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

Step by step solution

01

1. Calculate the expansion of fused silica tube for the given temperature range

To calculate the expansion, we will use the linear expansion formula: \(\delta L = L_0 \cdot \alpha \cdot \delta T\), where \(\delta L\) is the change in length, \(L_0\) is the original length, \(\alpha\) is the linear expansion coefficient, and \(\delta T\) is the change in temperature. Here, we are given the temperature range as \(20^\circ C\) to \(70^\circ C\). So, \(\delta T = 70 - 20 = 50^\circ C\). We don't have the initial length of the expansion tube, \(L_0\). However, as we are interested in the percentage difference in the readings, we can skip this value because it will get cancelled when we calculate the percentage difference. Now, let's calculate the expansion of the fused silica tube: \(\delta L_{\text{fused silica}} = \alpha_{\text{fused silica}} \cdot \delta T = 0.4 \cdot 10^{-6} \cdot 50 = 20 \cdot 10^{-6} \ \text{m}\).

02

2. Calculate the expansion of quartz tube for the given temperature range

Similarly, the expansion of the quartz tube can be calculated as: \(\delta L_{\text{quartz}} = \alpha_{\text{quartz}} \cdot \delta T = 12.3 \cdot 10^{-6} \cdot 50 = 615 \cdot 10^{-6} \ \text{m}\).

03

3. Calculate the percentage difference in readings

Now, we need to find the percentage difference in the readings: \(\text{Percentage difference} = \frac{\delta L_{\text{quartz}} - \delta L_{\text{fused silica}}}{\delta L_{\text{fused silica}}} \cdot 100\%\) \(\text{Percentage difference} = \frac{615 \cdot 10^{-6} - 20 \cdot 10^{-6}}{20 \cdot 10^{-6}} \cdot 100\% = 2975\%\)

04

4. Check if the percentage difference is within the acceptable range

Since the percentage difference in the readings is \(2975\%\), which is significantly higher than the acceptable error of \(5\%\). Therefore, the manufacturer must scrap the batch of thermometers made of quartz.

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