Chapter 13, Problem 21P

(a)

To determine

To calculate: The change in length of the mercury in the tube on the basis of information provided below.

Answer to Problem 21P

The change in the length of the mercury in the tube is $6.1\text{cm}$ .

Explanation of Solution

Given:

Original volume of the glass with Hg= $V_0{}^{\text{bulb}}$ = $0.255{\text{cm}}^{\text{3}}$

Change in volume = $\Delta V_{\text{glass}}=V_0{}^{\text{bulb}}{\beta }_{\text{glass}}\Delta T$

Change in Hg volume = $\Delta V_{\text{Hg}}=V_0{}^{\text{bulb}}{\beta }_{\text{Hg}}\Delta T$

Diameter of the Hg tube = $1.40\times {10}^{-2}\text{cm}$

Change in temperature = $\Delta T$ = $33°\text{C}-11.5°\text{C}$

Formula used:

Write the expression for the change in the volume.

  $\begin{array}{l}(\delta L)\pi r_0{}^2=\Delta V_{\text{Hg}}-\Delta V_{\text{glass}}\\ (\delta L)\pi r_0{}^2=V_0{}^{\text{bulb}}{\beta }_{\text{Hg}}\Delta T-V_0{}^{\text{bulb}}{\beta }_{\text{glass}}\Delta T\\ (\delta L)=\frac{V_0{}^{bulb}}{\pi r_0{}^2}\Delta T({\beta }_{\text{Hg}}-{\beta }_{\text{glass}})\end{array}$

Where $\delta L$ is the change in the length of the tube .

Calculation:

Substitute the values in the above expression.

  $\begin{array}{c}\delta L=\frac{V_0{}^{bulb}}{\pi r_0{}^2}\Delta T({\beta }_{\text{Hg}}-{\beta }_{\text{glass}})\\ =\frac{V_0{}^{\text{bulb}}}{\pi {(d_0/2)}^2}\Delta T({\beta }_{\text{Hg}}-{\beta }_{\text{glass}})\\ =\frac{4V_0{}^{\text{bulb}}}{\pi d_0{}^2}\Delta T({\beta }_{\text{Hg}}-{\beta }_{\text{glass}})\\ =\frac{4(0.255{\text{cm}}^3)}{\pi {(1.40\times {10}^{-2}\text{cm})}^2}(33°\text{C}-11.5°\text{C})[(180-9)\times {10}^{-6}/\text{C}°]\\ =6.1\text{cm}\end{array}$

Conclusion:

The expansion of the mercury in the tube is due to the heat. The change in the length of the mercury tube is $6.1\text{cm}$ .

(b)

To determine

To calculate: The formula for the change in the length of the mercury tube.

Answer to Problem 21P

The derived formula for the change in length due to expansion is $\frac{4V_0{}^{\text{bulb}}}{\pi d_0{}^2}\Delta T({\beta }_{Hg}-{\beta }_{\text{glass}})$ .

Explanation of Solution

Given:

Original volume of the glass with Hg= $V_0{}^{\text{bulb}}$

Change in volume = $\Delta V_{\text{glass}}=V_0{}^{\text{bulb}}{\beta }_{\text{glass}}\Delta T$

Change in Hg volume = $\Delta V_{Hg}=V_0{}^{\text{bulb}}{\beta }_{Hg}\Delta T$

Diameter of the Hg tube

Change in temperature = $\Delta T$

Calculation:

  $\begin{array}{c}{V}_0{}^{\text{Bulb}}\\ \Delta V_{\text{glass}}=V_0{}^{\text{bulb}}{\beta }_{\text{glass}}\Delta T\\ \Delta V_{Hg}=V_0{}^{\text{bulb}}{\beta }_{Hg}\Delta T\\ (\delta L)\pi r_0{}^2=\Delta V_{Hg}-\Delta V_{\text{glass}}\\ =V_0{}^{\text{bulb}}{\beta }_{Hg}\Delta T-V_0{}^{\text{bulb}}{\beta }_{\text{glass}}\Delta T\\ (\delta L)=\frac{V_0{}^{\text{bulb}}}{\pi r_0{}^2}\Delta T({\beta }_{Hg}-{\beta }_{\text{glass}})\\ =\frac{V_0{}^{\text{bulb}}}{\pi {(d_0/2)}^2}\Delta T({\beta }_{Hg}-{\beta }_{\text{glass}})\\ =\frac{4V_0{}^{\text{bulb}}}{\pi d_0{}^2}\Delta T({\beta }_{Hg}-{\beta }_{\text{glass}})\end{array}$

Conclusion:

The derived formula for the change in length due to expansion is $\frac{4V_0{}^{\text{bulb}}}{\pi d_0{}^2}\Delta T({\beta }_{Hg}-{\beta }_{\text{glass}})$ .