Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A copper cube of side length \(40 . \mathrm{cm}\) is heated from \(20 .{ }^{\circ} \mathrm{C}\) to \(120{ }^{\circ} \mathrm{C} .\) What is the change in the volume of the cube? The linear expansion coefficient of copper is \(17 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\)

Short Answer

Expert verified
Answer: The change in volume of the copper cube is approximately 106.196 cm³.

Step by step solution

01

Calculate the change in temperature

To find the change in temperature, simply subtract the initial temperature from the final temperature. In this case, the initial temperature is \(20^\circ \mathrm{C}\) and the final temperature is \(120^\circ \mathrm{C}\): \(\Delta T = T_{final} - T_{initial} = 120 - 20 = 100^\circ \mathrm{C}\)
02

Calculate the change in side length

To calculate the change in side length, use the linear expansion formula, which is given by: \(\Delta L = L_{initial}\cdot\alpha\cdot\Delta T\) Where: - \(\Delta L\) is the change in length - \(L_{initial}\) is the initial length - \(\alpha\) is the linear expansion coefficient - \(\Delta T\) is the change in temperature Plug in the given values for the side length, linear expansion coefficient, and the change in temperature: \(\Delta L = 40\,\mathrm{cm} \cdot (17\cdot 10^{-6}\,^{\circ} \mathrm{C}^{-1}) \cdot (100^{\circ} \mathrm{C})\) \(\Delta L \approx 0.068\,\mathrm{cm}\)
03

Calculate the final side length

Add the change in side length to the initial side length to find the final side length: \(L_{final} = L_{initial} + \Delta L = 40\,\mathrm{cm} + 0.068\,\mathrm{cm} \approx 40.068\,\mathrm{cm}\)
04

Calculate the initial and final volumes

Use the side lengths to calculate the initial and final volumes of the cube, using the formula: \(V = L^3\) Initial volume: \(V_{initial} = (40\,\mathrm{cm})^3 = 64000\,\mathrm{cm}^3\) Final volume: \(V_{final} = (40.068\,\mathrm{cm})^3 \approx 64106.196\,\mathrm{cm}^3\)
05

Calculate the change in volume

Subtract the initial volume from the final volume to find the change in volume: \(\Delta V = V_{final} - V_{initial} = 64106.196\,\mathrm{cm}^3 - 64000\,\mathrm{cm}^3 \approx 106.196\,\mathrm{cm}^3\) The change in volume of the copper cube is approximately \(106.196\,\mathrm{cm}^3\).

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 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

A solid cylinder and a cylindrical shell, of identical outer radius and length and made of the same material, experience the same temperature increase, \(\Delta T\). Which of the two will expand to a larger outer radius?

A piece of dry ice (solid carbon dioxide) sitting in a classroom has a temperature of approximately \(-79^{\circ} \mathrm{C}\). a) What is this temperature in kelvins? b) What is this temperature in degrees Fahrenheit?

An iron horseshoe at room temperature is dunked into a cylindrical tank of water (radius of \(10.0 \mathrm{~cm}\) ) and causes the water level to rise by \(0.250 \mathrm{~cm} .\) When the horseshoe is heated in the blacksmith's stove from room temperature to a temperature of \(7.00 \cdot 10^{2} \mathrm{~K},\) worked into its final shape, and then dunked back into the water, how much does the water level rise above the "no horseshoe" level (ignore any water that evaporates as the horseshoe enters the water)? Note: The linear expansion coefficient for iron is roughly that of steel: \(11.0 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\)

The main mirror of a telescope has a diameter of \(4.713 \mathrm{~m}\). The mirror is made of glass with a linear expansion coefficient of \(3.789 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\) By how much would the temperature of the mirror need to be raised to increase the area of the mirror by \(4.750 \cdot 10^{-3} \mathrm{~m}^{2}\) ?

At \(26.45^{\circ} \mathrm{C},\) a steel bar is \(268.67 \mathrm{~cm}\) long and a brass bar is \(268.27 \mathrm{~cm}\) long. At what temperature will the two bars be the same length? Take the linear expansion coefficient of steel to be \(13.00 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\) and the linear expansion coefficient of brass to be \(19.00 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\).

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.

Sign-up for free