Chapter 15, Problem 42P

To determine

To Estimate: Net change in entropy.

Answer to Problem 42P

  $18.73\text{J/K}$

Explanation of Solution

Given:

Mass of water, $m_1=m_2\text{= }30°\text{C}$

Temperature of water, $T_1=\text{ 30}°\text{C}$

Temperature of water in container, $T_2=\text{ 60}°\text{C}$

Formula used:

Change in entropy:

  $\Delta S=\frac{Q}{T}$

Here, Q is the heat energy and T is temperature

  $Q=mc\Delta T$

Here, m is the mass and cis the specific heat capacity.

Net change in entropy:

  $\Delta S_{net}=\frac{Q_1}{T_1}+\frac{Q_2}{T_2}$

Calculation:

Final temperature of water when they are mixed:

  $\begin{array}{l}{T}_f=\frac{\left(T_1+T_2\right)}{2}\\ T_f=\frac{\text{30°C+90°C}}{2}\\ T_f=45°\text{C}\end{array}$

  $\begin{array}{l}{Q}_1=\left(420\text{0}\text{J/kg°C}\right)\left(\text{1}\text{kg}\right)\left(\text{45°C-30}°\text{C}\right)\\ Q_1=6\text{3}\text{kJ}\\ Q_2=\left(420\text{0}\text{J/kg°C}\right)\left(\text{1}\text{kg}\right)\left(\text{45°C-60}°\text{C}\right)\\ Q_1=-6\text{3}\text{kJ}\end{array}$

  $\begin{array}{l}\Delta S_{net}=\frac{Q_1}{T_1}+\frac{Q_2}{T_2}\\ \Delta S_{net}=\frac{63\text{kJ}}{303\text{K}}-\frac{63\text{kJ}}{333\text{K}}\\ \Delta S_{net}=18.73\text{J/K}\end{array}$

Conclusion:

Thus, the net change in entropy is $18.73\text{J/K}$ .