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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 \(1.70 \times 10^{-6} \, \text{C}^{-1}\).

Step by step solution

01

Identify the Given Values

The problem provides the following information:- Temperature change, \( \Delta T = 75 \, \text{C}^{\circ} \)- Change in diameter, \( \Delta L = 2.3 \times 10^{-5} \, \text{m} \)- Original diameter, \( L_0 = 1.8 \times 10^{-2} \, \text{m} \)
02

Understand the Formula

The formula for linear expansion is given by: \[ \Delta L = \alpha \times L_0 \times \Delta T \]where:- \( \Delta L \) is the change in length,- \( \alpha \) is the coefficient of linear expansion,- \( L_0 \) is the original length,- \( \Delta T \) is the change in temperature.
03

Rearrange the Formula

Solve for \( \alpha \):\[ \alpha = \frac{\Delta L}{L_0 \times \Delta T} \]
04

Substitute Known Values

Substitute the given values into the rearranged formula:\[ \alpha = \frac{2.3 \times 10^{-5} \, \text{m}}{1.8 \times 10^{-2} \, \text{m} \times 75 \, \text{C}^{\circ}} \]
05

Calculate the Result

Perform the calculation:\[ \alpha = \frac{2.3 \times 10^{-5}}{1.8 \times 10^{-2} \times 75} \]\[ \alpha = \frac{2.3 \times 10^{-5}}{1.35} \]\[ \alpha \approx 1.70 \times 10^{-6} \, \text{C}^{-1} \]
06

Interpret the Result

The coefficient of linear expansion \( \alpha \) is approximately \( 1.70 \times 10^{-6} \, \text{C}^{-1} \). This means for every degree Celsius increase in temperature, the diameter of the coin increases by this fraction times its original diameter.

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Key Concepts

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

Temperature Change
Understanding how temperature change affects objects is key to many scientific and engineering tasks. When you increase an object's temperature, it expands, and when you decrease it, the object contracts. This change in dimensions due to temperature is universally observed. It is crucial to note that different materials react differently to the same temperature change, making it vital to know the particular material properties.
  • A temperature change ( \( \Delta T \)) is simply the difference between an object's initial and final temperatures.
  • In our exercise, \( \Delta T \) is \( 75 \, \text{C}^{\circ} \), meaning the coin's temperature increased by 75 degrees Celsius.
Not only is the amount of temperature change important, but also how rapidly it occurs. Rapid temperature changes can sometimes cause materials to crack or deform. In the linear expansion context, the temperature change is directly related to how much a material's dimensions will change.
Linear Expansion Formula
The linear expansion formula provides a way to calculate how much a material's length will change given a change in temperature and some intrinsic material property known as the coefficient of linear expansion. The formula is:\[ \Delta L = \alpha \times L_0 \times \Delta T \]Where:
  • \( \Delta L \) is the change in length.
  • \( \alpha \) is the coefficient of linear expansion.
  • \( L_0 \) is the original length of the material.
  • \( \Delta T \) is the temperature change.
This formula helps predict how materials expand or contract with temperature changes. In the provided exercise, this formula is used to determine the coefficient of linear expansion of a coin material. This involves rearranging the formula to solve for \( \alpha \) given the changes in temperature and diameter observed for the coin. Understanding this formula allows engineers to design structures and engines, considering how they might behave under temperature fluctuations.
Thermal Expansion
Thermal expansion is a material's tendency to change in size or volume due to temperature variations. This phenomenon is critical in engineering applications, as ignoring it can lead to structural failures. There are three types of thermal expansion coefficients - linear, area, and volumetric - focusing respectively on changes in one, two, and three dimensions. In our scenario, we are dealing with linear expansion because we are only interested in the change in length (diameter) of the coin.Key facets of thermal expansion:
  • Different materials have different thermal expansion properties, largely based on their molecular structure.
  • The coefficient of linear expansion ( \( \alpha \)) provides a measure of how much a material will expand per degree of temperature change.
In practice, designers and engineers must consider thermal expansion when selecting materials and designing components. Bridges, trains, and even ground surfaces account for thermal expansion to mitigate potential structural issues. Hence, knowing the material properties like the coefficient of linear expansion is critical for maintaining safety and functionality in many applications.

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Most popular questions from this chapter

A thermos contains \(150 \mathrm{cm}^{3}\) of coffee at \(85^{\circ} \mathrm{C}\). To cool the coffee, you drop two \(11-\mathrm{g}\) ice cubes into the thermos. The ice cubes are initially at \(0^{\circ} \mathrm{C}\) and melt completely. What is the final temperature of the coffee? Treat the coffee as if it were water.

The latent heat of vaporization of \(\mathrm{H}_{2} \mathrm{O}\) at body temperature \(\left(37.0^{\circ} \mathrm{C}\right)\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} .\) To cool the body of a \(75-\mathrm{kg}\) jogger [average specific heat capacity \(\left.=3500 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\right]\) by \(1.5 \mathrm{C}^{\circ},\) how many kilograms of water in the form of sweat have to be evaporated?

The heating element of a water heater in an apartment building has a maximum power output of \(28 \mathrm{kW}\). Four residents of the building take showers at the same time, and each receives heated water at a volume flow rate of \(14 \times 10^{-5} \mathrm{m}^{3} / \mathrm{s} .\) If the water going into the heater has a temperature of \(11^{\circ} \mathrm{C},\) what is the maximum possible temperature of the hot water that each showering resident receives?

At a fabrication plant, a hot metal forging has a mass of \(75 \mathrm{kg}\) and a specific heat capacity of \(430 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right) .\) To harden it, the forging is immersed in \(710 \mathrm{kg}\) of oil that has a temperature of \(32^{\circ} \mathrm{C}\) and a specific heat capacity of \(2700 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right) .\) The final temperature of the oil and forging at thermal equilibrium is \(47^{\circ} \mathrm{C}\). Assuming that heat flows only between the forging and the oil, determine the initial temperature of the forging.

A copper-constantan thermocouple generates a voltage of \(4.75 \times\) \(10^{-3}\) volts when the temperature of the hot junction is \(110.0^{\circ} \mathrm{C}\) and the reference junction is kept at a temperature of \(0.0^{\circ} \mathrm{C} .\) If the voltage is proportional to the difference in temperature between the junctions, what is the temperature of the hot junction when the voltage is \(1.90 \times 10^{-3}\) volts?

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