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Expert-verified Found in: Page 541 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # Find the change in volume of an aluminum sphere with an initial radius of 10 cm when the sphere is heated from ${\mathbf{0}}{\mathbf{.}}{{\mathbf{0}}}^{{\mathbf{o}}}{\mathbf{C}}{\mathbf{}}{\mathbf{to}}{\mathbf{}}{\mathbf{100}}{\mathbf{.}}{{\mathbf{0}}}^{{\mathbf{o}}}{\mathbf{C}}$.

The change in volume of an aluminum sphere is $29{\mathrm{cm}}^{3}$

See the step by step solution

## Step 1: The given data

1. Initial radius of sphere ${\mathrm{r}}_{1}=10\mathrm{cm}$ at ${\mathrm{T}}_{1}={0}^{\mathrm{o}}\mathrm{C}$
2. Final temperature at which sphere expands ${\mathrm{T}}_{2}={100}^{\mathrm{o}}\mathrm{C}$
3. Linear expansion coefficient for aluminum is $\mathrm{\alpha }=23×{10}^{-6}/{}^{\mathrm{o}}\mathrm{C}$

## Step 2: Understanding the concept of linear expansion

When an object's temperature changes, it expands and grows larger, a process known as thermal expansion. Thermal expansion can occur due to an increase in temperature. For the given problem, we have to use the formula for volume expansion. Change in volume expansion depends on the original volume, the temperature change, and the volume expansion coefficient.

Formula:

The linear expansion of a body, $∆\mathrm{L}=\mathrm{L\alpha }∆\mathrm{T}$ …(i)

Where is the coefficient of linear expansion of body.

The volume change in expansion of a body, $∆\mathrm{V}=\mathrm{V\beta }∆\mathrm{T}$ …(ii)

Where is the coefficient of volume expansion of the rod

The volume of the sphere, $\mathrm{V}=\frac{4}{3}{\mathrm{\pi r}}^{3}$ …(iii)

## Step 3: Calculation of the volume change

So, the coefficient of volume expansion of the rod is given as:

$\begin{array}{rcl}\mathrm{\beta }& =& 3\mathrm{\alpha }\\ & =& 3×23×{10}^{-6}/{}^{\mathrm{o}}\mathrm{C}\\ & =& 69×{10}^{-6}/{}^{\mathrm{o}}\mathrm{C}\end{array}$

Therefore, we can calculate the initial volume of the sphere by using the formula of equation (iii) as given:

$\begin{array}{rcl}\mathrm{V}& =& \frac{4}{3}×3.14×{\left(10\mathrm{cm}\right)}^{3}\\ & =& 4186.66{\mathrm{cm}}^{3}\end{array}$

We can now find the change in volume using equation (ii) as:

$\begin{array}{rcl}∆\mathrm{V}& =& 4186.66{\mathrm{cm}}^{3}×69×{10}^{-6}/{}^{\mathrm{o}}\mathrm{C}×{\left(100-0\right)}^{\mathrm{o}}\mathrm{C}\\ & =& 28.88{\mathrm{cm}}^{3}\\ & \approx & 29{\mathrm{cm}}^{3}\end{array}$

Hence, the change in volume is $29{\mathrm{cm}}^{3}$ ### Want to see more solutions like these? 