Chapter 20, Problem 81CP

(a)

To determine

The speed of rise of the piston when the water begins boiling.

Answer to Problem 81CP

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

Explanation of Solution

A piston-cylinder apparatus of cylinder radius $r=7.50\text{ cm}$ and piston mass of $m=3.00\text{ kg}$ sits on the surface of water at the bottom of the cylinder of $2.00\text{ kg}$ at temperature just below $100.0°\text{C}$. Now, an electric heater in the cylinder base is turned on and it transfers energy to the water at the rate of $100\text{ W}$.

The speed of rise of the piston is the increase in the height $h$ due to the boiling of water and the steam produced in the process. The volume of the steam or gas is the area $A$ of the cylinder times the height $h$

    $v = \frac{dh}{dt} = \frac{d}{dt}{\left(\frac{V}{A}\right)}_{{}_{{}_{{}_{}}}}$

Using ideal gas law, $PV=nRT$ in the above equation and solving

  $v = \frac{1}{A}\frac{d}{dt}\left(\frac{nRT}{P}\right)$

  $v= \frac{RT}{PA}\frac{dn}{dt}$                                                                       (I)

Here, $PA$ is the force applied by the gas on the piston, $v$ is the volume of the gas, $n$ is the number of moles, $T$ is the temperature and $R$ is the ideal gas constant.

Write the formula for the force when the piston is in equilibrium

  $F=P_{0}A+m_{p}g$                                                              (II)

Here, $P_0$ is the atmospheric pressure, $A$ is the area, $m_p$ is the mass of the piston and $g$ is the acceleration due to gravity.

Write the formula to find the number of moles

      $n=\frac{m_g}{M_w}$                                                              (III)

Here, $m_g$ is the mass of gas and $M_w$ is the molecular mass.

Write the formula to find energy required to change the phase

    $Q=L_v\Delta m$                                                                              (IV)

Here, $Q$ is the energy required to change the phase, $L_v$ is the latent heat of vaporization and $\Delta m$ is the change in mass.

Substitute equation (II) and (III) in (I) and solving

  $\begin{array}{c}v = \left(\frac{RT}{m_{p}g\text{} + P_{0}A}\right)\frac{d}{dt}\left(\frac{m_g}{M_w}\right)\\ = \left[\frac{RT}{\left(m_{p}g\text{} + P_{0}A\right)M_w}\right]\frac{d{m}_g}{dt}\end{array}$

Substitute equation (IV) in the above equation

$\begin{array}{c}v = \frac{RT}{\left(m_{p}g\text{} + P_{0}A\right)M_w}\frac{d}{dt}\left(\frac{Q}{L_v}\right)\\ = \frac{RT}{\left(m_{p}g\text{} + P_{0}A\right)M_w{L}_v}\frac{dQ}{dt}\end{array}$

Here, $\frac{dQ}{dt}$ is power. Therefore,

$v = \frac{RT\left(Power\right)}{\left(m_{p}g\text{} + P_{0}A\right)M_w{L}_v}$

Substitute $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $373\text{ K}$ for $T$, $100$ for $P$, $3.00\text{ kg}$ for $m_p$, $9.80{\text{m}/\text{s}}^2$ for $g$, $1.013\times {10}^5\text{ Pa}$ for $P_0$, $7.50\text{ cm}$ for $r$, $0.0180$ for $M_w$ and $2.26\times {10}^6\text{J}/\text{kg}$ for $L_v$ in the above equation to find the value of $v$

$\begin{array}{c}v = \frac{\left(8.314\right)\left(373\right)\left(100\right)}{\left[\left(3.00\right)\left(9.80\right)+\left(1.013\times {10}^5\right)\left(\pi \right){\left(0.0750\right)}^{\text{2}}\right]\left(0.0180\right)\left(2.26\times {10}^6\right)}\\ = 4.19 \times {10}^{-3} \text{m/s }\\ =4.19\text{ mm/s}\end{array}$

Thus, the speed of rise of the piston when the water begins boiling is $4.19\text{ mm/s}$.

(b)

To determine

The speed of rise of the piston when the water is completely turned into steam.

Answer to Problem 81CP

The speed of rise of the piston when the water is completely turned into steam is $12.6 \text{mm/s}$.

Explanation of Solution

When the water in the cylinder is completely turned into steam, the number of moles will be fixed and the temperature changes as the heater continues to transfer energy to the steam, equation (I) becomes

    $\begin{array}{l}v = \left(\frac{nR}{PA}\right)\frac{dT}{dt} \\ v= \frac{nR}{\left(m_{p}g\text{} + P_{0}A\right)}\frac{dT}{dt} \end{array}$

Substitute equation (IV) in the above equation

$\begin{array}{c}v = \frac{nR}{\left(m_{p}g\text{} + P_{0}A\right)}\frac{d}{dt}\left(\frac{Q}{m_{g}c}\right) \\ = \frac{m_{g}R}{\left(m_{p}g\text{} + P_{0}A\right)m_{g}c{M}_w}{\frac{dQ}{dt}}_{{}_{{}_{{}_{{}_{{}_{{}_{{}_{{}_{}}}}}}}}}\\ = \frac{R\left(Power\right)}{\left(m_{p}g\text{} + P_{0}A\right)c{M}_w}\end{array}$

Substitute $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $100$ for $P$, $3.00\text{ kg}$ for $m_p$, $9.80{\text{m}/\text{s}}^2$ for $g$, $1.013\times {10}^5\text{ Pa}$ for $P_0$, $7.50\text{ cm}$ for $r$, $0.0180$ for $M_w$ and $2.26\times {10}^6\text{J}/\text{kg}$ for $L_v$ in the above equation to find the value of $v$

$\begin{array}{c}v = \frac{\left(8.314\right)\left(100\right)}{\left[\left(3.00\right)\left(9.80\right)+\left(1.013\times {10}^5\right)\left(\pi \right){\left(0.0750\right)}^2\right]\left(2010\right)\left(0.0180\right)} \\ = 0.0126 \text{m/s }\\ =12.6 \text{mm/s}\end{array}$

Thus, the speed of rise of the piston when the water is completely turned into steam is $12.6 \text{mm/s}$.