Chapter 16, Problem 14P

(a)

To determine

The temperature the ring must reach so that it will just slip over the rod if only the ring is warmed.

Answer to Problem 14P

The temperature when only the ring is warmed is $\text{437}°\text{C}$.

Explanation of Solution

Given information:Initial temperature is $20.0°\text{C}$, inner diameter of the aluminium ring is $5.0000\text{cm}$, diameter of the brass rod is $5.0500\text{cm}$.

Formula to calculate the change in temperature when only the ring is warmed.

$L_{\text{Brass}}=L_{\text{Aluminium}}\left(1+\alpha \Delta T\right)$ (I)

The expression for the change in the temperature is,

$\Delta T=T_2-T_1$

Substitute $T_2-T_1$ for $\Delta T$ in equation (I).

$\begin{array}{c}\frac{L_{\text{Brass}}}{L_{\text{Aluminium}}}=\left(1+\alpha \left(T_2-T_1\right)\right)\\ T_2=\frac{1}{\alpha }\left(\frac{L_{\text{Brass}}}{L_{\text{Aluminium}}}-1\right)+T_1\end{array}$

• $T_2$ is the temperature when only the ring is warmed.
• $T_1$ is the initial temperature.
• $L_{\text{Brass}}$ is the diameter of the brass rod.
• $\alpha$ is the average linear expansion coefficient.
• $L_{\text{Aluminium}}$ is the inner diameter of the aluminium ring.
• $\Delta T$ is the change in temperature.

The value of average linear expansion coefficient for aluminium is $24\times {10}^{-6}{\left(\text{°C}\right)}^{-1}$.

Substitute $24\times {10}^{-6}{\left(\text{°C}\right)}^{-1}$ for $\alpha$, $5.0500\text{cm}$ for $L_{\text{Brass}}$, $5.0000\text{cm}$ for $L_{\text{Aluminium}}$, $20.0°\text{C}$ for $T_1$ in equation (I) to find $T_2$,

$\begin{array}{c}{T}_2=\frac{1}{24\times {10}^{-6}{\left(\text{°C}\right)}^{-1}}\left(\frac{5.0500\text{cm}}{5.0000\text{cm}}-1\right)+20.0°\text{C}\\ \text{=436}\text{.67}°\text{C}\\ \approx \text{437}°\text{C}\end{array}$

Conclusion:

Therefore, the temperature when only the ring is warmed is $\text{437}°\text{C}$.

(b)

To determine

The temperature when both the ring and the rod are warmed together.

Answer to Problem 14P

Thetemperature when both the ring and the rod are warmed together is $\text{2}\text{.1}\times {\text{10}}^3°\text{C}$.

Explanation of Solution

Given information:Initial temperature is $20.0°\text{C}$, inner diameter of the aluminium ring is $5.0000\text{cm}$, diameter of the brass rod is $5.0500\text{cm}$.

Condition at which both reach at some temperature, when both the ring and the rod are warmed together is:

$\begin{array}{c}{L}_{\text{Aluminium}}=L_{\text{Brass}}\\ L_{\text{Aluminium}}\left(1+{\alpha }_{\text{Aluminium}}\Delta T\right)=L_{\text{Brass}}\left(1+{\alpha }_{\text{Brass}}\Delta T\right)\end{array}$

Formula to calculate the change in temperature when both the ring and the rod are warmed together is,

$\begin{array}{c}\left(1\times \frac{L_{\text{Aluminium}}}{L_{\text{Brass}}}+{\alpha }_{\text{Aluminium}}\times \frac{L_{\text{Aluminium}}}{L_{\text{Brass}}}\times \Delta T\right)=\left(1+{\alpha }_{\text{Brass}}\Delta T\right)\\ \frac{L_{\text{Aluminium}}}{L_{\text{Brass}}}-1=\left[{\alpha }_{\text{Brass}}-{\alpha }_{\text{Aluminium}}\left(\frac{L_{\text{Aluminium}}}{L_{\text{Brass}}}\right)\right]\Delta T\\ \Delta T=\frac{\left(\frac{L_{\text{Aluminium}}}{L_{\text{Brass}}}-1\right)}{\left[{\alpha }_{\text{Brass}}-{\alpha }_{\text{Aluminium}}\left(\frac{L_{\text{Aluminium}}}{L_{\text{Brass}}}\right)\right]}\end{array}$ (I)

• $\Delta T$ is the change in temperature.
• $L_{\text{Brass}}$ is the diameter of the brass rod.
• ${\alpha }_{\text{Brass}}$ is the average linear expansion coefficient for brass.
• ${\alpha }_{\text{Aluminium}}$ is the average linear expansion coefficient for Aluminium.
• $L_{\text{Aluminium}}$ is the inner diameter of the aluminium ring.

The value of average linear expansion coefficient for brass is $19\times {10}^{-6}{\left(\text{°C}\right)}^{-1}$.

Substitute $24\times {10}^{-6}{\left(\text{°C}\right)}^{-1}$ for $\alpha$, $5.0500\text{cm}$ for $L_{\text{Brass}}$, $5.0000\text{cm}$ for $L_{\text{Aluminium}}$, $20.0°\text{C}$ for $T_1$ in equation (I) to find $T_2$,

$\begin{array}{c}\Delta T=\frac{\left(\frac{5.0000\text{cm}}{5.0500\text{cm}}-1\right)}{\left[\left(19\times {10}^{-6}{\left(\text{°C}\right)}^{-1}\right)-\left(24\times {10}^{-6}{\left(\text{°C}\right)}^{-1}\right)\left(\frac{5.0000\text{cm}}{5.0500\text{cm}}\right)\right]}\\ \Delta T=\frac{\left(-9.900\times {10}^{-3}\right)}{\left[\left(19\times {10}^{-6}\right)-\left(2.376\times {10}^{-5}\right)\right]}\\ \Delta T=2079.83\text{K}\end{array}$

Formula to calculate the temperature when both the ring and the rod are warmed together is,

$\Delta T=T_2-T_1$ (I)

• $T_2$ is the temperature whenboth the ring and the rod are warmed together.
• $T_1$ is the initial temperature.
• $\Delta T$ is the change in temperature.

Substitute $2079.83\text{K}$ for $\Delta T$, $20.0°\text{C}$ for $T_1$ in equation (I) to find $T_2$,

$\begin{array}{c}{T}_2-20.0°\text{C}=2079.83°\text{C}\\ T_2=2079.83°\text{C}+20.0°\text{C}\\ =2099.83°\text{C}\\ \approx 2.1\times {10}^3°\text{C}\end{array}$

Conclusion:

Therefore, the temperature when both the ring and the rod are warmed together is $\text{2}\text{.1}\times {\text{10}}^3°\text{C}$.

(c)

To determine

To Explain: This latter process works or not.

Answer to Problem 14P

Therefore, this latter process will not work.

Explanation of Solution

No, this latter process will not work as the melting point of aluminium is $660°\text{C}$ that’s why it melts at $660°\text{C}$ and the brass which is alloy of zinc and copper melts at $900°\text{C}$. Here, the temperature when both the ring and the rod are warmed together is more than the melting point of aluminium and zinc. So, this latter process will not work.

Conclusion:

Therefore, this latter process will not work.