At 20 degrees C an aluminum ring has an inner diameter of 5.000 cm, and a brass rod has a diameter of 5.050 cm. Keeping the brass rod at 20°C, which of the following temperatures of the ring will allow the ring to just slip over the brass rod?
(Al = 2.4 105 /C, brass = 1.9 105/C )
437
To determine the temperature at which the aluminum ring will just slip over the brass rod, we need to consider the coefficient of linear expansion of each material.
The formula for linear expansion is given by:
ΔL = αLΔT
where ΔL is the change in length, α is the coefficient of linear expansion, L is the original length, and ΔT is the change in temperature.
First, we need to find the change in diameter for both the aluminum ring and the brass rod. The change in diameter is twice the change in radius.
For the aluminum ring:
ΔD_aluminum = 2 * ΔR_aluminum
For the brass rod:
ΔD_brass = 2 * ΔR_brass
We can substitute the formula for linear expansion into the equations for ΔD_aluminum and ΔD_brass:
ΔD_aluminum = 2 * α_aluminum * D_aluminum * ΔT
ΔD_brass = 2 * α_brass * D_brass * ΔT
where α_aluminum and α_brass are the coefficients of linear expansion for aluminum and brass respectively, and D_aluminum and D_brass are the original diameters of the aluminum ring and brass rod respectively.
Now, we can determine the temperature at which the ring will just slip over the brass rod by setting the change in diameter for the aluminum ring equal to the change in diameter for the brass rod:
ΔD_aluminum = ΔD_brass
2 * α_aluminum * D_aluminum * ΔT = 2 * α_brass * D_brass * ΔT
Since the brass rod is kept at 20°C, ΔT = 0 for the brass rod.
2 * α_aluminum * D_aluminum * ΔT = 2 * α_brass * D_brass * 0
Simplifying the equation, we find:
α_aluminum * D_aluminum = α_brass * D_brass
Now we can solve for the temperature of the aluminum ring when the two diameters are equal. Rearranging the equation, we get:
T_aluminum = (α_brass * D_brass * T_brass) / (α_aluminum * D_aluminum)
Substituting the given values:
α_aluminum = 2.4 * 10^(-5) / °C
α_brass = 1.9 * 10^(-5) / °C
D_aluminum = 5.000 cm = 0.050 m
D_brass = 5.050 cm = 0.05050 m
T_brass = 20°C
T_aluminum = (1.9 * 10^(-5) / °C * 0.05050 m * 20°C) / (2.4 * 10^(-5) / °C * 0.050 m)
Calculating the expression:
T_aluminum ≈ 16.833 °C
Therefore, the ring will just slip over the brass rod at a temperature of approximately 16.833 °C.
To find the temperature at which the aluminum ring will just slip over the brass rod, we need to consider the difference in thermal expansion between the two materials. The condition for the ring to slip over the rod is when the inner diameter of the ring is equal to the outer diameter of the rod.
Given:
Inner diameter of the aluminum ring (at 20°C) = 5.000 cm
Diameter of the brass rod (at 20°C) = 5.050 cm
We can use the formula for thermal expansion:
ΔL = α * L * ΔT
Where:
ΔL is the change in length or diameter
α is the coefficient of linear expansion
L is the original length or diameter
ΔT is the change in temperature
For the aluminum ring:
ΔL_ring = α_al * D_ring * ΔT_ring
D_ring = 5.000 cm
α_al = 2.4 x 10^(-5) /°C (coefficient of linear expansion for aluminum)
For the brass rod:
ΔL_rod = α_brass * D_rod * ΔT_rod
D_rod = 5.050 cm
α_brass = 1.9 x 10^(-5) /°C (coefficient of linear expansion for brass)
As the ring slips over the rod, the increase in diameter of the ring should be equal to the increase in diameter of the rod, so we can equate the two ΔL values:
α_al * D_ring * ΔT_ring = α_brass * D_rod * ΔT_rod
Now, we can solve for ΔT_ring:
ΔT_ring = (α_brass * D_rod * ΔT_rod) / (α_al * D_ring)
Substituting the given values:
ΔT_ring = (1.9 x 10^(-5) /°C * 5.050 cm * ΔT_rod) / (2.4 x 10^(-5) /°C * 5.000 cm)
Simplifying:
ΔT_ring = (1.9 / 2.4) * (5.050 / 5.000) * ΔT_rod
ΔT_ring = 0.7917 * 1.010 * ΔT_rod
To find the temperature at which the ring will slip over the rod, we need to find ΔT_rod such that ΔT_ring = 0. ΔT_rod = 0 will give us the desired temperature of the ring.
0.7917 * 1.010 * ΔT_rod = 0
ΔT_rod = 0
Therefore, the required temperature of the ring to just slip over the brass rod is 0°C.