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Chapter 15: Problem 19

Calculate the increase in volume of \(100 \mathrm{~cm}^{3}\) of mercury when its temperature changes from \(10^{\circ} \mathrm{C}\) to \(35^{\circ} \mathrm{C}\). The volume coefficient of expansion of mercury is \(0.00018^{\circ} \mathrm{C}^{-1}\).

Short Answer

Expert verified
The volume of mercury increases by 0.45 cm鲁.

Step by step solution

01

Identify Initial Values

Initially, we have the initial volume \( V_i = 100 \text{ cm}^3 \), initial temperature \( T_i = 10^{\circ} \text{C}\), final temperature \( T_f = 35^{\circ} \text{C}\), and volume coefficient of expansion \( \beta = 0.00018^{\circ} \text{C}^{-1} \).
02

Calculate Temperature Change

Find the change in temperature, \( \Delta T \), by subtracting the initial temperature from the final temperature: \( \Delta T = T_f - T_i = 35^{\circ} \text{C} - 10^{\circ} \text{C} = 25^{\circ} \text{C} \).
03

Apply Volume Expansion Formula

The formula for volume expansion is \( \Delta V = V_i \times \beta \times \Delta T \). Substituting the known values, we have \( \Delta V = 100 \times 0.00018 \times 25 \).
04

Perform Calculations

Execute the multiplication to find the change in volume: \( \Delta V = 100 \times 0.00018 \times 25 = 0.45 \text{ cm}^3 \).
05

Interpret Results

The increase in volume when the mercury is heated from \(10^{\circ} \text{C}\) to \(35^{\circ} \text{C}\) is \(0.45 \text{ cm}^3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume Expansion
Volume expansion refers to the increase in volume of a substance as its temperature rises. This phenomenon occurs because most materials expand when heated.
The atoms within the material move more vigorously, requiring more space. This results in an expansion of volume.
For instance, in the example involving mercury, when its temperature rises from \(10^{\circ} \mathrm{C}\) to \(35^{\circ} \mathrm{C}\), we observe an increase in volume.
  • Volume expansion can be significant in liquids like mercury.
  • It's generally characterized using a volume coefficient of expansion.
  • This coefficient is crucial for predicting how much a given volume will expand as the temperature changes.
Temperature Change
Temperature change is a key driver of volume expansion. In essence, it refers to the difference between the initial and final temperatures of a material as it undergoes heating or cooling.
In the case of the mercury exercise, the temperature of the mercury changes from \(10^{\circ} \mathrm{C} \) to \(35^{\circ} \mathrm{C} \). This change of \(25^{\circ} \mathrm{C} \) is crucial for calculating volume expansion.
  • Temperature change is calculated as \( \Delta T = T_f - T_i \), where \( T_f \) and \( T_i \) are the final and initial temperatures, respectively.
  • Knowing the temperature change helps in determining how much the substance's volume will change.
  • It's a straightforward calculation but vital for accurate results in expansion problems.
Volume Coefficient of Expansion
The volume coefficient of expansion, denoted as \( \beta \), is a material-specific constant that quantifies how much a substance's volume will expand per degree of temperature change.
This coefficient provides insights into the thermal sensitivity of materials like mercury.
  • In the given example, mercury has a volume coefficient of expansion \( \beta = 0.00018^{\circ} \mathrm{C}^{-1} \).
  • The unit \(^{\circ} \mathrm{C}^{-1}\) indicates that this value allows us to predict volume changes for each degree Celsius of temperature increase.
  • The application of this coefficient in the formula \( \Delta V = V_i \times \beta \times \Delta T \) gives precise estimations of volume changes.
Using the correct coefficient is essential and assures that calculations reflect real-world scenarios accurately.

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Most popular questions from this chapter

A solid sphere of mass \(m\) and radius \(b\) is spinning freely on its axis with angular velocity \(\omega_{0}\). When heated by an amount \(\Delta T\), its angular velocity changes to \(\omega .\) Find \(\omega_{0} / \omega\) if the linear expansion coefficient for the material of the sphere is \(\alpha\).

A steel tape measures the length of a copper rod as \(90.00 \mathrm{~cm}\) when both are at \(10^{\circ} \mathrm{C}\), the calibration temperature for the tape. What would the tape read for the length of the rod when both are at \(30^{\circ} \mathrm{C}\) ? \(\alpha_{\text {stel }}=1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1} ; \alpha_{\text {copper }}=1.7 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\) At \(30^{\circ} \mathrm{C}\), the copper rod will be of length $$ L_{0}\left(1+\alpha_{c} \Delta T\right) $$ while adjacent "centimeter" marks on the steel tape will be separated by a distance of $$ (1.000 \mathrm{~cm})\left(1+\alpha_{s} \Delta T\right) $$ Therefore, the number of "centimeters" read on the tape will be $$ \frac{L_{0}\left(1+\alpha_{c} \Delta T\right)}{(1 \mathrm{~cm})\left(1+\alpha_{s} \Delta T\right)}=\frac{(90.00 \mathrm{~cm})\left[1+\left(1.7 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)\left(20^{\circ} \mathrm{C}\right)\right]}{(1.000 \mathrm{~cm})\left[1+\left(1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)\left(20^{\circ} \mathrm{C}\right)\right]}=90.00 \frac{1+3.4 \times 10^{-4}}{1+2.2 \times 10^{-4}} $$ Using the approximation $$ \frac{1}{1+x} \approx 1-x $$ for \(x\) small compared to 1 , we have $$ \begin{aligned} 90.00 \frac{1+3.4 \times 10^{-4}}{1+2.2 \times 10^{-4}} & \approx 90.00\left(1+3.4 \times 10^{-4}\right)\left(1-2.2 \times 10^{-4}\right) \approx 9000\left(1+3.4 \times 10^{-4}-2.2 \times 10^{-4}\right) \\ &=90.00+0.0108 \end{aligned} $$ The tape will read \(90.01 \mathrm{~cm}\).

A glass vessel is filled with exactly 1 liter of turpentine at \(20^{\circ} \mathrm{C}\). What volume of the liquid will overflow if the temperature is raised to \(86^{\circ} \mathrm{C}\) ? The coefficient of linear expansion of the glass is \(9.0 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\); the coefficient of volume expansion of turpentine is \(97 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\).

A rod \(3.0 \mathrm{~m}\) long is found to have expanded \(0.091 \mathrm{~cm}\) in length after a temperature rise of \(60{ }^{\circ} \mathrm{C}\). What is \(\alpha\) for the material of the rod?

A copper rod \(\left(\alpha=1.70 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)\) is \(20 \mathrm{~cm}\) longer than an aluminum \(\operatorname{rod}\left(\alpha=2.20 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)\) How long should the copper rod be if the difference in their lengths is to be independent of temperature? For their difference in lengths not to change with temperature, \(\Delta L\) must be the same for both rods under the same temperature change. That is, or $$ \begin{array}{c} \left(\alpha L_{0} \Delta T\right)_{\text {copper }}=\left(\alpha L_{0} \Delta T\right)_{\text {abminum }} \\ \left(1.70 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right) L_{0} \Delta T=\left(2.20 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)\left(L_{0}-0.20 \mathrm{~m}\right) \Delta T \end{array} $$ where \(L_{0}\) is the length of the copper rod, and \(\Delta T\) is the same for both rods. Solving for the original length yields \(L_{0}=0.88 \mathrm{~m}\).
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