Chapter 7.13, Problem 205RP

a)

To determine

The final equilibrium temperature in the room.

Answer to Problem 205RP

The final equilibrium temperature in the room is $78.4°\text{C}$.

Explanation of Solution

Write the expression for the energy balance equation for closed system.

$E_{in}-E_{out}=\Delta E_{system}$ (I)

Here, energy transfer into the control volume is $E_{in}$, energy transfer out of  control volume is $E_{out}$ and change in internal energy of system is $\Delta E_{system}$.

Write the expression to calculate the ideal gas equation, to find the mass of the air.

$m_{air}=\frac{P_1{\left({\nu }_1\right)}_{air}}{R{\left(T_1\right)}_{air}}$ (II)

Here, mass of the air is $m_{air}$, initial pressure is $P_1$, initial volume of air is ${\left({\nu }_1\right)}_{air}$, gas constant is R and initial temperature of air is ${\left(T_1\right)}_{air}$.

Conclusion:

Substitute 0 for $E_{in}$, 0 for $E_{out}$ and $\Delta U_{air}+\Delta U_{water}$ for $\Delta E_{system}$ in Equation (I).

$\begin{array}{l}0-0=\Delta U_{water}+\Delta U_{air}\\ \left[m_{water}{c}_{p{,}_{water}}\left(T_2-{\left(T_1\right)}_{water}\right)\right]+\left[m_{air}{c}_{v,air}\left(T_2-{\left(T_1\right)}_{air}\right)\right]=0\end{array}$ (III)

Here, change in internal energy in air is $\Delta U_{air}$, change in internal energy in water is $\Delta U_{water}$, mass of the air is $m_{air}$, mass of the water is $m_{water}$, initial temperature of water is ${\left(T_1\right)}_{water}$, specific heat at constant pressure is $c_{p,water}$ , specific heat at constant volume is $c_{v,air}$ , and initial temperature of air is ${\left(T_1\right)}_{air}$.

From the Table A-1, “the molar mass, gas constant and critical point properties table”, select the gas constant $\left(R\right)$ of air as $0.2870\text{kJ}/\text{kg}\cdot \text{K}$.

Substitute $\text{100}\text{kPa}$ for $P_1$, $\left(4\text{m}\times 5\text{m}\times 7\text{m}\right)$ for ${\left({\nu }_1\right)}_{air}$, $0.2870\text{kJ}/\text{kg}\cdot \text{K}$ for R and $22 °\text{C}$ for ${\left(T_1\right)}_{air}$ in Equation (II).

$\begin{array}{c}{m}_{air}=\frac{\left(100\text{ kPa}\right)\left(4\text{m}\times 5\text{m}\times 7\text{m}\right)}{\left(0.2870\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\frac{1\text{kPa}\cdot {\text{m}}^{\text{3}}}{1\text{kJ}}\right)22 °\text{C}}\\ =\frac{\left(100\text{ kPa}\right)\left(140{\text{m}}^{\text{3}}\right)}{\left(0.2870\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\frac{1\text{kPa}\cdot {\text{m}}^{\text{3}}}{1\text{kJ}}\right)\left(22+273\right)\text{K}}\\ =165.4\text{kg}\end{array}$

From the Table A-3, “the properties of common liquids, solids and foods table”, select the specific heat at constant pressure $\left(c_p\right)$ and select the specific heat at constant volume $\left(c_v\right)$ of water at room temperature as $4.18\text{kJ}/\text{kg}\cdot \text{K}$ and $0.718\text{kJ}/\text{kg}\cdot \text{K}$ respectively.

Substitute 1000 kg for $m_{water}$, $4.18\text{kJ}/\text{kg}\cdot \text{K}$ for $c_{p,water}$, $80 °\text{C}$ for ${\left(T_1\right)}_{water}$, 165.4 kg for $m_{air}$, $0.718\text{kJ}/\text{kg}\cdot \text{K}$ for $c_v$ and $22 °\text{C}$ for ${\left(T_1\right)}_{air}$ in Equation (III).

$\begin{array}{l}\left\{\begin{array}{l}\left[\left(1000\text{kg}\right)\left(4.18\text{kJ}/\text{kg}\cdot \text{K}\right)\left(T_2-80°\text{C}\right)\right]\\ +\left[\left(165.4\text{kg}\right)\left(0.718\text{kJ}/\text{kg}\cdot \text{K}\right)\left(T_2-22°\text{C}\right)\right]\end{array}\right\}=0\\ 4298.75T_2=337012.6\\ T_2=\frac{337012.6}{4298.75}\\ T_2=78.4°\text{C}\end{array}$

Thus, the final equilibrium temperature in the room is $78.4°\text{C}$.

b)

To determine

The entropy generation during the process.

Answer to Problem 205RP

The entropy generation during the process is $1.79\text{kJ}/\text{K}$.

Explanation of Solution

Write the expression for the entropy balance equation of the system.

$S_{in}-S_{out}+S_{gen}=\Delta S_{system}$ (IV)

Here, rate of net entropy in is $S_{in}$, rate of net entropy out is $S_{out}$, rate of entropy generation is $S_{gen}$ and change of entropy of the system is $\Delta S_{system}$

Conclusion:

Substitute 0 for $S_{in}$, 0 for $S_{out}$ and $\Delta S_{air}+\Delta S_{water}$ for $\Delta S_{system}$ in Equation (V).

$\begin{array}{l}0-0+S_{gen}=\Delta S_{air}+\Delta S_{water}\\ S_{gen}=\left[\begin{array}{l}\left(m_{air}{c}_{v,air}\ln \left(\frac{T_2}{{\left(T_1\right)}_{air}}\right)+m_{air}R\ln \left(\frac{{\left({\nu }_2\right)}_{air}}{{\left({\nu }_1\right)}_{air}}\right)\right)\\ +\left(m_{water}{c}_{p,water}\ln \left(\frac{T_2}{{\left(T_1\right)}_{water}}\right)\right)\end{array}\right]\end{array}$ (V)

Here, final volume of air is ${\left({\nu }_2\right)}_{air}$.

Substitute 165.4 kg for $m_{air}$, $0.718\text{kJ}/\text{kg}\cdot \text{K}$ for $c_{v,air}$, $22 °\text{C}$ for ${\left(T_1\right)}_{air}$, $78.4 °\text{C}$ for $T_2$, 1000 kg for $m_{water}$, $4.18\text{kJ}/\text{kg}\cdot \text{K}$ for $c_{p,water}$, $80 °\text{C}$ for ${\left(T_1\right)}_{water}$, $\left(4\text{m}\times 5\text{m}\times 7\text{m}\right)$ for ${\left({\nu }_2\right)}_{air}$, $\left(4\text{m}\times 5\text{m}\times 7\text{m}\right)$ for ${\left({\nu }_1\right)}_{air}$, and $0.2870\text{kJ}/\text{kg}\cdot \text{K}$ for R in Equation (V).

$S_{gen}=\left\{\begin{array}{l}\left(\begin{array}{l}\left(165.4\text{ kg}\right)\left(0.718\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{78.4 °\text{C}}{22°\text{C}}\right)\\ +\left(165.4\text{ kg}\right)\left(0.2870\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{\left(4\text{m}\times 5\text{m}\times 7\text{m}\right)}{\left(4\text{m}\times 5\text{m}\times 7\text{m}\right)}\right)\end{array}\right)\\ +\left(\left(1000\text{kg}\right)\left(4.18\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{78.4 °\text{C}}{22°\text{C}}\right)\right)\end{array}\right\}$

$\begin{array}{c}{S}_{gen}=\left\{\begin{array}{l}\left(\begin{array}{l}\left(165.4\text{ kg}\right)\left(0.718\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{\left(78.4+273\right)\text{K}}{\left(22+273\right)\text{K}}\right)\\ +\left(165.4kg\right)\left(0.2870\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{140{\text{m}}^{\text{3}}}{140{\text{m}}^{\text{3}}}\right)\end{array}\right)\\ +\left(\left(1000\text{kg}\right)\left(4.18\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{\left(78.4+273\right)\text{K}}{\left(80+273\right)\text{K}}\right)\right)\end{array}\right\}\\ =20.78+\left(-18.99\right)\\ =1.79\text{kJ}/\text{K}\end{array}$

Thus, the entropy generation during the process is $1.79\text{kJ}/\text{K}$.