/*! 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 26 The volume of a solid cube with ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 26

The volume of a solid cube with side \(s_{0}\) at temperature \(T_{0}\) is \(V_{0}=s_{0}^{3} .\) Show that if \(\Delta s < s_{0},\) the change in volume \(\Delta V\) due to a change in temperature \(\Delta T\) is given by $$ \frac{\Delta V}{V_{0}}=3 \alpha \Delta T $$ and therefore that \(\beta=3 \alpha .\) (Although we derive this relation for a cube, it applies to a solid of any shape.)

Short Answer

Expert verified
Answer: The change in volume due to a change in temperature for any solid (not just cubes) is given by the equation \(\frac{\Delta V}{V_0} = 3\alpha \Delta T\), which implies that the volume expansion coefficient, \(\beta\), is equal to three times the linear expansion coefficient, \(\alpha\) (i.e., \(\beta = 3\alpha\)).

Step by step solution

01

Understanding the problem

First, let's understand the problem. We are given a solid cube with side \(s_0\) and temperature \(T_0\). The volume of the cube at this temperature is \(V_0 = s_0^3\). It is important to note that the change in side length due to a change in temperature will be less than the original side length, meaning \(\Delta s < s_0\). We need to show that the change in volume due to a change in temperature follows the equation \(\frac{\Delta V}{V_0} = 3\alpha \Delta T\), and as a result, \(\beta = 3\alpha\) for any solid, not just cubes.
02

Finding the change in side length

To find the change in the side length of the cube due to a change in temperature, we'll use the linear expansion equation: $$ \Delta s = s_0\alpha\Delta T $$ Here, \(\alpha\) is the linear expansion coefficient, and \(\Delta T\) is the change in temperature. This equation gives us the change in side length, \(\Delta s\), when the temperature changes by \(\Delta T\).
03

Calculating the new side length and volume

Now that we have the change in side length, we can find the new side length and volume of the cube. The new side length \(s_1\) will be: $$ s_1 = s_0 + \Delta s $$ And the new volume \(V_1\) will be: $$ V_1 = s_1^3 $$
04

Expressing the change in volume

We want to find the change in volume, \(\Delta V\), which is defined as: $$ \Delta V = V_1 - V_0 $$ Using the equations for \(V_1\) and \(V_0\), we get: $$ \Delta V = s_1^3 - s_0^3 $$
05

Substituting the change in side length

Now, we can substitute our expressions for \(s_1\) and \(\Delta s\) into this equation. Doing this, we get: $$ \Delta V = (s_0 + s_0\alpha\Delta T)^3 - s_0^3 $$
06

Expanding the equation and simplifying

By expanding the equation and simplifying it, we will get: $$ \Delta V = s_0^3 (1 + \alpha\Delta T)^3 - s_0^3 $$
07

Using the binomial approximation

Since \(\Delta s < s_0\), we have \(\alpha\Delta T \ll 1\). Therefore, we can use the binomial approximation: $$ (1 + x)^n \approx 1 + nx $$ Applying this to our equation, we get: $$ \Delta V \approx s_0^3(1 + 3\alpha\Delta T) - s_0^3 $$ Simplifying further, we find: $$ \Delta V \approx 3s_0^3\alpha\Delta T $$
08

Finding the ratio between change in volume and initial volume

The problem asks us to find the ratio between the change in volume and the initial volume: $$ \frac{\Delta V}{V_0} $$ Since \(V_0 = s_0^3\), we can substitute our simplified equation for \(\Delta V\): $$ \frac{\Delta V}{V_0} = \frac{3s_0^3\alpha\Delta T}{s_0^3} $$
09

Simplifying the final expression

Finally, we can simplify the fraction, getting the desired expression: $$ \frac{\Delta V}{V_0} = 3\alpha\Delta T $$ This shows that \(\beta = 3\alpha\) for any solid, not just cubes, as the derivation is based on the general properties of linear expansion and volume.

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

Sand is composed of \(\mathrm{SiO}_{2} .\) Find the order of magnitude of the number of silicon (Si) atoms in a grain of sand. Approximate the sand grain as a sphere of diameter \(0.5 \mathrm{mm}\) and an \(\mathrm{SiO}_{2}\) molecule as a sphere of diameter \(0.5 \mathrm{nm}\)
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 \(95.0^{\circ} \mathrm{C} ?\)
The coefficient of linear expansion of brass is \(1.9 \times 10^{-5}\) $^{\circ} \mathrm{C}^{-1} .\( At \)20.0^{\circ} \mathrm{C},$ a hole in a sheet of brass has an area of \(1.00 \mathrm{mm}^{2} .\) How much larger is the area of the hole at $30.0^{\circ} \mathrm{C} ?(\text { Wile tutorial: loop around the equator })$
Agnes Pockels \((1862-1935)\) was able to determine Avogadro's number using only a few household chemicals, in particular oleic acid, whose formula is \(\mathrm{C}_{18} \mathrm{H}_{34} \mathrm{O}_{2}\) (a) What is the molar mass of this acid? (b) The mass of one drop of oleic acid is \(2.3 \times 10^{-5} \mathrm{g}\) and the volume is $2.6 \times 10^{-5} \mathrm{cm}^{3} .$ How many moles of oleic acid are there in one drop? (c) Now all Pockels needed was to find the number of molecules of oleic acid. Luckily, when oleic acid is spread out on water, it lines up in a layer one molecule thick. If the base of the molecule of oleic acid is a square of side \(d\), the height of the molecule is known to be \(7 d .\) Pockels spread out one drop of oleic acid on some water, and measured the area to be \(70.0 \mathrm{cm}^{2}\) Using the volume and the area of oleic acid, what is \(d ?\) (d) If we assume that this film is one molecule thick, how many molecules of oleic acid are there in the drop? (e) What value does this give you for Avogadro's number?

There are two identical containers of gas at the same temperature and pressure, one containing argon and the other neon. What is the ratio of the rms speed of the argon atoms to that of the neon atoms? The atomic mass of argon is twice that of neon. tutorial: RMS speed)
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Rotational Dynamics

Read Explanation

Work Energy and Power

Read Explanation

Electromagnetism

Read Explanation

Oscillations

Read Explanation

Solid State Physics

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.