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Chapter 17: Problem 36

Even though steel has a relatively low linear expansion coefficient \(\left(\alpha_{\text {steel }}=13 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\right),\) the expansion of steel railroad tracks can potentially create significant problems on very hot summer days. To accommodate for the thermal expansion, a gap is left between consecutive sections of the track. If each section is \(25.0 \mathrm{~m}\) long at \(20.0{ }^{\circ} \mathrm{C}\) and the gap between sections is \(10.0 \mathrm{~mm}\) wide, what is the highest temperature the tracks can take before the expansion creates compressive forces between sections?

Short Answer

Expert verified
Answer: The highest temperature the steel railroad tracks can take before the expansion creates compressive forces between sections is \(50.77 \, ^{\circ}C\).

Step by step solution

01

List the given information and what we want to find

- Initial length of each section, \(L_0 = 25.0 \, m\) - Gap between sections, Gap \(= 10.0 \, mm = 0.010 \, m\) - Linear expansion coefficient of steel, \(\alpha_{\text{steel}}=13 \cdot 10^{-6} \, K^{-1}\) - Initial temperature \(T_i = 20.0 \, ^{\circ}C = 293.15 \, K\) - Final temperature \(T_f\): unknown (what we want to find)
02

Calculate the change in temperature (\(\Delta T\))

We first need to find the change in temperature \(\Delta T\). To do this, we will use the linear expansion formula, rearranged to solve for \(\Delta T\): $$\Delta T = \frac{\Delta L}{L_0 \alpha}$$ We know that the change in length \(\Delta L\) must be equal to the gap between the sections for compressive forces to arise: $$\Delta L = \text{Gap} = 0.010 \, m$$ Now we can plug the known values into the formula and solve for \(\Delta T\): $$\Delta T = \frac{0.010 \, m}{25.0 \, m \cdot 13 \cdot 10^{-6} \, K^{-1}} = 30.77 \, K$$
03

Calculate the highest temperature (\(T_f\))

Now that we have found the change in temperature, we can calculate the highest temperature the tracks can take before compressive forces arise. We can do this by adding the change in temperature to the initial temperature: $$T_f = T_i + \Delta T = 293.15 \, K + 30.77 \, K = 323.92 \, K$$ Since the given initial temperature was in Celsius, we will also convert our answer back to Celsius: $$T_f = 323.92 \, K - 273.15 = 50.77 \, ^{\circ}C$$
04

State the final answer

The highest temperature the steel railroad tracks can take before the expansion creates compressive forces between sections is \(50.77 \, ^{\circ}C\).

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Most popular questions from this chapter

You are outside on a hot day, with the air temperature at \(T_{0}\). Your sports drink is at a temperature \(T_{\mathrm{d}}\) in a sealed plastic bottle. There are a few remaining ice cubes in the sports drink, which are at a temperature \(T_{\mathrm{i}}\), but they are melting fast. a) Write an inequality expressing the relationship among the three temperatures. b) Give reasonable values for the three temperatures in degrees Celsius.

In a thermometer manufacturing plant, a type of mercury thermometer is built at room temperature \(\left(20^{\circ} \mathrm{C}\right)\) to measure temperatures in the \(20^{\circ} \mathrm{C}\) to \(70^{\circ} \mathrm{C}\) range, with \(\mathrm{a}\) \(1-\mathrm{cm}^{3}\) spherical reservoir at the bottom and a \(0.5-\mathrm{mm}\) inner diameter expansion tube. The wall thickness of the reservoir and tube is negligible, and the \(20^{\circ} \mathrm{C}\) mark is at the junction between the spherical reservoir and the tube. The tubes and reservoirs are made of fused silica, a transparent glass form of \(\mathrm{SiO}_{2}\) that has a very low linear expansion coefficient \((\alpha=\) \(\left.0.4 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\right) .\) By mistake, the material used for one batch of thermometers was quartz, a transparent crystalline form of \(\mathrm{SiO}_{2}\) with a much higher linear expansion coefficient \(\left(\alpha=12.3 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\right) .\) Will the manufacturer have to scrap the batch, or will the thermometers work fine, within the expected uncertainty of \(5 \%\) in reading the temperature? The volume expansion coefficient of mercury is \(\beta=181 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\).

The Rankine temperature scale is an absolute temperature scale that uses Fahrenheit degrees; that is, temperatures are measured in Fahrenheit degrees, starting at absolute zero. Find the relationships between temperature values on the Rankine scale and those on the Fahrenheit, Kelvin, and Celsius scales.

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?

The solar corona has a temperature of about \(1 \cdot 10^{6} \mathrm{~K}\). However, a spaceship flying in the corona will not be burned up. Why is this?
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