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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 0.100 kg contains 0.160 kg of water and 0.0180 kg of ice in thermal equilibrium at atmospheric pressure. If 0.750 kg of lead at 255\(^\circ\)C is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings.

The rate at which radiant energy from the sun reaches the earth's upper atmosphere is about 1.50 kW/m\(^2\). The distance from the earth to the sun is \(1.50 \times 10{^1}{^1} m\), and the radius of the sun is \(6.96 \times 10{^8} m\). (a) What is the rate of radiation of energy per unit area from the sun's surface? (b) If the sun radiates as an ideal blackbody, what is the temperature of its surface?

One experimental method of measuring an insulating material's thermal conductivity is to construct a box of the material and measure the power input to an electric heater inside the box that maintains the interior at a measured temperature above the outside surface. Suppose that in such an apparatus a power input of 180 W is required to keep the interior surface of the box 65.0 C\(^\circ\) (about 120 F\(^\circ\)) above the temperature of the outer surface. The total area of the box is 2.18 m\(^2\), and the wall thickness is 3.90 cm. Find the thermal conductivity of the material in SI units.

Convert the following Celsius temperatures to Fahrenheit: (a) -62.8\(^\circ\)C, the lowest temperature ever recorded in North America (February 3, 1947, Snag, Yukon); (b) 56.7\(^\circ\)C, the highest temperature ever recorded in the United States (July 10, 1913, Death Valley, California); (c) 31.1\(^\circ\)C, the world’s highest average annual temperature (Lugh Ferrandi, Somalia).

BIO Conduction Through the Skin. The blood plays an important role in removing heat from the body by bringing this energy directly to the surface where it can radiate away. Nevertheless, this heat must still travel through the skin before it can radiate away. Assume that the blood is brought to the bottom layer of skin at 37.0\(^\circ\)C and that the outer surface of the skin is at 30.0\(^\circ\)C. Skin varies in thickness from 0.50 mm to a few millimeters on the palms and soles, so assume an average thickness of 0.75 mm. A 165-lb, 6-ft-tall person has a surface area of about 2.0 m\(^2\) and loses heat at a net rate of 75 W while resting. On the basis of our assumptions, what is the thermal conductivity of this person’s skin?

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