Chapter 13, Problem 89A

(a)

To determine

Volume of an aluminum cube at $270.0\text{K}$ , if it has a volume of $\text{0}\text{.350}{\text{m}}^3$ at $\text{350}\text{K}$ .

Answer to Problem 89A

Volume of cube at $270.0\text{K}$ will be $\text{0}\text{.348}{\text{m}}^3$ .

Explanation of Solution

Given:

Volume coefficient of expansion, $\beta \text{ = 75}\times {10}^{-6}\text{per}°\text{C}$

Volume, $V_1\text{= 0}\text{.350}{\text{m}}^3$

Initial temperature, $T_1\text{= 350}\text{K}$

Final Temperature, $T_2\text{= 270}\text{K}$

Formula used:

Change in volume

  $\text{Δ}V=\beta V\text{Δ}T$

Where, $\text{Δ}V$ is change in volume, $\beta$ is coefficient of volume expansion, $V$ is the volume, $\Delta T$ is the change in temperature.

Calculation:

Expand the formula of change in volume, $\text{Δ}V=\beta V\text{Δ}T$ .

$\left(V_2-V_1\right)=\beta V\Delta T$ or $V_2=V_1+V_1\beta \Delta T$

  $V_2=V_1\left(1+\beta \left(T_2-T_1\right)\right)$

Substitute all the given values in formula,

  $V_2=V_1\left(1+\beta \left(T_2-T_1\right)\right)$

  $V_2\text{=}\text{0}\text{.350}{\text{m}}^3\left(1+\text{75}\times {10}^{-6}\text{per}°\text{C}\left(\text{270}\text{K}-\text{350}\text{K}\right)\right)$

  $V_2=\text{0}\text{.348}{\text{m}}^3$

Conclusion:

Volume of aluminum cube at $270.0\text{K}$ cooled from $\text{350}\text{K}$ will be $\text{0}\text{.348}{\text{m}}^3$ .

(b)

To determine

Length of an aluminum cube at $270.0\text{K}$ , if it have a volume of $\text{0}\text{.350}{\text{m}}^3$ at $\text{350}\text{K}$ .

Answer to Problem 89A

Length of an aluminum cube will be $\text{0}\text{.703}\text{m}$ .

Explanation of Solution

Given:

Volume at $\text{270}\text{K}\text{= 0}\text{.348}{\text{m}}^3$

Formula used:

Formula to calculate the length, $L={\left(V_2\right)}^{\frac{1}{3}}$ .

Here, $\text{V}$ is volume.

Calculation:

Substitute all the given values in formula,

  $L={\left(V_2\right)}^{\frac{1}{3}}$

  $\begin{array}{l}L={\left(\text{0}\text{.348}{\text{m}}^3\right)}^{\frac{1}{3}}\\ L=\text{ 0}\text{.703}\text{m}\end{array}$

Conclusion:

Length of the aluminum cube will be $\text{0}\text{.703}\text{m}$ .