Chapter 4, Problem 1P

An ice-cube tray containing 250 mL of freshwater at 15°C is placed in a freezer at −5°C. Determine the change in internal energy (kJ) and entropy (kJ/K) of the water when it has frozen.

To determine

The change in internal energy and entropy of the water.

Explanation of Solution

Given:

The volume of the water $\left(V\right)$ is 250 ml.

The temperature of the water $\left(T_1\right)$ is $15°\text{C}$.

The temperature of the freezer $\left(T_2\right)$ is $5°\text{C}$.

Calculation:

Consider the density of the water $\left(\rho \right)$ is $999{\text{kg/m}}^3$ and the specific heat at constant volume $\left(c_P\right)$ is $1\frac{\text{k}\cdot \text{cal}}{\text{kg}\cdot \text{K}}$.

Calculate the change in entropy $\left(\Delta S\right)$.

  $\begin{array}{c}\Delta S=m\left(s_2-s_1\right)\\ \Delta S=\rho V{c}_v\ln \left(\frac{T_2}{T_1}\right)\\ =999{\text{kg/m}}^3\times \left(250\text{ml}\right)\times 1\frac{\text{kcal}}{\text{kg}\cdot \text{K}}\times \ln \left(\frac{-5°\text{C}}{15°\text{C}}\right)\\ =999{\text{kg/m}}^3\times \left(250\text{ml}\right)\times 1\frac{\text{kcal}}{\text{kg}\cdot \text{K}}\times \ln \left(\frac{268\text{ K}}{288\text{ K}}\right)\\ =-0.01797\frac{\text{kcal}}{\text{kg}}\times \frac{4190\text{ J}}{1\text{ kcal}}\\ =-75.32\text{ J/K}\\ =\text{0}\text{.0753 kJ/K}\end{array}$

Thus, the change in entropy is $\left(\Delta s\right)$ is $\underset{\_}{0.0753\text{ kJ/K}}$.

Calculate the change in internal energy $\left(\Delta U\right)$.

  $\begin{array}{c}\Delta U=m{c}_v\left(T_2-T_1\right)\\ \Delta U=\rho V{c}_v\left(T_2-T_1\right)\\ =999{\text{kg/m}}^3\times \left(250\text{ml}\times \frac{{10}^{-6}{\text{m}}^3}{1\text{ml}}\right)\times 1\frac{\text{kcal}}{\text{kg}\cdot \text{K}}\times \left(-5°\text{C}-15°\text{C}\right)\\ =999{\text{kg/m}}^3\times \left(250\times {10}^{-6}{\text{m}}^3\right)\times \text{1 }\frac{\text{kcal}}{\text{kg}\cdot \text{K}}\times \left(268\text{ K}-288\text{ K}\right)\\ =-4.995\frac{\text{kcal}}{\text{kg}}\times \frac{4190\text{ J}}{1\text{ kcal}}\\ =-20929.05\text{ J/K}\\ \text{=}-\text{20}\text{.929 kJ/K}\end{array}$

Thus, the change in internal energy is $\left(\Delta s\right)$ is $\underset{\_}{-\text{20}\text{.929 kJ/K}}$.