/*! 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 28 Calculate the increase in volume... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
The volume increase of mercury is \(0.45 \, \mathrm{cm}^3\).

Step by step solution

01

Understanding the Parameters

To solve this problem, we need to calculate the increase in volume of mercury as its temperature changes. We are given the initial volume \(V_0 = 100 \, \mathrm{cm}^3\), the initial temperature \(T_1 = 10^{\circ} \mathrm{C}\), the final temperature \(T_2 = 35^{\circ} \mathrm{C}\), and the volume coefficient of expansion \( \beta = 0.00018 \, {}^{\circ} \mathrm{C}^{-1} \).
02

Determining Temperature Change

Calculate the change in temperature using the formula: \( \Delta T = T_2 - T_1 \). Substitute the given values: \( \Delta T = 35^{\circ} \mathrm{C} - 10^{\circ} \mathrm{C} = 25^{\circ} \mathrm{C} \).
03

Using the Formula for Volume Expansion

The increase in volume \( \Delta V \) can be calculated using the formula:\[ \Delta V = V_0 \times \beta \times \Delta T \]Substitute the known values into the formula:\[ \Delta V = 100 \, \mathrm{cm}^3 \times 0.00018 \, {}^{\circ} \mathrm{C}^{-1} \times 25^{\circ} \mathrm{C} \].
04

Calculating the Increase in Volume

Perform the multiplication:\[ \Delta V = 100 \times 0.00018 \times 25 = 0.45 \]This means the volume increases by \(0.45 \, \mathrm{cm}^3\).

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Key Concepts

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

Volume Coefficient
The volume coefficient of expansion holds great importance when considering how materials like mercury expand when subjected to temperature changes. It is represented by the symbol \( \beta \) and is a measure indicating how much a substance's volume increases per degree Celsius rise in temperature. In the context of mercury in the original exercise, \( \beta \) is given as \(0.00018 \, {}^{\circ} \mathrm{C}^{-1}\). This coefficient effectively quantifies the fractional change in volume for every one-degree increase in temperature.
The concept of the volume coefficient might seem abstract, but it facilitates the prediction of volume changes in materials, ranging from everyday objects to industrial applications.
  • High volume coefficients imply a significant expansion with temperature increase.
  • Conversely, low volume coefficients suggest minimal change.
Understanding this coefficient is crucial for calculations involving thermal expansion, ensuring that predictions of volume change are accurate.
Temperature Change
Understanding temperature change is key to solving problems involving thermal expansion. In the exercise, the temperature change is calculated as the difference between the final and initial temperatures. The given temperatures are \(10^{\circ} \mathrm{C}\) and \(35^{\circ} \mathrm{C}\), resulting in a temperature change, \( \Delta T \), of \(25^{\circ} \mathrm{C}\).
The formula used is: \[\Delta T = T_2 - T_1\]Where:
  • \(T_1\) is the initial temperature.
  • \(T_2\) is the final temperature.
  • \(\Delta T\) represents the change in temperature.
Temperature change is a straightforward concept but serves as a critical factor when calculating changes in other properties, such as volume, in materials subjected to thermal expansion. It helps transition from what physical conditions initially were to what they become after heating or cooling.
Volume Expansion Formula
The volume expansion formula is used to calculate how much a given substance expands in volume when subjected to a temperature change. The formula applied in the given exercise is:\[\Delta V = V_0 \times \beta \times \Delta T\]Where:
  • \(\Delta V\) is the change in volume.
  • \(V_0\) is the initial volume.
  • \(\beta\) is the volume coefficient of expansion.
  • \(\Delta T\) is the change in temperature.
This formula highlights the direct relationship between volume change and temperature change, as well as the role of the volume coefficient. In the example, substituting the known values gives:\[ \Delta V = 100 \, \mathrm{cm}^3 \times 0.00018 \, {}^{\circ} \mathrm{C}^{-1} \times 25^{\circ} \mathrm{C} = 0.45 \, \mathrm{cm}^3\]The result indicates that the mercury's volume increases by \(0.45 \, \mathrm{cm}^3\) with the specified temperature change. Understanding this formula helps in predicting thermal behavior in various practical applications by accounting for changes in material volume with temperature fluctuations.

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Most popular questions from this chapter

Dry ice freezes at a temperature of \(-109.3^{\circ} \mathrm{F}\). What is that in Celsius?

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 cylinder of diameter \(1.00000 \mathrm{~cm}\) at \(30^{\circ} \mathrm{C}\) is to be slid into a hole in a steel plate. The hole has a diameter of \(0.99970 \mathrm{~cm}\) at 30 \({ }^{\circ} \mathrm{C}\). To what temperature must the plate be heated? For steel, \(\alpha=\) \(1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\) The plate will expand in the same way whether or not there is a hole in it. Hence, the hole expands in the same way a circle of steel filling it would expand. We want the diameter of the hole to change by $$ \Delta L=(1.00000-0.99970) \mathrm{cm}=0.00030 \mathrm{~cm} $$ Using \(\Delta L=\alpha L \Delta T\), $$ \Delta T=\frac{\Delta L}{\alpha L_{0}}=\frac{0.00030 \mathrm{~cm}}{\left(1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)(0.99970 \mathrm{~cm})}=27^{\circ} \mathrm{C} $$ The temperature of the plate must be \(30+27=57^{\circ} \mathrm{C}\).

At \(15.0^{\circ} \mathrm{C}\), a bare wheel has a diameter of \(30.000 \mathrm{~cm}\), and the inside diameter of its steel rim is \(29.930 \mathrm{~cm}\). To what temperature must the rim be heated so as to slip over the wheel? For this type of steel, \(\alpha=1.10 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\).

An iron ball has a diameter of \(6 \mathrm{~cm}\) and is \(0.010 \mathrm{~mm}\) too large to pass through a hole in a brass plate when the ball and plate are at a temperature of \(30^{\circ} \mathrm{C}\). At what temperature (the same for ball and plate) will the ball just pass through the hole? \(\alpha=1.2 \times 10^{-5}\) \({ }^{\circ} \mathrm{C}^{-1}\) and \(1.9 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\) for iron and brass, respectively.

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