Question #1508

At 21C, an aluminum ring has an inner diameter of 8 cm and a brass rod has a diameter of 8.06 cm. If the temperature coefficient of expansion for brass is 1.9 × 10[sup]−5[/sup] (degrees C)[sup]−1[/sup] and the temperature coefficient of expansion for aluminum is 2.4 × 10[sup]−5[/sup] (degrees C)[sup]−1[/sup], to what temperature must the ring be heated so that it will just slip overthe rod?

Answer in units of degrees C.

Answer in units of degrees C.

Expert's answer

Denote the diameter of the ring as D_{1} and the the diameter of the rod as D_{2}, the temperature coefficient of expansion for aluminium as α_{1}, for the brass as α_{2}.

The change in the linear dimension can be estimated by the formula:

Δ L = α * L * Δ T

In our case to slip over the rod the ring must have minimum diameter equal to the diameter of the rod:

D_{1}+ΔD_{1} = D_{2} + ΔD_{2}

D_{1} + D_{1}* α_{1} * ΔT = D_{2} + D_{2}* α_{2}* ΔT

Δ T = (D_{1}-D_{2})/(α_{2}D_{2}-α_{1}D_{1})= (8-8.06)/(1.9*8.06-2.4*8) x10^{5} = 154.4 C.

Δ T = T - T_{initial}

T = Δ T +T_{initial} = 154.4 + 21 = 175.4 C.

The change in the linear dimension can be estimated by the formula:

Δ L = α * L * Δ T

In our case to slip over the rod the ring must have minimum diameter equal to the diameter of the rod:

D

D

Δ T = (D

Δ T = T - T

T = Δ T +T

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