Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 21

A copper washer is to be fit in place over a steel bolt. Both pieces of metal are at \(20.0^{\circ} \mathrm{C} .\) If the diameter of the bolt is $1.0000 \mathrm{cm}\( and the inner diameter of the washer is \)0.9980 \mathrm{cm},$ to what temperature must the washer be raised so it will fit over the bolt? Only the copper washer is heated.

Short Answer

Expert verified
Answer: The copper washer must be raised to approximately \(192.3^{\circ} \mathrm{C}\) to fit over the steel bolt.

Step by step solution

01

Write down the formula for linear expansion

The formula for linear expansion is: \(\Delta L = L_0 \alpha \Delta T\) where: - \(\Delta L\) is the change in length (diameter) of the object, - \(L_0\) is the initial length (initial diameter) of the object, - \(\alpha\) is the coefficient of linear expansion for the material, - and \(\Delta T\) is the change in temperature.
02

Find the coefficients of thermal expansion for copper and steel

The coefficients of linear thermal expansion for copper and steel are: Copper: \(\alpha_{Cu} = 17 \times 10^{-6} \mathrm{K^{-1}}\) Steel: \(\alpha_{Steel} = 11 \times 10^{-6} \mathrm{K^{-1}}\)
03

Calculate the change in diameter required for the washer to fit over the bolt

From the problem statement, we know the inner diameter of the washer \(L_0 = 0.9980\ \mathrm{cm}\) and the diameter of the bolt \(D_{bolt} = 1.0000\ \mathrm{cm}\). To fit over the bolt, the washer needs to expand by: \(\Delta L_{req} = D_{bolt} - L_0 = 1.0000\ \mathrm{cm} - 0.9980\ \mathrm{cm} = 0.0020\ \mathrm{cm}\)
04

Calculate the change in temperature needed for the washer to expand enough to fit over the bolt

Using the formula for linear expansion, we now solve for the change in temperature \(\Delta T\): \(\Delta T = \frac{\Delta L_{req}}{L_0 \alpha_{Cu}} = \frac{0.0020\ \mathrm{cm}}{0.9980\ \mathrm{cm} \times 17 \times 10^{-6}\ \mathrm{K^{-1}}}\) \(\Delta T \approx 172.3 \ \mathrm{K}\)
05

Calculate the final temperature of the copper washer

Now, we add the change in temperature to the initial temperature to find the temperature at which the washer will fit over the bolt: \(T_{final} = T_{initial} + \Delta T = 20.0^{\circ} \mathrm{C} + 172.3 \ \mathrm{K}\) Convert temperature from Kelvin to Celsius: \(T_{final} \approx 192.3^{\circ} \mathrm{C}\) Therefore, the copper washer must be raised to approximately \(192.3^{\circ} \mathrm{C}\) to fit over the steel bolt.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

These data are from a constant-volume gas thermometer experiment. The volume of the gas was kept constant, while the temperature was changed. The resulting pressure was measured. Plot the data on a pressure versus temperature diagram. Based on these data, estimate the value of absolute zero in Celsius. $$\begin{array}{cc} \hline T\left(^{\circ} \mathrm{C}\right) & P(\mathrm{atm}) \\\ 0 & 1.00 \\\ 20 & 1.07 \\\ 100 & 1.37 \\\ -33 & 0.88 \\\ -196 & 0.28 \\\ \hline \end{array}$$
A cylindrical brass container with a base of \(75.0 \mathrm{cm}^{2}\) and height of \(20.0 \mathrm{cm}\) is filled to the brim with water when the system is at \(25.0^{\circ} \mathrm{C} .\) How much water overflows when the temperature of the water and the container is raised to \(95.0^{\circ} \mathrm{C} ?\)
During hibernation, an animal's metabolism slows down, and its body temperature lowers. For example, a California ground squirrel's body temperature lowers from \(40.0^{\circ} \mathrm{C}\) to \(10.0^{\circ} \mathrm{C}\) during hibernation. If we assume that the air in the squirrel's lungs is $75.0 \% \mathrm{N}_{2}\( and \)25.0 \% \mathrm{O}_{2},$ by how much will the rms speed of the air molecules in the lungs have decreased during hibernation?
Estimate the mean free path of a \(\mathrm{N}_{2}\) molecule in air at (a) sea level \((P \approx 100 \mathrm{kPa} \text { and } T \approx 290 \mathrm{K}),\) (b) the top of \(\quad\) Mt. Everest (altitude $=8.8 \mathrm{km}, P \approx 50 \mathrm{kPa},\( and \)T \approx 230 \mathrm{K}),\( and \)(\mathrm{c})$ an altitude of \(30 \mathrm{km}(P \approx 1 \mathrm{kPa}\) and \(T=230 \mathrm{K}) .\) For simplicity, assume that air is pure nitrogen gas. The diameter of a \(\mathrm{N}_{2}\) molecule is approximately \(0.3 \mathrm{nm}\)
The temperature at which liquid nitrogen boils (at atmospheric pressure) is \(77 \mathrm{K}\). Express this temperature in (a) \(^{\circ} \mathrm{C}\) and (b) \(^{\circ} \mathrm{F}\).
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Fundamentals of Physics

Read Explanation

Electromagnetism

Read Explanation

Nuclear Physics

Read Explanation

Rotational Dynamics

Read Explanation

Electric Charge Field and Potential

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.

Sign-up for free