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# A rigid 10-L vessel initially contains a mixture of liquid water and vapor at $100 ^ { \circ } \mathrm { C }$ with 12.3 percent quality. The mixture is then heated until its temperature is $150 ^ { \circ } \mathrm { C }$. Calculate the heat transfer required for this process.

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To calculate the $\textbf{heat}$ required for the given process $Q$ we will use the energy balance equation. There is no kinetic or potential energy change so the heat input to the system is converted only to $\textbf{internal energy}$.

$$$Q=\Delta U= m(u_2 - u_1)$$$

We will start the calculation with the $\textbf{mass}$ $m$ of the water. Using the given initial quality $x_1=0.123$ and the initial temperature $T_1=100\text{\textdegree}\text{C}$ we can determine the $\textbf{initial specific volume}$ $v_1$ of the mixture. For the calculation we will also need the specific volume of liquid $v_f=0.001043\text{ m}^3\text{ kg}^{-1}$ and water vapor $v_g=1.6720\text{ m}^3\text{ kg}^{-1}$.

\begin{align*} v_1&=v_f + x_1\cdot \left(v_g - v_f\right) \\ v_1&=0.001043\,\frac{\text{m}^3}{\text{kg}} + 0.123\cdot \left(1.6720\,\frac{\text{m}^3}{\text{kg}}- 0.001043\,\frac{\text{m}^3}{\text{kg}}\right) \\ v_1&=0.2066\,\frac{\text{m}^3}{\text{kg}} \end{align*}

The $\textbf{mass }$of the water can then be calculated using the given volume of the tank $V=10\text{ L}$. Before the calculation we need to express the volume in $m^3$ units.

$\begin{equation*} V=10\text{ L} \cdot \frac{1\text{ m}^3}{1000\text{ L}}= 0.01\text{ m}^3 \end{equation*}$

\begin{align*} m&=\frac{V}{v_1} \\ m&=\frac{0.01\text{ m}^3}{0.2066\,\dfrac{\text{m}^3}{\text{kg}}} \\ m&=\color{#4257b2}0.0484\text{ kg} \end{align*}

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