/*! 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 86 An aluminium sphere of \(20 \mat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 86

An aluminium sphere of \(20 \mathrm{~cm}\) diameter is heated from \(0^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\). Its volume changes by (given that the coefficient of linear expansion for aluminium \(\left.\alpha_{A l}=23 \times 10^{-6} /{ }^{\circ} \mathrm{C}\right)\) (A) \(28.9 \mathrm{cc}\) (B) \(2.89 \mathrm{cc}\) (C) \(9.28 \mathrm{cc}\) (D) \(49.8 \mathrm{cc}\)

Short Answer

Expert verified
The change in volume of the aluminium sphere when heated from \(0^{\circ}\) C to \(100^{\circ}\) C is approximately \(28.9 \mathrm{cm^3}\) (Option A).

Step by step solution

01

Find the initial volume of the sphere

To find the initial volume of the sphere, use the formula for the volume of a sphere, which is given by: \[V = \frac{4}{3}\pi r^3\] The diameter of the sphere is 20 cm, so the radius (r) is 10 cm. Plug the value of r into the formula and calculate the initial volume of the sphere: \[V = \frac{4}{3}\pi (10 \mathrm{cm})^3\] \[V = \frac{4000}{3}\pi \mathrm{cm^3}\]
02

Calculate the volume expansion factor

Use the formula for volume expansion, which is given by: \[V_{final} = V_{initial}(1+3\alpha \Delta T)\] Where \(\alpha\) is the coefficient of linear expansion, \(\Delta T\) is the change in temperature, and \(3\alpha \Delta T\) is the volume expansion factor. With the given values, find the volume expansion factor: \[3\alpha \Delta T = 3(23 \times 10^{-6}/{}^{\circ}C)(100^{\circ}C)\] \[3\alpha \Delta T = 6.9 \times 10^{-3}\]
03

Calculate the final volume of the sphere

Use the volume expansion factor to find the final volume of the sphere: \[V_{final} = V_{initial}(1+3\alpha \Delta T)\] \[V_{final} = \frac{4000}{3}\pi \mathrm{cm^3}(1+6.9\times10^{-3})\] \[V_{final} = \frac{4000}{3}\pi \mathrm{cm^3}(1.0069)\]
04

Determine the change in volume

Now, determine the change in volume by subtracting the initial volume from the final volume: \[\Delta V = V_{final} - V_{initial}\] \[\Delta V = (\frac{4000}{3}\pi \mathrm{cm^3})(1.0069) - \frac{4000}{3}\pi \mathrm{cm^3}\] \[\Delta V = \frac{4000}{3}\pi (0.0069)\mathrm{cm^3}\] Now, calculate the value of \(\Delta V\): \[\Delta V \approx 28.9 \mathrm{cm^3}\]
05

Choose the correct option

Compare the calculated value of \(\Delta V\) with the given options and choose the correct answer: (A) 28.9 cc is the correct answer

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!

Key Concepts

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

Aluminium Properties
Aluminium is a lightweight, silver-white metallic element found extensively in the Earth's crust. It is known for its prominent properties that make it a favorite in many applications including construction, transportation, and consumer goods.
  • Lightweight: Aluminium is about one-third of the weight of steel, making it easier to shape and transport.
  • Corrosion Resistant: Naturally forms a protective oxide coating that resists corrosion, making it durable.
  • Conductivity: Known for good electrical and thermal conductivity, ideal for heat sinks, and electronic components.
  • Reflective: Excellent reflective properties that make it suitable for heat and light barriers.
Its ability to undergo thermal expansion without cracking or losing strength makes aluminium crucial in applications requiring reliable dimensional stability under shifting temperatures.
Coefficient of Linear Expansion
The coefficient of linear expansion is a material property that quantifies how much a material expands per degree of temperature increase. Aluminium has a relatively high coefficient, making it expand more than other metals like steel when heated.
  • Definition: It is defined as the fractional change in length per degree Celsius change in temperature.
  • For aluminium, it is denoted as \( \alpha_{Al} = 23 \times 10^{-6} /\degree C \).
  • This value indicates that for every temperature degree increase, each length unit of aluminium increases slightly.
  • In calculations, its effects are often tripled to measure volumetric changes using \( 3\alpha \Delta T \).
This property is crucial for designing components that will experience temperature changes, ensuring that they fit and function correctly without causing stress or damage to the material or connected parts.
Sphere Volume Calculations
Calculating the volume of a sphere involves determining how much space it occupies. The formula for the volume of a sphere is derived from calculus principles and is represented as:\[ V = \frac{4}{3}\pi r^3 \]Where:
  • \( V \) is the volume,
  • \( \pi \approx 3.1415 \), a mathematical constant, and
  • \( r \) is the radius of the sphere.
This formula requires knowing the radius, which is half the diameter. For thermal expansion scenarios, the initial volume is recalculated by incorporating the temperature-based expansion factor to determine the final volume and subsequent volume change.

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

Two rods of length \(L_{1}\) and \(L_{2}\) are made of materials whose coefficients of linear expansion are \(\alpha_{1}\) and \(\alpha_{2}\). If the difference between the two lengths is independent of temperature. (A) \(\left(L_{1} / L_{2}\right)=\left(\alpha_{1} / \alpha_{2}\right)\) (B) \(\left(L_{1} / L_{2}\right)=\left(\alpha_{2} / \alpha_{1}\right)\) (C) \(L_{1}^{2} \alpha_{1}=L_{2}^{2} \alpha_{2}\) (D) \(\alpha_{1}^{2} L_{1}=\alpha_{2}^{2} L_{2}\)

Water flows at the rate of \(0.1500 \mathrm{~kg} / \mathrm{min}\) through a tube and is heated by a heater dissipating \(25.2 \mathrm{~W}\). The inflow and outflow water temperatures are \(15.2^{\circ} \mathrm{C}\) and \(17.4^{\circ} \mathrm{C}\), respectively. When the rate of flow is increased to \(0.2318 \mathrm{~kg} /\) min and the rate of heating to \(37.8 \mathrm{~W}\), the inflow and outflow temperatures are unaltered. Find the rate of loss of heat from the tube

If the degrees of freedom of a gas molecule be \(f\), then the ratio of two specific heats \(C_{p} / C_{v}\) is given by (A) \(\frac{2}{f}+1\) (B) \(1-\frac{2}{f}\) (C) \(1+\frac{1}{f}\) (D) \(1-\frac{1}{f}\)

A substance of mass \(m \mathrm{~kg}\) requires is power input of \(P\) watts to remain in the molten state at its melting point. When the power is turned off, the sample completely solidifies in time \(t\) second. What is the latent heat of fusion of the substance? (A) \(\frac{P m}{t}\) (B) \(\frac{P t}{m}\) (C) \(\frac{m}{P t}\) (D) \(\frac{t}{P m}\)

A non-conducting piston of mass \(m\) and area \(S_{0}\) divides a non-conducting, closed cylinder as shown in Fig. \(10.22\). Piston having mass \(m\) is connected with top wall of cylinder by a spring of force constant \(k\). Top part is evacuated and bottom part contains an ideal gas at pressure \(P_{0}\) in equilibrium position. Adiabatic constant \(\gamma\) and in equilibrium length of each part is \(l\). (neglect friction) Find the angular frequency of oscillation. If process is isothermal. Length of gas column at equilibrium position is \(l_{1}\) and gas pressure is \(P_{1}\) at equilibrium position. (A) \(\sqrt{\frac{P_{1} S_{0}}{4 m l_{1}}}\) (B) \(\sqrt{\frac{2 P_{1} S_{0}}{m l_{1}}}\) (C) \(\sqrt{\frac{P_{1} S_{0}}{m l_{1}}}\) (D) \(\sqrt{\frac{P_{1} S_{0}}{2 m l_{1}}}\)
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Waves Physics

Read Explanation

Fluids

Read Explanation

Classical Mechanics

Read Explanation

Scientific Method Physics

Read Explanation

Astrophysics

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.