Chapter 14, Problem 76QAP

To determine

(a)

The change in length when temperature of the solid aluminum cylinder increases from

  $5°\text{C}$ to $30°\text{C}$.

Answer to Problem 76QAP

The change in length when temperature of the solid aluminum cylinder increases from $5°\text{C}$ to $30°\text{C}$ is $\Delta L=0.033\text{m}$.

Explanation of Solution

Given:

Coefficient of liner expansion, $\alpha =22.2\times {10}^{-6}{\text{K}}^{\text{-1}}$

Original length, $L_0=5\text{m}$

Change in temperature, $\Delta T={\left(30-5\right)}^0\text{C}=25+273.15=298.15\text{K}$

Formula used:

The change in length is given by,

  $\Delta L=L_0\alpha \Delta T$

Where,

  $\Delta L$ =change in length

  $L_0$ = Original length

  $\Delta T$ =Change in temperature

  $\alpha$ = Coefficient of liner expansion

Calculation:

  $\begin{array}{l}\Delta L=L_0\alpha \Delta T\\ \Delta L=5\times 22.2\times {10}^{-6}\times 298.15\\ \Delta L=0.033m\end{array}$

Conclusion:

The change in length is

  $\Delta L=0.033\text{m}$

To determine

(b)

The percentage change in density of aluminum cylinder.

Answer to Problem 76QAP

The percentage change in density of aluminum cylinder is $2.48\%$

Explanation of Solution

Given:

Radius, $r=2\text{cm}=0.02\text{m}$

Original length, $L_0=5\text{m}$

Change in temperature, $\Delta T=298.15\text{K}$

Coefficient of volume expansion, $\beta =69\times {10}^{-6}{\text{K}}^{-1}$

Formula used:

The change in length is given by,

  $\Delta V=V_0\beta \Delta T$

Where,

  $\Delta V$ =change in volume

  $V_0$ = Original volume

  $\Delta T$ =Change in temperature

  $\beta$ = Coefficient of volume expansion

Calculation:

Original volume of the cylinder is,

  $\begin{array}{l}{V}_0=\pi r^2{L}_0\\ V_0=3.14\times 0.02\times 0.02\times 5\\ V_0=0.00628m^3\end{array}$

The change in volume is given by,

  $\begin{array}{l}\Delta V=V_0\beta \Delta T\\ \Delta V=0.00628\times 69\times {10}^{-6}\times 298.15\\ \Delta V=1.29\times {10}^{-4}{m}^3\end{array}$

Original density of the cylinder is,

  ${\rho }_0=\frac{m}{0.00628m^3}$

New density of the cylinder is,

  $\begin{array}{l}\rho =\frac{m}{0.00628+1.29\times {10}^{-4}}\\ \rho =\frac{m}{0.006409m^3}\end{array}$

Percentage change in density is,

  $\begin{array}{l}\frac{{\rho }_0-\rho }{{\rho }_0}\times 100\\ =\frac{\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$0.00628$}\right.-\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$0.006409$}\right.}{\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$0.00628$}\right.}\times 100\\ =2.48\%\end{array}$

Conclusion:

The percent change in density is $2.48\%$.

To determine

(c)

The percentage change in volume of cylinder.

Answer to Problem 76QAP

The percentage change in volume of cylinder is $2.05\%$

Explanation of Solution

Given:

Radius, $r=2\text{cm}=0.02\text{m}$

Original length, $L_0=5\text{m}$

Change in temperature, $\Delta T=298.15\text{K}$

Coefficient of volume expansion, $\beta =69\times {10}^{-6}{\text{K}}^{-1}$

Formula used:

The change in length is given by,

  $\Delta V=V_0\beta \Delta T$

Where,

  $\Delta V$ =change in volume

  $V_0$ = Original volume

  $\Delta T$ =Change in temperature

  $\beta$ = Coefficient of volume expansion

Calculation:

Original volume of the cylinder is,

  $\begin{array}{l}{V}_0=\pi r^2{L}_0\\ V_0=3.14\times 0.02\times 0.02\times 5\\ V_0=0.00628{\text{m}}^{\text{3}}\end{array}$

The change in volume is given by,

  $\begin{array}{l}\Delta V=V_0\beta \Delta T\\ \Delta V=0.00628\times 69\times {10}^{-6}\times 298.15\\ \Delta V=1.29\times {10}^{-4}{\text{m}}^{\text{3}}\end{array}$

Percentage change in volume of the cylinder is,

  $\begin{array}{l}\frac{\Delta V}{V_0}\times 100\\ =\frac{1.29\times {10}^{-4}}{0.00628}\times 100\\ =2.05\%\end{array}$

Conclusion:

The change in volume is $2.05\%$.