Chapter 20, Problem 20.81CP

Consider the piston— cylinder apparatus shown in Figure P20.81. The bottom of the cylinder contains 2.00 kg of water at just under 100.0ºc. The cylinder has a radius of r = 7.50 cm. The piston of mass m = 3.00 kg sits on the surface of the water. An electric heater in the cylinder base transfers energy into the water at a rate of 100 W. Assume the cylinder is much taller than shown in the figure, so we don’t need to be concerned about the piston reaching the top of the cylinder. (a) Once the water begins boiling, how fast is the piston rising? Model the steam as an ideal gas. (b) After the water has completely turned to steam and the heater continues to transfer energy to the steam at the same rate, how fast is the piston rising?

To determine

The speed of the rise of the piston.

Answer to Problem 20.81CP

The speed of the rise of the piston is $4.19\text{mm}/\text{s}$.

Explanation of Solution

Given info: The mass of the water is $3.00\text{kg}$, the temperature of the water is $100°\text{C}$, the radius of the cylinder is $7.50\text{cm}$, the mass of the piston is $3.00\text{kg}$ and the power is $100\text{W}$.

Write the expression for the volume of the piston.

$V=\pi r^{2}h$

Here,

$r$ is the radius of the piston.

$h$ is the height of the piston.

Write the expression for the density of the piston.

$\rho =\frac{m}{V}$

Here,

$m$ is the mass of the piston.

Substitute $\pi r^{2}h$ for $V$ in above equation.

$\begin{array}{c}\rho =\frac{m}{\pi r^{2}h}\\ m=\rho \pi r^{2}h\end{array}$

Write the expression for the rate of energy required during a phase change.

$\frac{Q}{\Delta t}=\frac{L\Delta m}{\Delta t}$

Here,

$L$ is the latent heat of the substance.

Substitute $\rho \pi r^{2}h$ for $m$ in above equation.

$\begin{array}{c}\frac{Q}{\Delta t}=\frac{L\Delta \left(\rho \pi r^{2}h\right)}{\Delta t}\\ =\frac{L\rho \pi r^2\left(\Delta h\right)}{\Delta t}\\ \frac{\Delta h}{\Delta t}=\left(\frac{1}{L\rho \pi r^2}\right)\frac{Q}{\Delta t}\end{array}$

Substitute $7.50\text{cm}$ for $r$, $2.26\times {10}^6\text{J}/\text{kg}$ for $L$, $100\text{W}$ for $\frac{Q}{\Delta t}$ and $0.598\text{kg}/{\text{m}}^3$ for $\rho$ in above equation.

$\begin{array}{c}\frac{\Delta h}{\Delta t}=\left(\frac{1}{2.26\times {10}^6\text{J}/\text{kg}\times 0.598\text{kg}/{\text{m}}^3\times \pi {\left(7.50\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2}\right)100\text{W}\\ \approx 4.19\times {10}^{-3}\text{m}/\text{s}\times \frac{1000\text{mm}}{1\text{m}}\\ =4.19\text{mm}/\text{s}\end{array}$

Conclusion:

Therefore, the rise of the piston is $4.19\text{mm}/\text{s}$.

To determine

The rise of the piston.

Answer to Problem 20.81CP

The rise of the piston is $12.6\text{mm}/\text{s}$.

Explanation of Solution

Given info: The mass of the water is $3.00\text{kg}$, the temperature of the water is $100°\text{C}$, the radius of the cylinder is $7.50\text{cm}$, the mass of the piston is $3.00\text{kg}$ and the power is $100\text{W}$.

Write the expression for the area of the piston.

$A=\pi r^2$

Write the expression for the force applied.

$F=mg$

Here,

$g$ is the acceleration due to gravity.

Write the expression for the pressure due to the weight of the piston.

$P_{\text{piston}}=\frac{F}{A}$

Substitute $mg$ for $F$ and $\pi r^2$ for $A$ in the above equation.

$P_{\text{piston}}=\frac{mg}{\pi r^2}$

Write the expression for the total pressure.

$P=P_{\text{piston}}+P_{\text{atm}}$

Here,

$P_{\text{atm}}$ is the atmospheric pressure.

Substitute $\frac{mg}{\pi r^2}$ for $P_{\text{piston}}$ in the above equation.

$P=\frac{mg}{\pi r^2}+P_{\text{atm}}$

Write the expression for the ideal gas equation.

$\begin{array}{c}PV=\left(\frac{m_{\text{w}}}{M}\right)RT\\ T=\frac{PVM}{m_{\text{w}}R}\end{array}$

Here,

$P$ is the pressure.

$T$ is the temperature.

$V$ is the volume.

$R$ is the ideal gas constant.

$M$ is the molar mass of water.

$m_{\text{w}}$ is the mass of the water.

Write the expression for the energy required to increase the temperature.

$Q=m_{\text{w}}c\Delta T$

Here,

$c$ is the specific heat capacity.

$\Delta T$ is the change in temperature.

Differentiate above equation with respect to time.

$\begin{array}{c}\frac{dQ}{dt}=\frac{d}{dt}\left(m_{\text{w}}c\Delta T\right)\\ \frac{dQ}{dt}=m_{\text{w}}c\frac{dT}{dt}\end{array}$

Substitute $\frac{PVM}{m_{\text{w}}R}$ for $T$ in above equation.

$\begin{array}{c}\frac{dQ}{dt}=m_{\text{w}}c\frac{d\left(\frac{PVM}{m_{\text{w}}R}\right)}{dt}\\ \frac{dQ}{dt}=\frac{cPM}{R}\frac{dV}{dt}\end{array}$

Substitute $\frac{mg}{\pi r^2}+P_{\text{atm}}$ for $P$ and $\pi r^{2}h$ for $V$ in above equation.

$\begin{array}{c}\frac{dQ}{dt}=\frac{c\left(\frac{mg}{\pi r^2}+P_{\text{atm}}\right)M}{R}\frac{d\left(\pi r^{2}h\right)}{dt}\\ =\left(\frac{mg}{\pi r^2}+P_{\text{atm}}\right)\frac{cM\pi r^2}{R}\frac{dh}{dt}\\ =\left(\frac{mg+P_{\text{atm}}\left(\pi r^2\right)}{\pi r^2}\right)\left(\frac{cM\pi r^2}{R}\right)\frac{dh}{dt}\\ \frac{dh}{dt}=\left(\frac{1}{mg+P_{\text{atm}}\left(\pi r^2\right)}\right)\left(\frac{R}{cM}\right)\left(\frac{dQ}{dt}\right)\end{array}$

Substitute $7.50\text{cm}$ for $r$, $8.314\text{g}\cdot \text{mol}\cdot \text{K}$ for $R$, $1.996\text{J}/\text{g}\cdot \text{K}$ for $c$, $18.0\text{g}/\text{mol}$ for $M$, $3.00\text{kg}$ for $m$, $9.81\text{m}/{\text{s}}^2$ for $g$, $1.01\times {10}^5\text{N}/{\text{m}}^2$ for $P_{\text{atm}}$ and $100\text{W}$ for $\frac{dQ}{dt}$ in above equation.

$\begin{array}{c}\frac{dh}{dt}=\left(\begin{array}{l}\left(\frac{1}{\left(3.00\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)+\left(1.01\times {10}^5\text{N}/{\text{m}}^2\right)\left(\pi {\left(7.50\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\right)}\right)\\ \left(\frac{8.314\text{g}\cdot \text{mol}\cdot \text{K}}{\left(1.996\text{J}/\text{g}\cdot \text{K}\right)\left(18.0\text{g}/\text{mol}\right)}\right)\left(100\text{W}\right)\end{array}\right)\\ \approx 1.26\times {10}^{-2}\text{m}/\text{s}\times \frac{1000\text{mm}}{1\text{m}}\\ =12.6\text{mm}/\text{s}\end{array}$

Conclusion:

Therefore, the rise of the piston is $12.6\text{mm}/\text{s}$.