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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

You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 190.0 J/s from a furnace at 400.0\(^\circ\)C to a container of boiling water under 1 atmosphere. What must the rod’s diameter be?

(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.

You have probably seen people jogging in extremely hot weather. There are good reasons not to do this! When jogging strenuously, an average runner of mass 68 kg and surface area 1.85 m\(^2\) produces energy at a rate of up to 1300 W, 80% of which is converted to heat. The jogger radiates heat but actually absorbs more from the hot air than he radiates away. At such high levels of activity, the skin's temperature can be elevated to around 33\(^\circ\)C instead of the usual 30\(^\circ\)C. (Ignore conduction, which would bring even more heat into his body.) The only way for the body to get rid of this extra heat is by evaporating water (sweating). (a) How much heat per second is produced just by the act of jogging? (b) How much net heat per second does the runner gain just from radiation if the air temperature is 40.0\(^\circ\)C (104\(^\circ\)F)? (Remember: He radiates out, but the environment radiates back in.) (c) What is the total amount of excess heat this runner's body must get rid of per second? (d) How much water must his body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is \(2.42 \times 10{^6} J/kg\). (e) How many 750-mL bottles of water must he drink after (or preferably before!) jogging for a half hour? Recall that a liter of water has a mass of 1.0 kg.

A thirsty nurse cools a 2.00-L bottle of a soft drink (mostly water) by pouring it into a large aluminum mug of mass 0.257 kg and adding 0.120 kg of ice initially at -15.0\(^\circ\)C. If the soft drink and mug are initially at 20.0\(^\circ\)C, what is the final temperature of the system, assuming that no heat is lost?

A 25,000-kg subway train initially traveling at 15.5 m/s slows to a stop in a station and then stays there long enough for its brakes to cool. The station's dimensions are 65.0 m long by 20.0 m wide by 12.0 m high. Assuming all the work done by the brakes in stopping the train is transferred as heat uniformly to all the air in the station, by how much does the air temperature in the station rise? Take the density of the air to be 1.20 kg/m\(^3\) and its specific heat to be 1020 J /kg \(\cdot\) K.

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