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Chapter 17: Problem 19

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

Short Answer

Expert verified
The penny's diameter would be approximately 1.9014 cm on a hot day in Death Valley and 1.8964 cm on a cold night in Greenland.

Step by step solution

01

Understand the formula for linear expansion

The formula for linear expansion due to temperature change is given by:\[ \Delta L = L_0 \times \alpha \times \Delta T \] where \( \Delta L \) is the change in length (or diameter in this case), \( L_0 \) is the original length, \( \alpha \) is the coefficient of linear expansion, and \( \Delta T \) is the change in temperature.
02

Calculate the change in temperature

First, calculate the temperature change for both scenarios:1. **Hot day in Death Valley**: The temperature increases from \(20.0^{\circ} \mathrm{C}\) to \(48.0^{\circ} \mathrm{C}\). \[ \Delta T_{\text{hot}} = 48.0 - 20.0 = 28.0 \text{ }^{\circ} \mathrm{C} \]2. **Cold night in Greenland**: The temperature decreases from \(20.0^{\circ} \mathrm{C}\) to \(-53^{\circ} \mathrm{C}\). \[ \Delta T_{\text{cold}} = -53 - 20.0 = -73.0 \text{ }^{\circ} \mathrm{C} \]
03

Calculate diameter change for hot day

Using the linear expansion formula for Death Valley:\[ \Delta L_{\text{hot}} = 1.9000 \times 2.6 \times 10^{-5} \times 28.0 \]Calculate \( \Delta L_{\text{hot}} \):\( \Delta L_{\text{hot}} = 0.001382 \text{ cm} \) approximately.
04

Calculate new diameter for hot day

Add the diameter change \( \Delta L_{\text{hot}} \) to the original diameter:\[ L_{\text{hot}} = 1.9000 + 0.001382 = 1.901382 \text{ cm} \]So, the diameter on a hot day is approximately \(1.9014 \text{ cm}\).
05

Calculate diameter change for cold night

Use the linear expansion formula for the cold condition:\[ \Delta L_{\text{cold}} = 1.9000 \times 2.6 \times 10^{-5} \times (-73.0) \]Calculate \( \Delta L_{\text{cold}} \):\( \Delta L_{\text{cold}} = -0.003598 \text{ cm} \) approximately.
06

Calculate new diameter for cold night

Subtract the diameter change \( \Delta L_{\text{cold}} \) from the original diameter:\[ L_{\text{cold}} = 1.9000 - 0.003598 = 1.896402 \text{ cm} \] So, the diameter on a cold night is approximately \(1.8964 \text{ cm}\).

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Key Concepts

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

Coefficient of Linear Expansion
The Coefficient of Linear Expansion (\(\alpha\)) is a measure of how a material's size changes with temperature. It reflects the fractional change in length per degree of temperature change. For most materials, this value remains constant over a moderate range of temperatures, allowing for predictable calculations.
  • This coefficient is usually expressed in units of \(\mathrm{K}^{-1}\), indicating the change per Kelvin (or Celsius, since the scales have equal degree intervals).
  • The value is specific to each material, affected by its molecular structure and bonding. In the penny exercise, \(\alpha = 2.6 \times 10^{-5} \mathrm{K}^{-1}\), primarily due to its zinc content.
  • Materials with higher coefficients expand and contract more significantly with temperature changes.
This information is crucial when designing anything from bridges to thermometers, where dimensional stability is important in varying thermal environments.
Temperature Change
Temperature Change (\(\Delta T\)) is a critical component in understanding how materials expand or contract as conditions fluctuate. It is simply the difference between the final and initial temperatures. In most practical applications, positive \(\Delta T\) denotes warming, causing expansion, while a negative \(\Delta T\) indicates cooling and contraction.

In the penny example:
  • On a hot day in Death Valley,\(\Delta T\) is calculated as \(48.0^{\circ} \mathrm{C} - 20.0^{\circ} \mathrm{C} = 28.0^{\circ} \mathrm{C}\).
  • On a cold night in Greenland,\(\Delta T\) becomes \(-53^{\circ} \mathrm{C} - 20.0^{\circ} \mathrm{C} = -73.0^{\circ} \mathrm{C}\).
Understanding temperature change helps us predict material behavior under different climate conditions, which is essential in fields ranging from construction to electronics.
Linear Expansion Formula
The Linear Expansion Formula provides a mathematical method to determine how an object's size changes with temperature. It is written as:\[\Delta L = L_0 \times \alpha \times \Delta T\]where:
  • \(\Delta L\) is the change in length or size (diameter for the penny).
  • \(L_0\) is the original length.
  • \(\alpha\) is the coefficient of linear expansion.
  • \(\Delta T\) is the change in temperature.
This formula is invaluable in anticipating dimensional changes. For the penny, applying the formula shows:
  • On a hot day, it expands to accommodate the increased temperature.
  • On a cold night, it contracts due to lower temperatures.
This guidance is necessary for anyone working with materials liable to experience thermal changes in their applications, ensuring structural integrity and efficiency.

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Most popular questions from this chapter

Suppose that a steel hoop could be constructed to fit just around the earth's equator at a temperature of \(20.0^{\circ} \mathrm{C}\) . What would be the thickness of space between the hoop and the earth if the temperature of the hoop were increased by 0.500 \(\mathrm{C}^{\circ} ?\)

A metal rod is 40.125 \(\mathrm{cm}\) long at \(20.0^{\circ} \mathrm{C}\) and 40.148 \(\mathrm{cm}\) long at \(45.0^{\circ} \mathrm{C}\) . Calculate the average coefficient of linear expansion of the rod for this temperature range.

A vessel whose walls are thermally insulated contains 2.40 \(\mathrm{kg}\) of water and 0.450 \(\mathrm{kg}\) of ice, all at a temperature of \(0.0^{\circ} \mathrm{C}\) The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to \(28.0^{\circ} \mathrm{C}\) ? You can ignore the heat transferred to the container.

In an effort to stay awake for an all-night study session, a student makes a cup of coffee by first placing a \(200-\mathrm{W}\) electric immersion heater in 0.320 \(\mathrm{kg}\) of water. (a) How much heat must be added to the water to raise its temperature from \(20.0^{\circ} \mathrm{C}\) to \(80.0^{\circ} \mathrm{C}\) ? (b) How much time is required? Assume that all of the heater's power goes into heating the water.

(a) You feel sick and are told that you have a temperature of \(40.2^{\circ} \mathrm{C}\) . What is your temperature in "F? Should you be concerned? (b) The morning weather report in Sydney gives a current temperature of \(12^{\circ} \mathrm{C}\) . What is this temperature in \(^{\circ} \mathrm{F} ?\)
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