Need help on Thermodynamics problem: thermal expansion to be more specific

[Juniorphysics]

Need help on Thermodynamics problem: thermal expansion to be more specific!

A steel ring with a 2.5000in- inside diameter at 20.0°C is to be warmed and slipped over a brass shaft with a 2.5015in outside diameter at 20.0°C.
Part A:

To what temperature should the ring be warmed?

Part B:

If the ring and the shaft together are cooled by some means such as liquid air, at what temperature will the ring just slip off the shaft?

Attempt at solution:
using dL=alphaL_idT
i converted the diameters to meters because i was given alpha in m/m°C. so i got d₁=0.0635m d₂=0.0635381m and dL=0.0000381m
i used alpha=13*10^-7m/m°C
and got dT to be 45.9 but that gave me the incorrect answer

 

[Doc Al]

and got dT to be 45.9 but that gave me the incorrect answer

Double check your arithmetic. (FYI: No need to change units for the lengths.)

 

[Juniorphysics]

It gave me the same answer

 

[Doc Al]

Show your calculation. You are off by about a factor of 10.

 

[asteriatic]

Part A:

To determine the temperature at which the ring should be warmed, we can use the formula for thermal expansion: ΔL = αLΔT, where ΔL is the change in length, α is the coefficient of thermal expansion, L is the original length, and ΔT is the change in temperature.

In this case, we can rearrange the formula to solve for ΔT: ΔT = ΔL/(αL).

We already have the values for ΔL and L (the change in diameter and original diameter, respectively), but we need to determine the coefficient of thermal expansion (α) for both steel and brass.

According to engineeringtoolbox.com, the coefficient of thermal expansion for steel is 11.7*10^-6 m/m°C and for brass it is 18.7*10^-6 m/m°C.

Plugging in these values, we get:

ΔT = (0.0635381m - 0.0635m)/(11.7*10^-6 m/m°C * 0.0635m) = 0.0381m/(0.000007335m)

= 5188.62°C

Therefore, the ring should be warmed to a temperature of 5188.62°C for it to fit over the brass shaft at 20.0°C.

Part B:

To determine the temperature at which the ring will just slip off the shaft, we can use the same formula as in Part A, but this time we need to solve for ΔT when the change in length is equal to the diameter of the brass shaft (0.0635381m).

ΔT = ΔL/(αL)

ΔT = (0.0635381m)/(18.7*10^-6 m/m°C * 0.0635m)

= 0.0003386m/(0.000011845m)

= 28.65°C

Therefore, the ring and the shaft will just slip off at a temperature of 28.65°C when cooled by liquid air.