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Chapter 17: Problem 69

You are building a device for monitoring ultracold environments. Because the device will be used in environments where its temperature will change by \(200 .{ }^{\circ} \mathrm{C}\) in \(3.00 \mathrm{~s}\), it must have the ability to withstand thermal shock (rapid temperature changes). The volume of the device is \(5.00 \cdot 10^{-5} \mathrm{~m}^{3}\), and if the volume changes by \(1.00 \cdot 10^{-7} \mathrm{~m}^{3}\) in a time interval of \(5.00 \mathrm{~s}\), the device will crack and be rendered useless. What is the maximum volume expansion coefficient that the material you use to build the device can have?

Short Answer

Expert verified
Answer: The maximum volume expansion coefficient is \(6.00 \cdot 10^{-6} \mathrm{m}^{3}/\mathrm{m}^{3}\).

Step by step solution

01

Understand the Physics of the Problem

In this problem, we have a device that will experience rapid temperature changes (thermal shock). The device's volume should not change by more than \(1.00 \cdot 10^{-7} \mathrm{~m}^{3}\) in a time interval of \(5.00 \mathrm{~s}\). We are tasked with determining the maximum volume expansion coefficient that the material used to build the device can have to avoid cracking.
02

Calculate Volume Change due to Temperature Change

We know that the volume change due to temperature change is given by: \(\Delta V = \beta V_{0} \Delta T\) where \(\Delta V\) is the volume change, \(\beta\) is the volume expansion coefficient, \(V_{0}\) is the initial volume, and \(\Delta T\) is the temperature change. We are given the values for \(V_{0}\) and \(\Delta T\), and want to find the maximum value of \(\beta\) given the constraint on the volume change.
03

Use Constraints to Determine Maximum Volume Expansion Coefficient

Let's use the constraint that the volume change must be less than or equal to \(1.00 \cdot 10^{-7} \mathrm{~m}^{3}\) in \(5.00 \mathrm{~s}\). Since we have a temperature change of \(200.{ }^{\circ} \mathrm{C}\) occurring in \(3.00 \mathrm{~s}\), we can estimate that the device's volume will change in the same proportion during the \(5.00 \mathrm{~s}\) interval. Thus we have: \(\Delta V_{5s} = \frac{5.00}{3.00} \cdot \Delta V_{3s}\) \(\Delta V_{5s} = \frac{5.00}{3.00} \cdot \beta V_{0} \Delta T\) Now, we can plug in the given values and constraint for \(\Delta V_{5s}\): \(1.00 \cdot 10^{-7} \mathrm{~m}^{3} = \frac{5.00}{3.00} \cdot \beta (5.00 \cdot 10^{-5} \mathrm{~m}^{3})(200.{ }^{\circ} \mathrm{C})\)
04

Solve for the Maximum Volume Expansion Coefficient

Now, we can solve for \(\beta\): \(\beta = \frac{1.00 \cdot 10^{-7} \mathrm{~m}^{3}}{\frac{5.00}{3.00} \cdot (5.00 \cdot 10^{-5} \mathrm{~m}^{3})(200.{ }^{\circ} \mathrm{C})}\) \(\beta = \frac{1.00 \cdot 10^{-7} \mathrm{~m}^{3}}{\frac{5.00}{3.00} \cdot 10^{-2} \mathrm{~m}^{3}}\) \(\beta = \frac{1.00 \cdot 10^{-7} \mathrm{~m}^{3}}{ \frac{5.00 \cdot 10^{-2} \mathrm{~m}^{3}}{3.00}}\) \(\beta = \frac{3.00}{5.00} \cdot 10^{-5} \mathrm{m}^{3}/\mathrm{m}^{3}\) \(\beta = 6.00 \cdot 10^{-6} \mathrm{m}^{3}/\mathrm{m}^{3}\) So, the maximum volume expansion coefficient that the material used to build the device can have is \(6.00 \cdot 10^{-6} \mathrm{m}^{3}/\mathrm{m}^{3}\).

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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?

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?

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.

Thermal expansion seems like a small effect, but it can engender tremendous, often damaging, forces. For example, steel has a linear expansion coefficient of \(\alpha=1.2 \cdot 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\) and a bulk modulus of \(B=160\) GPa. Calculate the pressure engendered in steel by a \(1.0^{\circ} \mathrm{C}\) temperature increase.

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 \(40.0^{\circ} \mathrm{C}\).
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