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When the temperature of a coin is raised by \(75 \mathrm{C}^{\circ},\) the coin's diameter increases by \(2.3 \times 10^{-5} \mathrm{~m}\). If the original diameter of the coin is \(1.8 \times 10^{-2} \mathrm{~m}\), find the coefficient of linear expansion.

Short Answer

Expert verified
The coefficient of linear expansion is \( 1.7 \times 10^{-6} \mathrm{m/}\mathrm{m}\cdot\mathrm{C}^{\circ} \).

Step by step solution

01

Understanding the Coefficient of Linear Expansion

The coefficient of linear expansion, denoted as \( \alpha \), measures how much a material's length changes per degree change in temperature. The formula to calculate it is given by: \( \Delta L = \alpha L_0 \Delta T \), where \( \Delta L \) is the change in length, \( L_0 \) is the original length, and \( \Delta T \) is the temperature change.
02

Identifying Given Values

From the problem, we have: \( \Delta L = 2.3 \times 10^{-5} \) m, \( L_0 = 1.8 \times 10^{-2} \) m, and \( \Delta T = 75 \mathrm{C}^{\circ} \). We need to use these values to find the coefficient of linear expansion \( \alpha \).
03

Rearranging the Formula

Rearrange the formula for the coefficient of linear expansion to solve for \( \alpha \):\[\alpha = \frac{\Delta L}{L_0 \Delta T}\] Substituting the given values will allow us to find \( \alpha \).
04

Substituting Values and Calculating

Substitute the known values into the equation:\[\alpha = \frac{2.3 \times 10^{-5}}{1.8 \times 10^{-2} \times 75}\]Calculate the coefficient: \[\alpha = \frac{2.3 \times 10^{-5}}{1.35} = 1.7 \times 10^{-6} \mathrm{~m/}\mathrm{m}\cdot\mathrm{C}^{\circ}\]
05

Final Result

The coefficient of linear expansion for the coin is \( 1.7 \times 10^{-6} \mathrm{m/}\mathrm{m}\cdot\mathrm{C}^{\circ} \). This means that for every degree Celsius change, the coin's diameter changes by this coefficient relative to its original size.

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Key Concepts

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

Thermal Expansion
Thermal expansion is an important concept in physics and engineering. It describes how a material expands when its temperature increases. When a substance is heated, its particles begin to move more vigorously. This causes the substance to take up more space and results in the material expanding.
There are three types of thermal expansion: linear expansion, area expansion, and volume expansion.
  • Linear expansion is focused on changes in length.
  • Area expansion applies to changes in surface area.
  • Volume expansion considers changes in overall volume.
In the context of the original problem, we are dealing with linear expansion. This is because we are interested in how the diameter of the coin changes with temperature. The degree of expansion depends on the material's coefficient of linear expansion. Understanding this property helps in designing objects that will undergo temperature changes, like bridges and thermostats.
Linear Expansion Formula
The linear expansion formula is an essential tool for calculating how much a material will increase in size with a temperature change. It's given by:\[\Delta L = \alpha L_0 \Delta T\]where:
  • \( \Delta L \) is the change in length or size.
  • \( \alpha \) is the coefficient of linear expansion for the material.
  • \( L_0 \) is the original length or size.
  • \( \Delta T \) is the change in temperature.
In our exercise, \( \Delta L \) is the increase in the coin's diameter, \( L_0 \) is the original diameter of the coin, and \( \Delta T \) is the temperature change. Rearranging this formula allows one to solve for any of the variables, provided the other values are known. This flexibility makes it incredibly useful in practical applications, such as determining necessary material clearances and tolerances in manufacturing.
Temperature Change Effect
The temperature change effect is the principle that the size of materials will change when exposed to different temperatures. When a material is heated, its particles gain energy and move more, causes it to expand. Conversely, when cooled, it contracts as particle movement decreases.

In the example of the coin, heating by \(75 \mathrm{C}^{\circ}\) leads to an increase in the diameter. The extent of this effect depends on the coefficient of linear expansion, \(\alpha\). A higher \(\alpha\) implies a larger change for the same temperature change.
  • Materials with tightly packed atomic structures have lower expansion coefficients.
  • Materials like metals usually have higher coefficients, making them more susceptible to noticeable size changes with temperature fluctuations.
Understanding the temperature change effect and having accurate \(\alpha\) values is crucial in construction, as buildings must accommodate thermal expansion without compromising structural integrity.

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Most popular questions from this chapter

At a temperature of \(0{ }^{\circ} \mathrm{C}\), the mass and volume of a fluid are \(825 \mathrm{~kg}\) and \(1.17 \mathrm{~m}^{3}\). The coefficient of volume expansion is \(1.26 \times 10^{-3}\left(\mathrm{C}^{\circ}\right)^{-1}\). (a) What is the density of the fluid at this temperature? (b) What is the density of the fluid when the temperature has risen to \(20.0^{\circ} \mathrm{C}\) ?

Concept Questions An aluminum can is filled to the brim with a liquid. The can and the liquid are heated so their temperatures change by the same amount. (a) In general, what factors determine how the volume of an object changes when it is heated? (b) Aluminum has a smaller coefficient of volume expansion than the liquid does. Which, if either, will expand more, the can or the liquid? Provide a reason for your answer. (c) How is the volume of liquid that spills over related to the changes in the volume of the can and the liquid?

Multiple-Concept Example 4 reviews the concepts that are involved in this problem. A ruler is accurate when the temperature is \(25^{\circ} \mathrm{C}\). When the temperature drops to \(-14^{\circ} \mathrm{C}\), the ruler shrinks and no longer measures distances accurately. However, the ruler can be made to read correctly if a force of magnitude \(1.2 \times 10^{3} \mathrm{~N}\) is applied to each end so as to stretch it back to its original length. The ruler has a cross-sectional area of \(1.6 \times 10^{-5} \mathrm{~m}^{2},\) and it is made from a material whose coefficient of linear expansion is \(2.5 \times 10^{-5}\left(\mathrm{C}^{0}\right)^{-1}\). What is Young's modulus for the material from which the ruler is made?

A locomotive wheel is \(1.00 \mathrm{~m}\) in diameter. A \(25.0-\mathrm{kg}\) steel band has a temperature of \(20.0^{\circ} \mathrm{C}\) and a diameter that is \(6.00 \times 10^{-4} \mathrm{~m}\) less than that of the wheel. What is the smallest mass of water vapor at \(100^{\circ} \mathrm{C}\) that can be condensed on the steel band to heat it, so that it will fit onto the wheel? Do not ignore the water that results from the condensation.

A \(0.200-\mathrm{kg}\) piece of aluminum that has a temperature of \(-155^{\circ} \mathrm{C}\) is added to \(1.5 \mathrm{~kg}\) of water that has a temperature of \(3.0^{\circ} \mathrm{C}\). At equilibrium the temperature is \(0.0^{\circ} \mathrm{C}\). Ignoring the container and assuming that the heat exchanged with the surroundings is negligible, determine the mass of water that has been frozen into ice.

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