A rigid, well-insulated tank contains a two-phase mixture consisting of .005 ft^3 saturated liquid water and OSK1.2 ft^3 of saturated water vapor, ini

The energy transfer by work can be calculated as:
$\Delta U=Q-W$
where:
$\Delta U$ represents the change in internal energy,
$Q$ represents the heat added to the system,
$W$ represents the work done by the system.
Since the paddlewheel stirs the mixture until only saturated vapor remains in the tank, we can assume that the process occurs at constant volume (isochoric process).
For an isochoric process, the work done is given by:
$W=P\Delta V$
where:
$P$ represents the pressure,
$\Delta V$ represents the change in volume.
In this case, the volume changes from the initial volume to zero, as only saturated vapor remains in the tank. Therefore, $\Delta V=-V_i$, where $V_i$ represents the initial volume of the mixture.
The work done can be expressed as:
$W=-P{V}_i$
Now, we need to determine the pressure $P$ in order to calculate the work done. The initial pressure of the system is given as 14.7 lb$\frac{f}{i{n}^2}$, which needs to be converted to the appropriate unit system.
Since we want the answer in British thermal units (Btu), we need to convert the pressure from lb$\frac{f}{i{n}^2}$ to Btu.
The conversion factor is:
1 Btu = 778 ft lb$\frac{f}{ft}$
To convert lb$\frac{f}{i{n}^2}$ to ft lb$\frac{f}{ft}$, we can use the conversion factor:
1 ft = 12 inches
Therefore, the conversion can be done as follows:
$P_{\text{converted}}=P_{\text{initial}}\times {\left(\frac{1}{12}\right)}^2\times 778$
Substituting the values, we get:
$P_{\text{converted}}=14.7\times {\left(\frac{1}{12}\right)}^2\times 778$
Finally, we can calculate the work done:
$W=-P_{\text{converted}}\times V_i$
Substituting the given initial volumes, we have:
$W=-P_{\text{converted}}\times (0.05+1.2)$
Simplifying this expression will give us the amount of energy transfer by work, in Btu.

Step 1: Let's start by finding the initial state and the final state of the mixture.
Given:
Initial volume of saturated liquid water ($V_{\text{liq}}$) = 0.05 ft${}^3$
Initial volume of saturated water vapor ($V_{\text{vap}}$) = 1.2 ft${}^3$
Initial pressure ($P_{\text{initial}}$) = 14.7 lb$\frac{f}{{\text{in}}^2}$
We know that the total volume of the mixture is the sum of the volumes of the liquid water and the water vapor:
$V_{\text{total, initial}}=V_{\text{liq}}+V_{\text{vap}}$
Substituting the given values:
$V_{\text{total, initial}}=0.05+1.2=1.25{\text{ ft}}^3$
Since the tank is rigid, the volume remains constant throughout the process. Therefore, the final volume ($V_{\text{total, final}}$) is also 1.25 ft${}^3$.
To find the final pressure ($P_{\text{final}}$), we use the fact that only saturated vapor remains in the tank. At saturated conditions, the pressure and temperature are related by the saturation pressure-temperature relationship.
Step 2: Now, we can determine the initial temperature ($T_{\text{initial}}$) corresponding to the initial pressure ($P_{\text{initial}}$) using the saturation temperature-pressure table or steam tables. Let's assume the initial temperature is $T_{\text{initial}}$.
After stirring, only saturated vapor remains in the tank, so the final temperature ($T_{\text{final}}$) is equal to the saturation temperature at the final pressure ($P_{\text{final}}$). Let's assume the final temperature is $T_{\text{final}}$.
Now, we need to calculate the energy transfer by work. For an ideal gas process, the work done can be calculated using the equation:
$W=P·\Delta V$
Since the volume remains constant, the change in volume ($\Delta V$) is zero, and therefore the work done is zero.
Hence, the amount of energy transfer by work, in Btu, is $0$ Btu.
Therefore, $W=0$ Btu.