/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 Even though steel has a relative... [FREE SOLUTION] | 91影视

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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

A 25.01 -mm-diameter brass ball sits at room temperature on a 25.00 - mm- diameter hole made in an aluminum plate. The ball and plate are heated uniformly in a furnace, so both are at the same temperature at all times. At what temperature will the ball fall through the plate?

Suppose a bimetallic strip is constructed of two strips of metals with linear expansion coefficients \(\alpha_{1}\) and \(\alpha_{2}\), where \(\alpha_{1}>\alpha_{2}\) a) If the temperature of the bimetallic strip is reduced by \(\Delta T\), what way will the strip bend (toward the side made of metal 1 or the side made of metal 2)? Briefly explain. b) If the temperature is increased by \(\Delta T\), which way will the strip bend?

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.

For a class demonstration, your physics instructor uniformly heats a bimetallic strip that is held in a horizontal orientation. As a result, the bimetallic strip bends upward. This tells you that the coefficient of linear thermal expansion for metal T, on the top is _____ that of metal B, on the bottom. a) smaller than b) larger than c) equal to

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 \(\beta=181 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\).
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