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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 4.00-kg silver ingot is taken from a furnace, where its temperature is 750.0\(^\circ\)C, and placed on a large block of ice at 0.0\(^\circ\)C. Assuming that all the heat given up by the silver is used to melt the ice, how much ice is melted?

A picture window has dimensions of 1.40 m \(\times\) 2.50 mand is made of glass 5.20 mm thick. On a winter day, the temperature of the outside surface of the glass is -20.0\(^\circ\)C, while the temperature of the inside surface is a comfortable 19.5\(^\circ\)C. (a) At what rate is heat being lost through the window by conduction? (b) At what rate would heat be lost through the window if you covered it with a 0.750-mm-thick layer of paper (thermal conductivity 0.0500 W/m \(\cdot\) K)?

You are given a sample of metal and asked to determine its specific heat. You weigh the sample and find that its weight is 28.4 N. You carefully add \(1.25 \times 10{^4}\) J of heat energy to the sample and find that its temperature rises 18.0 C\(^\circ\). What is the sample's specific heat?

You put a bottle of soft drink in a refrigerator and leave it until its temperature has dropped 10.0 K. What is its temperature change in (a) F\(^\circ\) and (b) C\(^\circ\)?

To measure the specific heat in the liquid phase of a newly developed cryoprotectant, you place a sample of the new cryoprotectant in contact with a cold plate until the solution's temperature drops from room temperature to its freezing point. Then you measure the heat transferred to the cold plate. If the system isn't sufficiently isolated from its room-temperature surroundings, what will be the effect on the measurement of the specific heat? (a) The measured specific heat will be greater than the actual specific heat; (b) the measured specific heat will be less than the actual specific heat; (c) there will be no effect because the thermal conductivity of the cryoprotectant is so low; (d) there will be no effect on the specific heat, but the temperature of the freezing point will change.

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