Chapter 19, Problem 17P

(a)

To determine

The temperature at which the ring will slip over the rod.

Answer to Problem 17P

The temperature at which the ring will slip over the rod is $\underset{\_}{437\text{°C}}$.

Explanation of Solution

Write the equation of new length.

  $L=L_i\left(1+\alpha \left(T_2-T_1\right)\right)$                                                                           (I)

Here, $L$ is the new length, $L_i$ is the initial length, $\alpha$ is the expansion coefficient, $T_1$ is the initial temperature and $T_2$ is the final temperature.

Rewrite the equation (I) to express the temperature.

  $T_2=T_1+\frac{1}{\alpha }\left[\frac{L}{L_i}-1\right]$                                                                                               (II)

Conclusion:

Substitute, $20\text{°C}$ for $T_1$, $24.0\times {10}^{-6}\text{/°C}$ for $\alpha$, $5.000\text{cm}$ for $L_i$ and $5.050\text{cm}$ for $L$ in equation (II) to find $T_2$.

  $\begin{array}{c}{T}_2=\left(20\text{°C}\right)+\frac{1}{\left(24.0\times {10}^{-6}\text{/°C}\right)}\left[\frac{\left(5.050\text{cm}\right)}{\left(5.000\text{cm}\right)}-1\right]\\ =\left(20\text{°C}\right)+\left(417\text{°C}\right)\\ =437°\text{C}\end{array}$

Thus, the temperature at which the ring will slip over the rod is $\underset{\_}{437\text{°C}}$.

(b)

To determine

The temperature at which the ring barely slips over the rod.

Answer to Problem 17P

The temperature at which the ring barely slips over the rod is $\underset{\_}{2.1\times {10}^3\text{°C}}$.

Explanation of Solution

Write the equation of new length at equilibrium for the aluminium and Brass.

  $L_{Al}=L_{Brass}$                                                                                             (III)

Here, $L_{Al}$ is the new length of the aluminium, $L_{Brass}$ is the new length of the Brass.

Write the expression for the new length of the aluminium is,

  $L_{Al}=L_{i,Al}\left(1+{\alpha }_{Al}\left(\Delta T\right)\right)$                                                                                   (IV)

Here, $L_{i,Al}$ is the initial length of the aluminium, ${\alpha }_{Al}$ is the expansion coefficient of aluminium and $\left(\Delta T\right)$ is the temperature

Write the expression for the new length of the Brass is,

  $L_{Brass}=L_{i,Brass}\left(1+{\alpha }_{Brass}\left(\Delta T\right)\right)$                                                               (V)

Here, $L_{i,Brass}$ is the initial length of the Brass, ${\alpha }_{Brass}$ is the expansion coefficient of aluminium and $\left(\Delta T\right)$ is the temperature

Rewrite the expression from equation (III) by using (IV) and (V).

  $L_{i,Al}\left(1+{\alpha }_{Al}\left(\Delta T\right)\right)=L_{i,Brass}\left(1+{\alpha }_{Brass}\left(\Delta T\right)\right)$                                                  (VI)

Conclusion:

Substitute, $19.0\times {10}^{-6}\text{/°C}$ for ${\alpha }_{Brass}$, $24.0\times {10}^{-6}\text{/°C}$ for ${\alpha }_{Al}$, $5.000\text{cm}$ for $L_{i,Al}$ and $5.050\text{cm}$ for $L_{i,Brass}$ in equation (II) to find $\left(\Delta T\right)$.

  $\begin{array}{l}\left(5.000\text{cm}\right)\left(1+\left(24.0\times {10}^{-6}\text{/°C}\right)\left(\Delta T\right)\right)=\left(5.050\text{cm}\right)\left(1+\left(19.0\times {10}^{-6}\text{/°C}\right)\left(\Delta T\right)\right)\\ \left(\Delta T\right)=2.1\times {10}^3\text{°C}\end{array}$

Thus, the temperature at which the ring barely slips over the rod is $\underset{\_}{2.1\times {10}^3\text{°C}}$.

(c)

To determine

The latter process will work or not.

Answer to Problem 17P

The latter process will not work.

Explanation of Solution

The latter process will not happen. The melting point of Aluminium is $660\text{°C}$. And the melting point of the Brass is $900\text{°C}$.

Thus, at this calculated value of temperature, both aluminium and Brass will melt.

Therefore, the latter process will not work.