Question

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

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

Step-by-step solution

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.

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.

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 \(\Delta V_{5s}\): \(1.00 \cdot 10^{-7} \mathrm{~m}^{3} = \frac{5.00}{3.00} \cdot \beta (5.00 \cdot 10^{-5} \mathrm{~m}^{3})(200.{ }^{\circ} \mathrm{C})\)

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 \(6.00 \cdot 10^{-6} \mathrm{m}^{3}/\mathrm{m}^{3}\).