Question

An iron ring is to fit snugly on a cylindrical iron rod. At 20°C, the diameter of the rod is 6.445 cm and the inside diameter of the ring is 6.420 cm. To slip over the rod, the ring must be slightly larger than than the diameter by about 0.008 cm. To what temperature must the ring be brought if it's hole is to be large enough so it will slip over the rod?

Answer 1

The hole in the ring must be increased from a diameter of 6.420 cm to 6.445 cm + 0.008 cm = 6.453 cm. The ring must be heated since the hole diameter will increase linearly. Solve for ΔT.

$∆T = \frac{∆L}{αL_0}$

Coefficient of linear expansion $\alpha = 12 \times 10^{-6}\;ºC^{-1}$

$∆T = \frac{6.453 – 6.420}{12 \times 10^{-6}\times 6.420} = 430\;ºC$

Temperature must be raised to T = 20 ºC + 430 ºC = 450 ºC

Answer: 450 ºC.