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A U.S. penny has a diameter of 1.9000 cm at 20.0\(^\circ\)C. The coin is made of a metal alloy (mostly zinc) for which the coefficient of linear expansion is \(2.6 \times 10{^-}{^5} K{^-}{^1}\). What would its diameter be on a hot day in Death Valley (48.0\(^\circ\)C)? On a cold night in the mountains of Greenland (-53\(^\circ\)C)?

Short Answer

Expert verified
Diameter at 48.0°C: 1.90138 cm, at -53.0°C: 1.89639 cm.

Step by step solution

01

Understand the Problem

We need to find the change in diameter of a penny when it's exposed to different temperatures. Given is the initial diameter of 1.9000 cm and the temperatures: 20.0°C as the initial temperature, 48.0°C and -53.0°C as the final temperatures, with a linear expansion coefficient of \(2.6 \times 10^{-5} \, K^{-1}\).
02

Formula for Linear Expansion

The formula to find the change in length (or diameter in this case) due to temperature change is:\[ \Delta L = \alpha \times L_0 \times \Delta T \]where \(\alpha\) is the coefficient of linear expansion, \(L_0\) is the original length or diameter, and \(\Delta T\) is the change in temperature.
03

Calculate Diameter at 48.0°C

Use the formula to calculate the diameter at 48.0°C.- \( \Delta T = 48.0 - 20.0 = 28.0 \, °C \)- \( \Delta L = (2.6 \times 10^{-5}) \times 1.9000 \times 28.0 \)Calculate \( \Delta L \):\[ \Delta L = 0.00138 \, cm \]Thus, the new diameter is:\[ D = 1.9000 + 0.00138 = 1.90138 \, cm \]
04

Calculate Diameter at -53.0°C

Use the formula to calculate the diameter at -53.0°C.- \( \Delta T = -53.0 - 20.0 = -73.0 \, °C \)- \( \Delta L = (2.6 \times 10^{-5}) \times 1.9000 \times (-73.0) \)Calculate \( \Delta L \):\[ \Delta L = -0.00361 \, cm \]Thus, the new diameter is:\[ D = 1.9000 - 0.00361 = 1.89639 \, cm \]

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Key Concepts

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

Coefficient of Linear Expansion
When discussing the expansion or contraction of materials with changes in temperature, the coefficient of linear expansion is crucial. It's a number that tells us how much a material will expand or contract per degree of temperature change. This coefficient is specific to each material. In the case of the penny made from a zinc alloy, the coefficient is given as \(2.6 \times 10^{-5} \, K^{-1}\).
This means for every degree Celsius (or Kelvin) the temperature changes, the alloy will expand or contract by \(2.6 \times 10^{-5}\) times its original size. It's important to note that this coefficient always relates to linear dimensions, such as length or diameter, rather than area or volume.
  • The coefficient of linear expansion is represented by the Greek letter alpha (\(\alpha\)).
  • It is a small value for most materials, signifying that only minor changes in size usually occur with temperature changes.
  • An understanding of \(\alpha\) aids in predicting how structures might behave under different temperatures, ensuring safety and functionality.
Temperature Change Effects
Temperature change significantly impacts the dimensions of materials due to their atomic structure. When heated, most materials expand because particles move faster and take up more space. Conversely, cooling might make these materials contract as particles slow down and move closer together.
For the penny scenario:
  • When the temperature rises to 48.0°C, the diameter of the penny increases. This is due to the kinetic energy increase in particles causing them to spread apart.
  • When the temperature drops to -53.0°C, the penny contracts as the kinetic energy decreases, allowing the particles to settle more tightly.
Understanding these effects is pivotal in engineering and construction where materials may undergo large temperature shifts. It ensures that designs accommodate possible expansion or contraction, preventing stress, damage, or malfunction.
Linear Expansion Formula
The linear expansion formula is used to calculate the change in a material's dimension when temperature changes. The formula is:\[\Delta L = \alpha \times L_0 \times \Delta T\]Using this formula:
  • \(\Delta L\) represents the change in length (or diameter for the penny).
  • \(\alpha\) is the coefficient of linear expansion.
  • \(L_0\) is the original length or diameter of the object.
  • \(\Delta T\) is the change in temperature (final temperature minus initial temperature). This term might be positive or negative, affecting whether \(\Delta L\) represents expansion or contraction.
For the penny, this formula calculates how much its diameter would change when subjected to different temperatures—an essential tool for understanding and predicting the behavior of materials in various thermal conditions.

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Most popular questions from this chapter

A copper calorimeter can with mass 0.446 kg contains 0.0950 kg of ice. The system is initially at 0.0\(^\circ\)C. (a) If 0.0350 kg of steam at 100.0\(^\circ\)C and 1.00 atm pressure is added to the can, what is the final temperature of the calorimeter can and its contents? (b) At the final temperature, how many kilograms are there of ice, how many of liquid water, and how many of steam?

In a container of negligible mass, 0.0400 kg of steam at 100\(^\circ\)C and atmospheric pressure is added to 0.200 kg of water at 50.0\(^\circ\)C. (a) If no heat is lost to the surroundings, what is the final temperature of the system? (b) At the final temperature, how many kilograms are there of steam and how many of liquid water?

A typical doughnut contains 2.0 g of protein, 17.0 g of carbohydrates, and 7.0 g of fat. Average food energy values are 4.0 kcal/g for protein and carbohydrates and 9.0 kcal/g for fat. (a) During heavy exercise, an average person uses energy at a rate of 510 kcal/h. How long would you have to exercise to 'work off' one doughnut? (b) If the energy in the doughnut could somehow be converted into the kinetic energy of your body as a whole, how fast could you move after eating the doughnut? Take your mass to be 60 kg, and express your answer in m/s and in km/h.

A long rod, insulated to prevent heat loss along its sides, is in perfect thermal contact with boiling water (at atmospheric pressure) at one end and with an ice-water mixture at the other (Fig. E17.62). The rod consists of a 1.00-m section of copper (one end in boiling water) joined end to end to a length \(L_2\) of steel (one end in the ice-water mixture). Both sections of the rod have crosssectional areas of 4.00 cm\(^2\). The temperature of the copper- steel junction is 65.0\(^\circ\)C after a steady state has been set up. (a) How much heat per second flows from the boiling water to the ice-water mixture? (b) What is the length \(L_2\) of the steel section?

(a) Calculate the one temperature at which Fahrenheit and Celsius thermometers agree with each other. (b) Calculate the one temperature at which Fahrenheit and Kelvin thermometers agree with each other.

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