluminum = α (assumed constant)
Initial temperature (T_initial) = 20.0°C
We can calculate the change in diameter for both the brass ring and the aluminum rod using the initial values:
Change in diameter of brass ring: Δd_brass = α * d_brass_initial * ΔT
Change in diameter of aluminum rod: Δd_aluminum = α * d_aluminum_initial * ΔT
To separate the two, the change in diameter of the brass ring must be greater than the change in diameter of the aluminum rod. Therefore:
Δd_brass > Δd_aluminum
Simplifying,
α * d_brass_initial * ΔT > α * d_aluminum_initial * ΔT
We can cancel out α and ΔT from both sides of the equation, as they are assumed to be constant:
d_brass_initial > d_aluminum_initial
Substituting in the given values,
10.00 cm > 10.01 cm
Since this is not true, it means that the rings cannot be separated by cooling them down. No temperature will be able to separate the brass ring from the aluminum rod.
(b) If the diameter of the aluminum rod were increased to 10.02 cm, we can repeat the same calculations as above:
Initial diameter of brass ring (d_brass_initial) = 10.00 cm
Initial diameter of aluminum rod (d_aluminum_initial) = 10.02 cm
Coefficient of linear expansion for brass (α_brass) = α_aluminum = α (assumed constant)
Initial temperature (T_initial) = 20.0°C
Following the same steps as in part (a):
Δd_brass = α * d_brass_initial * ΔT
Δd_aluminum = α * d_aluminum_initial * ΔT
And setting the conditions for separation:
α * d_brass_initial * ΔT > α * d_aluminum_initial * ΔT
Simplifying,
d_brass_initial > d_aluminum_initial
Substituting in the given values,
10.00 cm > 10.02 cm
Again, this is not true, so the rings cannot be separated by cooling them down even if the diameter of the aluminum rod is increased to 10.02 cm.
To answer these questions, we need to consider the concept of thermal expansion and the linear expansion coefficients of brass and aluminum.
(a) To separate the brass ring from the aluminum rod, we want to cool the combination to a temperature where the brass ring contracts more than the aluminum rod. The formula for linear expansion is given by:
ΔL = α * L * ΔT
Where ΔL is the change in length, α is the coefficient of linear expansion, L is the initial length, and ΔT is the change in temperature.
Assuming the average coefficients of linear expansion are constant, we can set up an equation to find the temperature at which the lengths of the two objects are equal:
ΔL_brass = ΔL_aluminum
α_brass * L_brass * ΔT = α_aluminum * L_aluminum * ΔT
Dividing both sides by ΔT and rearranging the equation, we get:
α_brass * L_brass = α_aluminum * L_aluminum
Given that α_brass = 19 x 10^-6 / °C and α_aluminum = 23 x 10^-6 / °C, and the initial lengths (diameters) of the brass ring and aluminum rod are 10.00 cm and 10.01 cm respectively, we can calculate the temperature at which the lengths are equal:
α_brass * L_brass = α_aluminum * L_aluminum
(19 x 10^-6 / °C) * (10.00 cm) = (23 x 10^-6 / °C) * (10.01 cm)
Solving for ΔT, we find ΔT ≈ -0.0199 °C
So, the combination must be cooled by approximately 0.0199 °C to separate the brass ring from the aluminum rod.
(b) If the aluminum rod were 10.02 cm in diameter instead, we can repeat the calculations using the new diameter:
α_brass * L_brass = α_aluminum * L_aluminum
(19 x 10^-6 / °C) * (10.00 cm) = (23 x 10^-6 / °C) * (10.02 cm)
Solving for ΔT, we find ΔT ≈ -0.0174 °C
Therefore, if the aluminum rod were 10.02 cm in diameter, the combination would need to be cooled by approximately 0.0174 °C to separate the brass ring from the aluminum rod.
In both cases, it is theoretically possible to attain the desired temperature by cooling the combination.