Linear Expansion Problems

Question

1.) A brass ring has an inside diameter of 3.98 centimeters at 20.0°C. To what temperature should it be heated to fit exactly on a shaft 4.00 centimeters in diameter?

2.) A steel pipeline is 364.0 kilometers long at a temperature of 20.00°C.

a. Determine the pipeline's length at a temperature of -35.00°C

b. Determine the pipeline's length at a temperature of 35.00°C

c. Determine the total difference in length of the pipeline between -35.00°C and 35.00°C

Answer 1

To determine thermal expansion we will use the following equation ΔL = αLΔT

Where: -ΔL: the change in length of the object.

-α: Coefficient of linear expansion (depends on the type of material, brass: 19x10^-6 °C^-1).

-L: Original length of the object.

-ΔT: Change in temperature of the object.

1.) We want to use heat to expand a brass ring with a diameter of d₀ = 3.98 cm to have a diameter of d = 4.00 cm

this means we must change the diameter by ΔL = Δd = 4.00 cm - 3.98 cm = 0.02 cm. the coefficient of linear expansion for brass (found online) is 19x10^-6 °C^-1. the original diameter of the ring L = d = 3.98cm. So what temperature change ΔT = T - T₀ will yield this diameter change of 0.02 cm? We know T₀ = 20.0 °C (the original temperature of the ring) so we must use our equation to solve for T.

Δd = αdΔT

Δd/(αd) = ΔT = T - T₀

Δd/(αd) + T₀ = T

T = (0.02 cm)/(19x10^-6 °C^-1 * 3.98 cm) + 20.0 °C

T = 284 °C

So to change the diameter of our brass ring by just 0.02 cm, we must increase the temperature to 284 °C!

2.a.) We have a pipeline with an original length L₀ = 364.0 km and an original temperature of T₀ = 20.00 °C. The linear coefficient of expansion of steel (found online) is α = 13 x10^-6 °C^-1. We would like to determine the new length of the pipeline (L) if its temperature is changed to T = -35.00°C, this will yield a temperature change of ΔT = -35.00°C - 20.00 °C = -55.00 °C. We will use our equation to solve for L:

ΔL = αLΔT

L - L₀ = αL₀ΔT

L = αLΔT + L₀

L = (13 x10^-6 °C^-1)(364.0 km)(-55.00 °C) + 364.0 km

L = 363.7 km

2.b.) Everything is the same as in 2.a. but now the final temperature is T = 35.00 °C. yielding a change in temperature of ΔT = 35.00°C - 20.00 °C = 15.00 °C

L = αLΔT + L₀

L = (13 x10^-6 °C^-1)(364.0 km)(15.00 °C) + 364.0 km

L = 364.1 km

2.c.) Now we simply subtract our 2.a. answer from our 2.b. answer:

364.1 km - 363.7 km = 0.4 km (which is 1/4 a mile)