/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 A metal rod is \(40.125 \mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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's material for this temperature range.

Short Answer

Expert verified
The coefficient of linear expansion is approximately \( 2.29 \times 10^{-5} \mathrm{~^{\circ}C^{-1}} \).

Step by step solution

01

Understand the Concept

The coefficient of linear expansion, denoted by \( \alpha \), is a measure of how much a material expands per degree change in temperature. The formula to calculate \( \alpha \) is \[ \alpha = \frac{\Delta L}{L_0 \Delta T} \] where \( \Delta L \) is the change in length, \( L_0 \) is the original length, and \( \Delta T \) is the temperature change.
02

Identify Known Values

Given in the problem: \( L_0 = 40.125 \) cm at \( 20.0^{\circ} \mathrm{C} \), \( L = 40.148 \) cm at \( 45.0^{\circ} \mathrm{C} \). So, \( \Delta L = 40.148 - 40.125 \), and \( \Delta T = 45.0 - 20.0 \).
03

Calculate Change in Length, \( \Delta L \)

Calculate \( \Delta L \), which is the difference in length: \( \Delta L = 40.148 \mathrm{~cm} - 40.125 \mathrm{~cm} = 0.023 \mathrm{~cm} \).
04

Calculate Change in Temperature, \( \Delta T \)

Calculate \( \Delta T \) as the difference in temperature: \( \Delta T = 45.0^{\circ} \mathrm{C} - 20.0^{\circ} \mathrm{C} = 25.0^{\circ} \mathrm{C} \).
05

Substitute Values into the Formula

Substitute the known values into the coefficient of linear expansion formula: \[ \alpha = \frac{\Delta L}{L_0 \Delta T} = \frac{0.023 \mathrm{~cm}}{40.125 \mathrm{~cm} \times 25.0^{\circ} \mathrm{C}} \]
06

Perform the Calculation

Calculate \( \alpha \): \[ \alpha = \frac{0.023}{40.125 \times 25} \approx 2.29 \times 10^{-5} \mathrm{~^{\circ}C^{-1}} \]
07

Interpret the Result

The average coefficient of linear expansion for the rod's material over the given temperature range is \( 2.29 \times 10^{-5} \mathrm{~^{\circ}C^{-1}} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Thermal Expansion
Thermal expansion refers to the change in size of a material when its temperature changes. When heated, most materials expand, and when cooled, they contract. This phenomenon is due to the increase in the vibration of atoms as the temperature rises, which causes them to take up more space. The concept is crucial in understanding how materials will behave under different temperature conditions.

The amount a material expands is usually characterized by the coefficient of linear expansion, denoted by \( \alpha \). This coefficient tells you how much the length of a material will change for each degree change in temperature. It's typically measured in units of \( \mathrm{^{\circ} C^{-1}} \).

Understanding thermal expansion is important for practical applications, as it can affect the integrity of structural components and everyday items such as pipes and bridges. For instance, gaps between sections of metal rails on train tracks can be explained by thermal expansion theory, as it allows the rails to expand in the heat without bending.
Physics Problem Solving
Physics problem-solving involves applying established principles and formulas to find a solution to a given problem. It's a vital skill that requires understanding the underlying physics concepts and being able to manipulate mathematical equations to reach a logical conclusion.

When tackling a problem related to thermal expansion, like calculating the coefficient of linear expansion, here are some steps to consider:
  • Identify the known variables such as initial length, change in length, and temperature change.
  • Understand and write down the formula that applies to the problem at hand.
  • Substitute the known values into the formula correctly to ensure precision.
  • Perform calculations carefully, checking each step to avoid errors.
  • Interpret the result meaningfully to align with the context of the problem.
Solving physics problems can be challenging, but with practice and a systematic approach, it becomes more manageable and rewarding.
Materials Science
Materials science is the study of the properties and applications of materials of all types. By understanding materials science, one gains insight into why certain materials behave the way they do, especially under varying conditions like temperature changes.

Key factors considered in materials science, particularly with regard to thermal properties, include:
  • The structure of the material at atomic or molecular scales, which influences how it reacts to temperature changes.
  • The type of bonding between atoms, be it metallic, covalent, or ionic, as this affects expansion characteristics.
  • The material's thermal expansion properties, which help in selecting the appropriate material for specific applications.
By studying these elements, engineers and scientists can predict how different materials will behave in diverse environments, ensuring that the best materials are chosen for safety, efficiency, and longevity in various applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Global warming. As the earth warms, sea level will rise due to melting of the polar ice and thermal expansion of the oceans. Estimates of the expected temperature increase vary, but \(3.5 \mathrm{C}^{\circ}\) by the end of the century has been plausibly suggested. If we assume that the temperature of the oceans also increases by this amount, how much will sea level rise by the year 2100 due only to the thermal expansion of the water? Assume, reasonably, that the ocean basins do not expand appreciably. The average depth of the ocean is \(4000 \mathrm{~m}\), and the coefficient of volume expansion of water at \(20^{\circ} \mathrm{C}\) is \(0.207 \times 10^{-3}\left(\mathrm{C}^{\circ}\right)^{-1}\)

A picture window has dimensions of \(1.40 \mathrm{~m} \times 2.50 \mathrm{~m}\) and is made of glass \(5.20 \mathrm{~mm}\) thick. On a winter day, the outside temperature is \(-20.0^{\circ} \mathrm{C},\) while the inside temperature is a comfortable \(19.56^{\circ} \mathrm{C}\). (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 \(0.750-\mathrm{mm}\) -thick layer of paper (thermal conductivity \(0.0500 \mathrm{~W} /(\mathrm{m} \cdot \mathrm{K})) ?\)

for this temperature? (b) Elevated body temperature. During very vigorous exercise, the body's temperature can go as high as \(40^{\circ} \mathrm{C}\). What would Kelvin and Fahrenheit thermometers read for this temperature? (c) Temperature difference in the body. The surface temperature of the body is normally about \(7 \mathrm{C}^{\circ}\) lower than the internal temperature. Express this temperature difference in kelvins and in Fahrenheit degrees. (d) Blood storage. Blood stored at \(4.0^{\circ} \mathrm{C}\) lasts safely for about 3 weeks, whereas blood stored at \(-160^{\circ} \mathrm{C}\) lasts for 5 years. Express both temperatures on the Fahrenheit and Kelvin scales. (e) Heat stroke. If the body's temperature is above \(105^{\circ} \mathrm{F}\) for a prolonged period, heat stroke can result. Express this temperature on the Celsius and Kelvin scales.

A spherical pot of hot coffee contains \(0.75 \mathrm{~L}\) of liquid (essentially water) at an initial temperature of \(95^{\circ} \mathrm{C}\). The pot has an emissivity of \(0.60,\) and the surroundings are at a temperature of \(20.0^{\circ} \mathrm{C}\). Calculate the coffee's rate of heat loss by radiation.

In very cold weather, a significant mechanism for heat loss by the human body is energy expended in warming the air taken into the lungs with each breath. (a) On a cold winter day when the temperature is \(-20^{\circ} \mathrm{C},\) what is the amount of heat needed to warm to internal body temperature \(\left(37^{\circ} \mathrm{C}\right)\) the \(0.50 \mathrm{~L}\) of air exchanged with each breath? Assume that the specific heat of air is \(1020 \mathrm{~J} /(\mathrm{kg} \cdot \mathrm{K})\) and that \(1.0 \mathrm{~L}\) of air has a mass of \(1.3 \mathrm{~g}\). (b) How much heat is lost per hour if the respiration rate is 20 breaths per minute?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.