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Chapter 17: Problem 37

A medical device used for handling tissue samples has two metal screws, one \(20.0 \mathrm{~cm}\) long and made from brass \(\left(\alpha_{\mathrm{b}}=18.9 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\right)\) 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

Expert verified
The linear expansion coefficients for brass and aluminum are 18.9 x 10^-6 ºC^-1 and 23.0 x 10^-6 ºC^-1 respectively. Answer: The two screws will touch when the temperature reaches approximately 22.02ºC.

Step by step solution

01

Identify the variables given

In this exercise, we are given: - The initial length of brass screw (\(L_{\mathrm{b}}=20.0 \mathrm{~cm}\)). - The initial length of aluminum screw (\(L_{\mathrm{a}}=30.0 \mathrm{~cm}\)). - The linear expansion coefficient for brass (\(\alpha_{\mathrm{b}}=18.9 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\)). - The linear expansion coefficient for aluminum (\(\alpha_{\mathrm{a}}=23.0 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\)). - The gap between the ends of the screws at \(22.0^{\circ} \mathrm{C}\) (\(d=1.00 \mathrm{~mm}\)). We need to find the temperature at which the two screws will touch.
02

Convert all measurements to meters

To make calculations easier, convert all measurements to meters: - Brass screw length: \(L_{\mathrm{b}}=0.20 \mathrm{~m}\) - Aluminum screw length: \(L_{\mathrm{a}}=0.30 \mathrm{~m}\) - Gap: \(d=0.001 \mathrm{~m}\)
03

Calculate the expansion of each screw

The expansion of an object due to temperature change can be calculated using the formula: $$\Delta L = L \cdot \alpha \cdot \Delta T$$ Where \(\Delta L\) is the change in length, \(L\) is the initial length, \(\alpha\) is the linear expansion coefficient, and \(\Delta T\) is the temperature change. For the brass screw, we have: $$\Delta L_{\mathrm{b}} = L_{\mathrm{b}} \cdot \alpha_{\mathrm{b}} \cdot \Delta T$$ For the aluminum screw, we have: $$\Delta L_{\mathrm{a}} = L_{\mathrm{a}} \cdot \alpha_{\mathrm{a}} \cdot \Delta T$$
04

Set up an equation for when the screws will touch

When the screws touch, the sum of their expansions will be equal to the initial gap between them: $$\Delta L_{\mathrm{b}} + \Delta L_{\mathrm{a}} = d$$ Substituting the expansion formulas from step 3, we get: $$L_{\mathrm{b}} \cdot \alpha_{\mathrm{b}} \cdot \Delta T + L_{\mathrm{a}} \cdot \alpha_{\mathrm{a}} \cdot \Delta T = d$$
05

Solve for the temperature change \(\Delta T\)

Factor out \(\Delta T\) from the equation in step 4: $$\Delta T (L_{\mathrm{b}} \cdot \alpha_{\mathrm{b}} + L_{\mathrm{a}} \cdot \alpha_{\mathrm{a}}) = d$$ Now, divide both sides by \((L_{\mathrm{b}} \cdot \alpha_{\mathrm{b}} + L_{\mathrm{a}} \cdot \alpha_{\mathrm{a}})\) to solve for \(\Delta T\), using the values in meters from step 2: $$\Delta T = \frac{d}{L_{\mathrm{b}} \cdot \alpha_{\mathrm{b}} + L_{\mathrm{a}} \cdot \alpha_{\mathrm{a}}} = \frac{0.001}{(0.20 \cdot 18.9 \cdot 10^{-6}) + (0.30 \cdot 23.0 \cdot 10^{-6})} \approx 0.019668 \mathrm{~K}$$
06

Calculate the final temperature

Now that we have the temperature change, we can add it to the initial temperature of \(22.0^{\circ}\)C to find the final temperature: $$T_{\mathrm{final}} = T_{\mathrm{initial}} + \Delta T = 22.0 + 0.019668 = 22.01967^{\circ} \mathrm{C}$$ The two screws will touch when the temperature reaches approximately \(22.02^{\circ} \mathrm{C}\).

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Most popular questions from this chapter

A steel rod of length \(1.0000 \mathrm{~m}\) and cross-sectional area \(5.00 \cdot 10^{-4} \mathrm{~m}^{2}\) is placed snugly against two immobile end points. The rod is initially placed when the temperature is \(0^{\circ} \mathrm{C}\). Find the stress in the rod when the temperature rises to \(40.0^{\circ} \mathrm{C}\).

A piece of dry ice (solid carbon dioxide) sitting in a classroom has a temperature of approximately \(-79^{\circ} \mathrm{C}\) a) What is this temperature in kelvins? b) What is this temperature in degrees Fahrenheit?

Would it be possible to have a temperature scale defined in such a way that the hotter an object or system got, the lower (less positive or more negative) its temperature was?

At what temperature do the Celsius and Fahrenheit temperature scales have the same numeric value? a) -40 degrees b) 0 degrees c) 40 degrees d) 100 degrees

On a hot summer day, a cubical swimming pool is filled to within \(1.0 \mathrm{~cm}\) of the top with water at \(21{ }^{\circ} \mathrm{C} .\) When the water warms to \(37^{\circ} \mathrm{C}\), the pool overflows. What is the depth of the pool?
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