A brass ring of diameter 10.00 cm at 19.9°C is heated and slipped over an aluminum rod of diameter 10.01 cm at 19.9°C. Assume the average coefficients of linear expansion are constant.
(a) To what temperature must the combination be cooled to separate the two metals?
(c) What if the aluminum rod were 10.96 cm in diameter?
i used the equation r + alpha(deltaT) = r +ralpha(deltaT)
and then i solved for T and i ended up getting 19.9 degrees C...which is what I started with
I don't know what I am supposed to do I guess. :|
To determine the temperature at which the combination needs to be cooled to separate the two metals, we need to make use of the equation you mentioned: r + α(ΔT) = r + rα(ΔT).
Considering the equation, r represents the initial radius of the brass ring (which is half of its diameter), α is the coefficient of linear expansion for both the brass and aluminum, ΔT is the change in temperature, and rα represents the coefficient of linear expansion for the aluminum rod.
To solve for ΔT, we need to find a value that results in a difference between the expanded radius of the brass ring and the expanded radius of the aluminum rod.
For simplicity in the calculations, let's assume the coefficient of linear expansion for both the brass and aluminum remains constant and is equal to α.
(a) To solve for the temperature at which they separate, we can equate the expanded radius of the brass ring and the aluminum rod:
r + α(ΔT) = r + rα(ΔT)
Here, r represents the initial radius of the brass ring.
Substituting the given values, r = 5.00 cm and ΔT = T - 19.9°C, we get:
5.00 cm + α(T - 19.9°C) = 5.005 cm + 5.005α(T - 19.9°C)
Now, we can solve for T:
5.00 cm + αT - α(19.9°C) = 5.005 cm + 5.005αT - 5.005α(19.9°C)
αT - α(19.9°C) = 0.005 cm + 5.005αT - 5.005α(19.9°C)
αT - 5.005αT = 0.005 cm + 5.005α(19.9°C) - α(19.9°C)
-4.005αT = 0.005 - α(99.5°C)
Dividing both sides by -4.005α:
T = (0.005 - α(99.5°C))/(-4.005α)
Since α is a small value, we can assume the proportionality of ΔT to the radii, and consequently, α. Thus, we can approximate this as:
T ≈ (0.005)/(-4.005α) = -0.00125/α
Substituting α = 19 × 10^(-6) C^(-1) - coefficient of linear expansion for both brass and aluminum - we can calculate the value of T:
T ≈ -0.00125/(19 × 10^(-6) C^(-1)) = -65.79°C
Therefore, the combination of metals needs to be cooled to approximately -65.79°C to separate them.
(c) Now, considering the aluminum rod diameter of 10.96 cm, we can use a similar approach:
r + α(ΔT) = r + rα(ΔT)
Assuming the same coefficient of linear expansion, α, for both the brass and aluminum, we can substitute the values:
5.48 cm + α(ΔT) = 5.48 cm + 5.48α(ΔT)
Simplifying the equation, we get:
α(ΔT) = 5.48α(ΔT)
This equation indicates that the expanded radii of the brass ring and aluminum rod are equal; therefore, they will not separate at any temperature.
Hope this explanation helps you understand the problem and solution better. Let me know if you have any further questions!