/*! 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 15 A brass ring of diameter \(10.00... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 15

A brass ring of diameter \(10.00 \mathrm{~cm}\) at \(20.0^{\circ} \mathrm{C}\) is heated and slipped over an aluminum rod of diameter \(10.01 \mathrm{~cm}\) at \(20.0^{\circ} \mathrm{C}\). Assuming the average coefficients of linear expansion are constant, (a) to what temperature must the combination be cooled to separate the two metals? Is that temperature attainable? (b) What if the aluminum rod were \(10.02 \mathrm{~cm}\) in diameter?

Short Answer

Expert verified
a) It is found that for the two metals to separate, the temperature has to be cooled down to -202.6 掳C. This is not an attainable temperature in ordinary conditions. b) If the diameter of the aluminum rod were \(10.02 \mathrm{~cm}\), the temperature required to separate the two metals would be even less attainable.

Step by step solution

01

Understanding thermal expansion

Thermal expansion depends on three factors - initial size, change in temperature, and type of material. Linear expansion can be expressed in the formula \(螖L = 伪L_0 螖T\), where \(螖L\) is the change in length (which in this case is the diameter), \(伪\) is the coefficient of linear expansion, \(L_0\) is the initial length (diameter), and \(螖T\) is the change in temperature
02

Applying the thermal expansion theory to brass

Initially, the brass ring has a diameter of \(10.00cm\) at \(20.0掳C\). As it cools and contracts, the diameter decreases. The brass ring should contract enough to just slip off the aluminum rod. Using the thermal expansion formula, we can set \(L_0 = 10.00 cm\), \(螖L = (10.00 cm - 10.01 cm) = -0.01 cm\) for contraction, and \(伪 = 19脳10^{-6}掳C^{-1} \(for brass).
03

Calculating the new temperature for brass

We now use these values to solve for \(螖T\) using the formula \(螖T = 螖L /(伪L_0) \). This gives us the change in temperature needed for the brass ring to slip off the aluminum. By subtracting this change from the initial temperature of \(20掳C\), we can find the final temperature at which the metals will separate.
04

Testing attainability and applying expansion theory to an aluminum rod with different diameter

Given the final temperature, we assess if it's realistic or attainable. If it falls within ordinary temperature ranges, it is attainable. In the second part of the problem, the same steps are repeated using a larger diameter for the aluminum rod. In this case, the brass ring must contract more - i.e., cool down more - to slip off the rod. Applying the same linear thermal expansion theory, we obtain a new \(螖L\) equalling the difference between the diameter of the brass ring and the larger aluminum rod, and we then calculate the new final temperature.

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 91影视!

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

S The pressure gauge on a cylinder of gas registers the gauge pressure, which is the difference between the interior and exterior pressure, \(P_{0}\). When the cylinder is full, the mass of the gas in it is \(m_{i}\) at a gauge pressure of \(P_{i}\). Assuming the temperature of the cylinder remains constant, use the ideal gas law and a relationship between moles and mass to show that the mass of the gas remaining in the cylinder when the gauge pressure reading is \(P_{f}\) is given by

The boiling point of liquid hydrogen is \(20.3 \mathrm{~K}\) at atmospheric pressure. What is this temperature on (a) the Celsius scale and (b) the Fahrenheit scale?

Use Avogadro's number to find the mass of a helium atom.

Before beginning a long trip on a hot day, a driver inflates an automobile tire to a gauge pressure of \(1.80 \mathrm{~atm}\) at \(300 \mathrm{~K}\). At the end of the trip, the gauge pressure has increased to \(2.20\) atm. (a) Assuming the volume has remained constant, what is the tempera- ture of the air inside the tire? (b) What percentage of the original mass of air in the tire should be released so the pressure returns to its original value? Assume the temperature remains at the value found in part (a) and the volume of the tire remains constant as air is released.

Death Valley holds the record for the highest recorded temperature in the United States. On July 10,1913 , at a place called Furnace Creek Ranch, the temperature rose to \(134^{\circ} \mathrm{F}\). The lowest U.S. temperature ever recorded occurred at Prospect Creek Camp in Alaska on January 23,1971 , when the temperature plummeted to \(-79.8^{\circ} \mathrm{F}\). (a) Convert these temperatures to the Celsius scale. (b) Convert the Celsius temperatures to Kelvin.
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Energy Physics

Read Explanation

Force

Read Explanation

Fluids

Read Explanation

Dynamics

Read Explanation

Famous Physicists

Read Explanation
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.