Chapter 19, Problem 19.77CP

(a)

To determine

The expression for the final length of the rod.

Answer to Problem 19.77CP

The expression for the final length of the rod is $L_{\text{i}}{e}^{\alpha \Delta T}$.

Explanation of Solution

Given Info: The length of the rod is $1.00\text{m}$, the temperature change of the rod is

$100.0\text{°C}$.

The relation of the coefficient of the linear expansion with the changing length with temperature is,

$\frac{dL}{dT}=\alpha L$

Here,

$L$ is the length of the rod.

To get the expression for the final length at high temperature, integrate the above expression.

$\begin{array}{c}{\displaystyle \int \frac{dL}{dT}}={\displaystyle \int \alpha L}\\ {\displaystyle \underset{L_{\text{i}}}{\overset{L_{\text{f}}}{\int }}\frac{dL}{L}}={\displaystyle \underset{T_{\text{1}}}{\overset{T_{\text{2}}}{\int }}\alpha dT}\\ \ln \left(\frac{L_{\text{f}}}{L_{\text{i}}}\right)=\alpha \left(T_{\text{2}}-T_{\text{1}}\right)\\ L_{\text{f}}=L_{\text{i}}{e}^{\alpha \Delta T}\end{array}$

Here,

$L_{\text{i}}$ is the initial length of the rod.

$L_{\text{f}}$ is the final length of the rod.

Conclusion:

Therefore, the expression for the final length of the rod is $L_{\text{i}}{e}^{\alpha \Delta T}$.

(b)

To determine

The error caused by the approximation.

Answer to Problem 19.77CP

The error caused by the approximation is $2.00\times {10}^{-4}\%$.

Explanation of Solution

Given Info: The length of the rod is $1.00\text{m}$, the temperature change of the rod is

$100.0\text{°C}$, the coefficient of the linear expansion is $2.00\times {10}^{-5}{\left(\text{°C}\right)}^{-1}$.

Formula to calculate the new length of the rod at higher temperature is,

$L_{\text{f}}=L_{\text{i}}\left(1+\alpha \Delta T\right)$

Substitute $1\text{m}$ for $L_{\text{i}}$, $100\text{°C}$ for $\Delta T$ and $2.00\times {10}^{-5}{\left(\text{°C}\right)}^{-1}$ for $\alpha$ in the above expression.

$\begin{array}{c}{L}_{\text{f}}=\left(1\text{m}\right)\left(1+\left(2.00\times {10}^{-5}{\left(\text{°C}\right)}^{-1}\right)\left(100\text{°C}\right)\right)\\ =1.002\text{m}\end{array}$

Thus, the length of the rod at higher temperature is $1.002\text{m}$.

The formula used in part (a) to calculate the new length of the rod at higher temperature is,

$L{\text{'}}_{\text{f}}^{}=L_{\text{i}}\exp \left(\alpha \Delta T\right)$

Substitute $1\text{m}$ for $L_{\text{i}}$, $100\text{°C}$ for $\Delta T$ and $2.00\times {10}^{-5}{\left(\text{°C}\right)}^{-1}$ for $\alpha$ in the above expression.

$\begin{array}{c}{L}_{\text{f}}^{\text{'}}=1\cdot \exp \left(\left(2.00\times {10}^{-5}{\left(\text{°C}\right)}^{-1}\right)\left(100\text{°C}\right)\right)\\ =1.002002\text{m}\end{array}$

Formula to calculate the percentage difference of the new length is,

$E=\left(\frac{L_{\text{f}}^{\text{'}}-L_{\text{i}}}{L_{\text{f}}^{\text{'}}}\right)\times 100\%$

Substitute $1.002002\text{m}$ for $L_{\text{f}}^{\text{'}}$, $1.002\text{m}$ for $L_{\text{f}}$, $100\text{°C}$ for $\Delta T$ and $2.00\times {10}^{-5}{\left(\text{°C}\right)}^{-1}$ for $\alpha$ in the above expression.

$\begin{array}{c}E=\left(\frac{\left(1.002002\text{m}\right)-\left(1.002\text{m}\right)}{1.002002\text{m}}\right)\times 100\%\\ =2.00\times {10}^{-4}\%\end{array}$

Conclusion:

Therefore, error caused by the approximation is $2.00\times {10}^{-4}\%$.

(c)

To determine

The error caused by the approximation.

Answer to Problem 19.77CP

The error caused by the approximation is $59.4\%$.

Explanation of Solution

Given Info: The length of the rod is $1.00\text{m}$, the temperature change of the rod is

$100.0\text{°C}$, the coefficient of the linear expansion is $0.0200{\left(\text{°C}\right)}^{-1}$.

Formula to calculate the new length of the rod at higher temperature is,

$L_{\text{f}}=L_{\text{i}}\left(1+\alpha \Delta T\right)$

Substitute $1\text{m}$ for $L_{\text{i}}$, $100\text{°C}$ for $\Delta T$ and $0.0200{\left(\text{°C}\right)}^{-1}$ for $\alpha$ in the above expression.

$\begin{array}{c}{L}_{\text{f}}=\left(1\text{m}\right)\left(1+\left(0.0200{\left(\text{°C}\right)}^{-1}\right)\left(100\text{°C}\right)\right)\\ =3\text{m}\end{array}$

Thus, the new length of the rod at higher temperature is $3\text{m}$.

The formula used in part (a) to calculate the new length of the rod at higher temperature is,

$L{\text{'}}_{\text{f}}^{}=L_{\text{i}}\exp \left(\alpha \Delta T\right)$

Substitute $1\text{m}$ for $L_{\text{i}}$, $100\text{°C}$ for $\Delta T$ and $0.0200{\left(\text{°C}\right)}^{-1}$ for $\alpha$ in the above expression.

$\begin{array}{c}{L}_{\text{f}}^{\text{'}}=1\cdot \exp \left(\left(0.0200{\left(\text{°C}\right)}^{-1}\right)\left(100\text{°C}\right)\right)\\ =7.389\text{m}\end{array}$

Formula to calculate the percentage difference of the new length is,

$E=\left(\frac{L_{\text{f}}^{\text{'}}-L_{\text{i}}}{L_{\text{f}}^{\text{'}}}\right)\times 100\%$

Substitute $7.389\text{m}$ for $L_{\text{f}}^{\text{'}}$, $3\text{m}$ for $L_{\text{f}}$, $100\text{°C}$ for $\Delta T$ and $0.0200{\left(\text{°C}\right)}^{-1}$ for $\alpha$ in the above expression.

$\begin{array}{c}E=\left(\frac{\left(7.389\text{m}\right)-\left(3\text{m}\right)}{7.389\text{m}}\right)\times 100\%\\ =59.4\%\end{array}$

Conclusion:

Therefore, the error caused by the approximation is $59.4\%$.

(d)

To determine

The receding level of the turpentine.

Answer to Problem 19.77CP

The receding level of the turpentine is $0.969\text{cm}$.

Explanation of Solution

Given info: The length of the rod is $1.00\text{m}$, the initial temperature is $20.0\text{°C}$, the initial temperature is $80.0\text{°C}$, the coefficient of the linear expansion is $2.00\times {10}^{-5}{\left(\text{°C}\right)}^{-1}$, the area of their cylinder is $0.02\text{m}$, the volume of the turpentine is $2\text{L}$ at temperature $20\text{°C}$.

The formula used in part (a) to calculate the new length of the rod at higher temperature is,

$L{\text{'}}_{\text{f}}^{}=L_{\text{i}}\exp \left(\alpha \Delta T\right)$

The volume varies correspondingly as the above expression.

$V_2^{\text{'}}=V_{\text{1}}\exp \left(\alpha \Delta T\right)$ (1)

Here,

$V_1$ is the initial volume of the turpentine.

$V_2^{\text{'}}$ is the final volume of the turpentine.

The change in the temperature is,

$\Delta T=T_2-T_1$

Substitute $20\text{°C}$ for $T_1$ and $80.0\text{°C}$ for $T_2$ in the above expression.

$\begin{array}{c}\Delta T=80.0\text{°C}-20.0\text{°C}\\ =\text{6}0.0\text{°C}\end{array}$

Thus, the change in temperature is $60.0°\text{C}$.

Substitute $2\text{L}$ for $V_1$, $60\text{°C}$ for $\Delta T$ and $72.00\times {10}^{-6}{\left(\text{°C}\right)}^{-1}$ for $\alpha$ in the equation (1).

$\begin{array}{c}{V}_{2\text{C}}^{\text{'}}=\left(2\text{L}\right)\exp \left(\left(72.00\times {10}^{-6}{\left(\text{°C}\right)}^{-1}\right)\left(60\text{°C}\right)\right)\\ =2.00866\text{L}\end{array}$

Thus, the final volume of the turpentine is $2.00866\text{L}$.

Substitute $2\text{L}$ for $V_1$, $60\text{°C}$ for $\Delta T$ and $9\times {10}^{-4}{\left(\text{°C}\right)}^{-1}$ for $\alpha$ in the equation (1) to find the volume of the turpentine.

$\begin{array}{c}{V}_{\text{2T}}^{\text{'}}=\left(2\text{L}\right)\exp \left(\left(9\times {10}^{-4}{\left(\text{°C}\right)}^{-1}\right)\left(60\text{°C}\right)\right)\\ =2.110\text{L}\end{array}$

Write the expression to calculate the volume of the turpentine that overflows.

$V_{\text{f}}=V_{\text{2T}}^{\text{'}}-V_{\text{2C}}^{\text{'}}$

Substitute $2.110\text{L}$ for $V_{\text{2T}}^{\text{'}}$ and $2.00866\text{L}$ for $V_{\text{2C}}^{\text{'}}$.

$\begin{array}{c}{V}_{\text{f}}=\left(2.110\text{L}\right)-\left(2.00866\text{L}\right)\\ =0.102\text{L}\times \frac{{10}^{-3}\text{mL}}{1\text{L}}\\ =102\text{mL}\end{array}$

To calculate the volume of the turpentine remaining is,

$\begin{array}{c}{V}_{\text{R}}=2.110\text{L}-0.102\text{L}\\ =\text{2}\text{.01}\text{L}\end{array}$

Thus, the volume of the turpentine remaining is $\text{2}\text{.01}\text{L}$.

Receding level of turpentine is,

$\begin{array}{l}\frac{\left(2.001-1.904\right)\times {10}^{-3}{\text{m}}^{\text{3}}}{{10}^{-2}{\text{m}}^{\text{2}}}\\ =0.00969\text{m}\times \frac{100\text{cm}}{1\text{m}}\\ =0.969\text{cm}\end{array}$

Conclusion:

Therefore, receding level of the turpentine is $0.969\text{cm}$.