Question

A medical device used for handling tissue samples has two metal screws, one \(20.0 \mathrm{~cm}\) long and made from brass \(\left(\alpha_{\mathrm{h}}=18.9 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\right)\) and the other \(30.0 \mathrm{~cm}\) long and made from aluminum \(\left(\alpha_{\mathrm{a}}=23.0 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\right) .\) A gap of \(1.00 \mathrm{~mm}\) exists between the ends of the screws at \(22.0^{\circ} \mathrm{C}\). At what temperature will the two screws touch?

Short answer

Answer: The two metal screws will touch at 85.9 °C.

Step-by-step solution

Determine the change in length for the brass and aluminum screws

First, we need to find the change in length for each screw individually: ΔL = L₀ * α * ΔT For the brass screw, L₀ (brass) = 20.0 cm α (brass) = 18.9 * 10^(-6) 1/°C For the aluminum screw, L₀ (aluminum) = 30.0 cm α (aluminum) = 23.0 * 10^(-6) 1/°C We will use these values in the later steps.

Write the equation to represent the gap closing

Since the sum of the lengths changes equal the initial gap, we can write the equation: ΔL(brass) + ΔL(aluminum) = 1.00 mm = 0.1 cm Substituting the expressions for ΔL from step 1, we have: (20.0 * 18.9 * 10^(-6) * ΔT) + (30.0 * 23.0 * 10^(-6) * ΔT) = 0.1

Solve the equation for ΔT and find the new temperature

Now, we can simplify the equation to solve for ΔT: (20.0 * 18.9 * 10^(-6) * ΔT) + (30.0 * 23.0 * 10^(-6) * ΔT) = 0.1 (20.0 * 18.9 + 30.0 * 23.0) * 10^(-6) * ΔT = 0.1 Now, solve for ΔT: ΔT = 0.1 / ((20.0 * 18.9 + 30.0 * 23.0) * 10^(-6)) ΔT = 63.9 °C Finally, since we know the initial temperature was 22.0 °C, the new temperature is: T_new = T_initial + ΔT T_new = 22.0 + 63.9 T_new = 85.9 °C The two screws will touch when the temperature reaches 85.9 °C.