At 20 $\degree$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 $\degree$C,which of the following temperatures of the ring will allow the ring to just slip over the brass rod? (a_Al = 2.4 x 10^-5 /C $\degree$,a_brass = 1.9 x 10^-5/C $\degree$) A)111 $\degree$C B)236 $\degree$C C)384 $\degree$C D)437 $\degree$C

Question

Quizplus · Accepted Answer

To determine the temperature at which the ring will just slip over the brass rod, we need to calculate the coefficient of thermal expansion for both aluminum and brass:

αAluminum = 2.4 x 10^-5 /°C 
αBrass = 1.9 x 10^-5 /°C

We can use the following equation to calculate the temperature difference needed for the ring to expand enough to fit over the brass rod:

ΔT = (dBrass - dAluminum) / (αAluminum - αBrass)

Substituting the given values, we get:

ΔT = (5.050 cm - 5.000 cm) / ((2.4 x 10^-5 /°C) - (1.9 x 10^-5 /°C))

ΔT ≈ 437°C

Therefore, the temperature of the aluminum ring needs to be 437°F + 20°F = 457°F or 236°C + 111°C = 347°C (option D) for it to just slip over the brass rod at room temperature.

At 20 °\degree° C an Aluminum Ring Has an Inner Diameter of 5

At 20 $\degree$ C an Aluminum Ring Has an Inner Diameter of 5