In order to create a tight fit between two metal parts, machinists sometimes make the interior part larger than the hole into which it will fit and then either cool the interior part or heat the exterior part until they fogether. Suppose an aluminum rod with diameter \(D_{1}\) (at \(\left.2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C}\right)\) is to be fit into a hole in a brass plate that has a diameter \(D_{2}=10.000 \mathrm{~mm}\) (at \(\left.2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C}\right) .\) The machinists can cool the rod to \(77.0 \mathrm{~K}\) by immersing it in liquid nitrogen. What is the largest possible diameter that the rod can have at \(2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C}\) and just fit into the hole if the rod is cooled to \(77.0 \mathrm{~K}\) and the brass plate is left at \(2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C} ?\) The linear expansion coefficients for aluminum and brass are \(22 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\) and \(19 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\), respectively.

Step by step solution

01

Calculate the change in diameter for the aluminum rod

First, we need to find the change in diameter for the aluminum rod when cooled to 77 K. The change in diameter is given by: ΔD = αD₀ΔT ΔD = (22 * 10^{-6} °C^{-1})(D_{1})(20°C - 77K) Note that we need to make sure we have the same units for temperature. Since 0°C is equal to 273.15 K, the given temperatures can be converted as: 20°C = 293.15 K Now we can write the change in diameter equation as: ΔD = (22 * 10^{-6} °C^{-1})(D_{1})(293.15K - 77K)

02

Calculate the change in diameter for the brass hole

Next, we need to find the change in diameter for the brass hole when the plate is heated to 20°C. The change in diameter is given by: ΔD = αD₀ΔT ΔD = (19 * 10^{-6} °C^{-1})(10mm)(293.15K - 293.15K) Since the brass plate is not changing temperature (remaining at 20°C), the change in diameter for the brass hole is 0. D_{2} = D_{2 initial} + ΔD = 10mm + 0 = 10mm Note that D_{2} remains constant at 10mm.

03

Determine the expanded diameter of the aluminum rod

Now we will find the expanded diameter of the aluminum rod (D_{1 final}) at 20°C. When the rod returns from 77K to 20°C, it will expand back to its original size D_{1}, which should fit into the brass hole. D_{1 final} = D_{1 initial} - ΔD Now, we substitute the expression for ΔD for aluminum from Step 1: D_{1 final} = D_{1} - (22 * 10^{-6} °C^{-1})(D_{1})(293.15K - 77K)

04

Determine the maximum diameter of the aluminum rod

Finally, we can set the expanded aluminum rod diameter equal to the brass hole diameter, and solve for D_{1}: D_{1} - (22 * 10^{-6} °C^{-1})(D_{1})(293.15K - 77K) = 10mm D_{1} - (22 * 10^{-6})(D_{1})(216.15) = 10mm (1 - 22 * 10^{-6}(216.15))D_{1} = 10mm D_{1} = 10mm / (1 - 22 * 10^{-6}(216.15)) D_{1} ≈ 10.048 mm The largest possible diameter of the aluminum rod at 20°C is approximately 10.048 mm, so it can fit into the brass hole if the rod is cooled to 77 K and the brass plate is left at 20°C.

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