Chapter 8, Problem 61P

(a)

To determine

The entropy change of the computer chips.

Explanation of Solution

Given:

The mass of the computer chips $\left(m_{\text{chips}}\right)$ is $10\text{ g}$.

The specific heat of the computer chips $\left(c\right)$ is $0.3\text{kJ}/\text{kg}\cdot \text{K}$.

The initial temperature of the computer chips $\left(T_1\right)$ is $20°\text{C}$.

The final temperature of the computer chips $\left(T_2\right)$ is $-40°\text{C}$.

The mass of the refrigerant liquid $\left(m_R\right)$ is 5g.

Calculation:

Calculate the heat released by the computer chips.

  $Q_{\text{chips}}=mc\left(T_1-T_2\right)$

  $\begin{array}{c}{Q}_{\text{chips}}=\left(10\text{g}\right)\left(0.3\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\left(20°\text{C}\right)-\left(-40°\text{C}\right)\right)\\ =\left(10\text{g}\times \left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)\left(0.3\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\left(20°\text{C+273}\right)-\left(-40°\text{C+273}\right)\right)\\ =\left(0.010\text{kg}\right)\left(0.3\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\left(293\text{K}\right)-\left(233\text{K}\right)\right)\\ =0.18\text{kJ}\end{array}$

Refer the Table A-11 “Saturated refrigerant-134a—Temperature table”, to obtain the below properties for the at final temperature of $-40°\text{C}$ as

  $\begin{array}{l}{h}_{fg}=225.86\text{kJ}/\text{kg}\\ s_{g,2}=0.96866\text{kJ}/\text{kg}\cdot \text{K}\\ s_{f,1}=0.00000\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Calculate the mass of the refrigerant vaporized during this heat exchange process.

  $\begin{array}{c}{m}_{g,2}=\frac{Q_{\text{chips}}}{h_g-h_f}\\ =\frac{Q_{\text{chips}}}{h_{fg@-40°\text{C}}}\end{array}$

  $\begin{array}{c}{m}_{g,2}=\frac{\left(0.18\text{kJ}\right)}{\left(225.86\text{kJ}/\text{kg}\right)}\\ =0.000797\text{kg}\end{array}$

Write the expression for the energy balance equation.

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

Here, the total energy entering the system is $E_{\text{in}}$, the total energy leaving the system is $E_{\text{out}}$, and the change in the total energy of the system is $\Delta E_{\text{system}}$.

Substitute $E_{\text{in}}=0$, $E_{\text{out}}=0$, and $\Delta E_{\text{system}}=\Delta U$ in Equation (I)

  $\begin{array}{l}\left(0\right)-\left(0\right)=\Delta U\\ \left(0\right)={\left[m\left(u_2-u_1\right)\right]}_{\text{Chips}}+{\left[m\left(u_2-u_1\right)\right]}_{\text{R-134a}}\\ {\left[m\left(u_2-u_1\right)\right]}_{\text{Chips}}={\left[m\left(u_2-u_1\right)\right]}_{\text{R-134a}}\end{array}$        (II)

Here, the mass is $m$, the initial specific internal energy is $u_1$, and the final specific internal energy is $u_2$.

Small fraction of the refrigerant us vaporized during the process so the the mass of the refrigerant liquid at state 1 and state 2 are same $\left(m_{f,1}=m_{f,2}\right)$.

Calculate the change in the entropy of the R-134a.

  $\begin{array}{c}\Delta S_{\text{R-134a}}=m_{g,2}{s}_{g,2}+m_{f,2}{s}_{f,2}-m_{f,1}{s}_{f,1}\\ \Delta S_{\text{R-134a}}=\left[\begin{array}{l}\left(0.000797\text{kg}\right)\times \left(0.96866\text{kJ}/\text{kg}\cdot \text{K}\right)\\ +\left(5\text{g}\right)\times \left(0.00000\text{kJ}/\text{kg}\cdot \text{K}\right)\\ -\left(5\text{g}\right)\times \left(0.00000\text{kJ}/\text{kg}\cdot \text{K}\right)\end{array}\right]\\ =0.00772\text{kJ}/\text{K}\end{array}$

Thus, the entropy change of the computer chips is $\underset{\_}{0.000772\text{kJ}/\text{K}}$.

Calculate the entropy change of the computer chips.

  $\Delta S_{\text{chips}}=mc\ln \frac{T_2}{T_1}$

  $\begin{array}{c}\Delta S_{\text{chips}}=\left(10\text{g}\right)\left(0.3\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{\left(-40°\text{C}\right)}{\left(20°\text{C}\right)}\right)\\ =\left(10\text{g}\times \left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)\left(0.3\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{\left(-40°\text{C+273}\right)}{\left(20°\text{C+273}\right)}\right)\\ =\left(0.010\text{kg}\right)\left(0.3\text{kJ}/\text{kg}\cdot \text{K}\right)\left(-0.22913\right)\\ =-0.00069\text{kJ}/\text{K}\end{array}$

Thus, the entropy change of the R-134 is $\underset{\_}{-0.00069\text{kJ}/\text{K}}$.

Calculate the total entropy change of the entire system.

  $\begin{array}{c}\Delta S_{\text{total}}=\Delta S_{\text{R-134a}}+\Delta S_{\text{chips}}\\ \Delta S_{\text{total}}=\left(0.000772\text{kJ}/\text{K}\right)+\left(-0.00069\text{kJ}/\text{K}\right)\\ =0.00008457\text{kJ}/\text{K}\end{array}$

Thus, the entropy change of the entire system is $\underset{\_}{0.00008457\text{kJ}/\text{K}}$.

(b)

To determine

The entropy change of the R-134.

Explanation of Solution

Calculate the entropy change of the computer chips.

  $\Delta S_{\text{chips}}=mc\ln \frac{T_2}{T_1}$

  $\begin{array}{c}\Delta S_{\text{chips}}=\left(10\text{g}\right)\left(0.3\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{\left(-40°\text{C}\right)}{\left(20°\text{C}\right)}\right)\\ =\left(10\text{g}\times \left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)\left(0.3\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{\left(-40°\text{C+273}\right)}{\left(20°\text{C+273}\right)}\right)\\ =\left(0.010\text{kg}\right)\left(0.3\text{kJ}/\text{kg}\cdot \text{K}\right)\left(-0.22913\right)\\ =-0.00069\text{kJ}/\text{K}\end{array}$

Thus, the entropy change of the R-134 is $\underset{\_}{-0.00069\text{kJ}/\text{K}}$.

(c)

To determine

The entropy change of the entire system.

Explanation of Solution

Calculate the total entropy change of the entire system.

  $\begin{array}{c}\Delta S_{\text{total}}=\Delta S_{\text{R-134a}}+\Delta S_{\text{chips}}\\ \Delta S_{\text{total}}=\left(0.000772\text{kJ}/\text{K}\right)+\left(-0.00069\text{kJ}/\text{K}\right)\\ =0.00008457\text{kJ}/\text{K}\end{array}$

Thus, the entropy change of the entire system is $\underset{\_}{0.00008457\text{kJ}/\text{K}}$.