Chapter 19, Problem 77CP

(a)

To determine

The expression for the final length of the rod.

Answer to Problem 77CP

The expression for the final length of the rod is $\underset{\_}{L_f=L_i{e}^{\alpha \left(\Delta T\right)}}$.

Explanation of Solution

Write the equation for the elementary change in length.

  $dL=\alpha LdT$                                                                                     (I)

Here, $dL$ is the infinitesimal change in length, $\alpha$ is the expansion coefficient, $L$ is the initial length and $dT$ is the change in temperature.

Conclusion:

Rearrange the equation (I) and integrating both sides.

  $\begin{array}{l}{\displaystyle {\int }_{L_i}^{L_f}\frac{dL}{L}}={\displaystyle {\int }_{T_i}^{T_f}\alpha dT}\\ \ln \left(\frac{L_f}{L_i}\right)=\alpha \left(T_f-T_i\right)\\ \ln \left(\frac{L_f}{L_i}\right)=\alpha \left(\Delta T\right)\end{array}$

Rewrite the expression for $L_f$.

  $L_f=L_i{e}^{\alpha \left(\Delta T\right)}$

Here, $L_i$ is the initial length, $L_f$ is the final length, $T_f$ is the final temperature, $T_i$ is the initial temperature and $\left(\Delta T\right)$ is the change in temperature.

Thus, the expression for the final length of the rod is $\underset{\_}{L_f=L_i{e}^{\alpha \left(\Delta T\right)}}$.

(b)

To determine

The error because of approximation.

Answer to Problem 77CP

The error because of approximation is $\underset{\_}{2.00\times {10}^{-4}\%}$.

Explanation of Solution

Write the expression for the error due to approximation.

$e=\left[\frac{L_f-{L^{\prime }}_f}{L_f}\right]\times 100\%$ (II)

Write the equation for final length.

  $L_f=L_i{e}^{\alpha \left(\Delta T\right)}$                                                                                         (III)

Write the equation for final length with approximation.

  ${L^{\prime }}_f=L_i+\alpha L_i\left(\Delta T\right)$                                                                                         (IV)

Here, ${L^{\prime }}_f$ is the final length with approximation.

Rewrite the expression for the error due to approximation.

$e=\left[\frac{\left(L_i{e}^{\alpha \left(\Delta T\right)}\right)-\left[L_i+\alpha L_i\left(\Delta T\right)\right]}{\left(L_i{e}^{\alpha \left(\Delta T\right)}\right)}\right]\times 100\%$ (V)

Conclusion:

Substitute $1.00\text{m}$ for $L_i$, $2.00\times {10}^{-5}\text{/°C}$ for $\alpha$, $100\text{°C}$ for $\left(\Delta T\right)$ in Equation (V) to find $\left(\Delta T\right)$.

  $\begin{array}{c}e=\left[\frac{\left(\left(1.00\text{m}\right)e^{\left(2.00\times {10}^{-6}\text{/°C}\right)\left(100\text{°C}\right)}\right)-\left[\left(1.00\text{m}\right)+\left(2.00\times {10}^{-5}\text{/°C}\right)\left(1.00\text{m}\right)\left(100\text{°C}\right)\right]}{\left(\left(1.00\text{m}\right)e^{\left(2.00\times {10}^{-6}\text{/°C}\right)\left(100\text{°C}\right)}\right)}\right]\times 100\%\\ =\left(2.00\times {10}^{-6}\right)\times 100\%\\ =2.00\times {10}^{-4}\%\end{array}$

Thus, the error because of approximation is $\underset{\_}{2.00\times {10}^{-4}\%}$.

(c)

To determine

The error because of approximation.

Answer to Problem 77CP

The error because of approximation is $\underset{\_}{59.4\%}$.

Explanation of Solution

Write the expression for the error due to approximation.

$e=\left[\frac{L_f-{L^{\prime }}_f}{L_f}\right]\times 100\%$ (II)

Write the equation for final length.

  $L_f=L_i{e}^{\alpha \left(\Delta T\right)}$                                                                                          (III)

Write the equation for final length with approximation.

  ${L^{\prime }}_f=L_i+\alpha L_i\left(\Delta T\right)$                                                                                          (IV)

Here, ${L^{\prime }}_f$ is the final length with approximation.

Rewrite the expression for the error due to approximation.

$e=\left[\frac{\left(L_i{e}^{\alpha \left(\Delta T\right)}\right)-\left[L_i+\alpha L_i\left(\Delta T\right)\right]}{\left(L_i{e}^{\alpha \left(\Delta T\right)}\right)}\right]\times 100\%$ (V)

Conclusion:

Substitute $1.00\text{m}$ for $L_i$, $0.0200\text{/°C}$ for $\alpha$, $100\text{°C}$ for $\left(\Delta T\right)$ in Equation (V) to find $\left(\Delta T\right)$.

  $\begin{array}{c}e=\left[\frac{\left(\left(1.00\text{m}\right)e^{\left(2.00\times {10}^{-6}\text{/°C}\right)\left(100\text{°C}\right)}\right)-\left[\left(1.00\text{m}\right)+\left(0.0200\text{/°C}\right)\left(1.00\text{m}\right)\left(100\text{°C}\right)\right]}{\left(\left(1.00\text{m}\right)e^{\left(2.00\times {10}^{-6}\text{/°C}\right)\left(100\text{°C}\right)}\right)}\right]\times 100\%\\ =\left(0.594\right)\times 100\%\\ =59.4\%\end{array}$

Thus, the error because of approximation is $\underset{\_}{59.4\%}$.

(d)

To determine

Solve problem 21 with more accurate result.

Answer to Problem 77CP

Part (a)

The amount of turpentine overflows is $\underset{\_}{102\text{mL}}$.

Part (b)

The remaining amount of turpentine in the cylinder is $\underset{\_}{2.01\text{L}}$.

Part (c)

The distance below the cylinder’s rim is $\underset{\_}{0.969\text{cm}}$.

Explanation of Solution

Write the expression for the final volume for turpentine.

$V_{\text{t,final}}=V_t{e}^{{\beta }_t\left(T_2-T_1\right)}$ (VI)

Here, $V_{\text{t,final}}$ is the final volume of the turpentine, $V_t$ is the initial volume of turpentine, ${\beta }_t$ is the volume expansion coefficient of turpentine, $T_1$ is the initial temperature and $T_2$ is the final temperature.

Write the equation for final volume for the aluminium.

  $V_{\text{Al,final}}=V_{\text{Al}}{e}^{3{\alpha }_{\text{Al}}\left(T_2-T_1\right)}$                                                                                       (VII)

Here, $V_{\text{Al,final}}$ is the final volume of the aluminium, $V_{\text{Al}}$ is the initial volume of aluminium, ${\alpha }_{\text{Al}}$ is the linear expansion coefficient of aluminium.

Write the equation for amount of turpentine overflows.

  $\left(\Delta V\right)=V_{\text{t,final}}-V_{\text{Al,final}}$                                                                              (VIII)

Here, $\left(\Delta V\right)$ is the amount of overflow volume of turpentine.

Rewrite the expression for the amount of turpentine overflows.

$\left(\Delta V\right)=\left[V_t{e}^{{\beta }_t\left(T_2-T_1\right)}\right]-\left[V_{\text{Al}}{e}^{3{\alpha }_{\text{Al}}\left(T_2-T_1\right)}\right]$ (IX)

Write the expression for the remaining amount of turpentine in the cylinder.

$V_{\text{remaining}}=V_{\text{Al}}{e}^{3{\alpha }_{\text{Al}}\left(\Delta T\right)}$ (X)

Here, $V_{\text{remaining}}$ is the remaining amount of turpentine in the cylinder.

Write the expression for the height below the cylinder’s rim.

$h=h_i\left[\frac{\left(V_{\text{Al}}\right)-\left[V_{\text{Al}}{e}^{\left(3{\alpha }_{\text{Al}}-{\beta }_t\right)\left(T_2-T_1\right)}\right]}{V_{\text{Al}}}\right]$ (XI)

Here, $h$ is the height below the cylinder’s rim and $h_i$ is the initial height aluminium.

Conclusion:

Part (a)

Substitute $2.000\text{L}$ for $V_t$, $2.000\text{L}$ for $V_{\text{Al}}$, $9.00\times {10}^{-4}\text{/°C}$ for ${\beta }_t$, $24.0\times {10}^{-6}\text{/°C}$ for ${\alpha }_{\text{Al}}$ , $20\text{°C}$ for $T_1$ and $80\text{°C}$ for $T_2$ in Equation (IX) to find $\left(\Delta V\right)$.

  $\begin{array}{c}\left(\Delta V\right)=\left[\left(2.000\text{L}\right)e^{\left(9.00\times {10}^{-4}\text{/°C}\right)\left(80\text{°C}-20\text{°C}\right)}\right]-\left[\left(2.000\text{L}\right)e^{3\left(24.0\times {10}^{-4}\text{/°C}\right)\left(80\text{°C}-20\text{°C}\right)}\right]\\ =\left(2.000\text{L}\right)\left[e^{\left(9.00\times {10}^{-4}\text{/°C}\right)\left(80\text{°C}-20\text{°C}\right)}\right]-\left[e^{3\left(24.0\times {10}^{-4}\text{/°C}\right)\left(80\text{°C}-20\text{°C}\right)}\right]\\ =102\times {10}^{-3}\text{L}\\ \text{=102 mL}\end{array}$

Thus, the amount of turpentine overflows is $\underset{\_}{102\text{mL}}$.

Part (b)

Substitute  $2.000\text{L}$ for $V_{\text{Al}}$, $24.0\times {10}^{-6}\text{/°C}$ for ${\alpha }_{\text{Al}}$ , $20\text{°C}$ for $T_1$ and $80\text{°C}$ for $T_2$ in Equation (X) to find $V_{\text{remaining}}$.

  $\begin{array}{c}{V}_{\text{remaining}}=\left[\left(2.000\text{L}\right)e^{3\left(24.0\times {10}^{-4}\text{/°C}\right)\left(80\text{°C}-20\text{°C}\right)}\right]\\ =2.01\text{L}\end{array}$

Thus, the remaining amount of turpentine in the cylinder is $\underset{\_}{2.01\text{L}}$.

Part (c)

Substitute $20.0\text{cm}$ for $h_i$, $2.000\text{L}$ for $V_{\text{Al}}$, $9.00\times {10}^{-4}\text{/°C}$ for ${\beta }_t$, $24.0\times {10}^{-6}\text{/°C}$ for ${\alpha }_{\text{Al}}$ , $20\text{°C}$ for $T_1$ and $80\text{°C}$ for $T_2$ in Equation (XI) to find $h$.

$\begin{array}{c}h=\left(20.0\text{cm}\right)\left[\frac{\left(2.000\text{L}\right)-\left[\left(2.000\text{L}\right)e^{\left(3\left(24.0\times {10}^{-6}\text{/°C}\right)-\left(9.00\times {10}^{-4}\text{/°C}\right)\right)\left(80.0\text{°C}-20.0\text{°C}\right)}\right]}{\left(2.000\text{L}\right)}\right]\\ =\left(20.0\text{cm}\right)\left(0.0485\right)\\ =0.969\text{cm}\end{array}$

Thus, the distance below the cylinder’s rim is $\underset{\_}{0.969\text{cm}}$.