Problem 58 - University Physics with Modern Physics

Question

20.58. A \(0.0500-\mathrm{kg}\) cube of ice at an initial temperature of \(-15.0^{\circ} \mathrm{C}\) is placed in 0.600 \(\mathrm{kg}\) of water at \(T=45.0^{\circ} \mathrm{C}\) in an insulated container of negligible mass. (a) Calculate the final temperature of the water once the ice has melted. (b) Calculate the change in entropy of the system.

Step-by-Step Solution

Answer

(a) Final temperature is 37.72°C. (b) Change in entropy is 22.73 J/K.

Step 1: Calculate the heat needed to warm the ice to 0°C

First, determine the heat required to increase the temperature of the ice from \(-15.0^{\circ}\text{C}\) to \(0^{\circ}\text{C}\).The specific heat of ice is \(c_{ice} = 2.09 \, \text{J/g°C}\).The heat required is given by:\[ q_1 = m_{ice} \, c_{ice} \, \Delta T \]Substitute the values:\[ q_1 = 50 \, ext{g} \, \times \, 2.09 \, \text{J/g°C} \, \times \, (0 - (-15)) \, \text{°C} = 50 \, \text{g} \, \times \, 2.09 \, \times \, 15 = 1567.5 \, \text{J} \]

Step 2: Calculate the heat needed to melt the ice

Next, determine the heat required to melt the ice at \(0^{\circ}\text{C}\).The latent heat of fusion for ice is \(L_{f} = 334 \, \text{J/g}\).The heat required is given by:\[ q_2 = m_{ice} \, L_{f} \]Substitute the values:\[ q_2 = 50 \, \text{g} \, \times \, 334 \, \text{J/g} = 16700 \, \text{J} \]

Step 3: Calculate the total heat absorbed by ice

The total heat absorbed by the ice is the sum of the heat to warm it to \(0^{\circ}\text{C}\) and the heat to melt it:\[ q_{ice} = q_1 + q_2 = 1567.5 \, \text{J} + 16700 \, \text{J} = 18267.5 \, \text{J} \]

Step 4: Calculate the heat lost by the water to reach equilibrium

The specific heat of water is \(c_{water} = 4.18 \, \text{J/g°C}\).The mass of the water is \(600 \, \text{g}\) and its initial temperature is \(45^{\circ}\text{C}\).The change in temperature \(\Delta T\) when the water reaches equilibrium is unknown.The heat lost by the water (\(q_{water}\)) is equal to the heat gained by the ice (\(q_{ice}\)):\[ q_{water} = m_{water} \, c_{water} \, \Delta T \]Substitute the values and solve for \(\Delta T\):\[ 600 \, \text{g} \, \times \, 4.18 \, \text{J/g°C} \, \times \, (45 - T_f) = 18267.5 \, \text{J} \]Solving,\[ (45 - T_f) = \frac{18267.5}{2508} \]\[ 45 - T_f = 7.28 \]\[ T_f = 45 - 7.28 = 37.72^{\circ}C \]

Step 5: Calculate change in entropy for ice

The change in entropy for the ice as it warms from \(-15^{\circ}\text{C}\) to \(0^{\circ}\text{C}\) is calculated using:\[ \Delta S_1 = \frac{q_1}{T_{avg}} \]Average temperature \(T_{avg}\) in Kelvin:\[ T_{avg} = \frac{273.15 - 15 + 273.15}{2} = 265.65 \text{K}\]\[ \Delta S_1 = \frac{1567.5}{265.65} = 5.90 \text{J/K}\]The change in entropy for melting:\[ \Delta S_2 = \frac{q_2}{273.15} = \frac{16700}{273.15} = 61.15 \text{J/K} \]

Step 6: Calculate change in entropy for water

The change in entropy for the water as it changes temperature is given by:\[ \Delta S_{water} = m_{water} \, c_{water} \, \ln\left(\frac{T_f}{T_{i}}\right) \]Where:- \(T_{i} = 318.15 \, \text{K}\)- \(T_f = 310.87 \, \text{K}\) (because \(37.72\degree C = 310.87 \, \text{K}\))\[ \Delta S_{water} = 600 \, \times \, 4.18 \, \times \, \ln\left(\frac{310.87}{318.15}\right) = -44.32 \text{J/K} \]

Step 7: Calculate total entropy change

The total change in entropy of the system is obtained by adding the changes calculated for ice and water:\[ \Delta S_{total} = \Delta S_1 + \Delta S_2 + \Delta S_{water} \]Substituting the values:\[ \Delta S_{total} = 5.90 + 61.15 - 44.32 = 22.73 \text{J/K} \]

Key Concepts

Entropy

Entropy is a measure of disorder or randomness in a system. In thermodynamics, it helps us to understand how energy is distributed and how processes evolve. When a process occurs, the total entropy of a closed system increases or stays the same, in accordance with the second law of thermodynamics.

In the exercise, we calculated the change in entropy during the process of ice melting in water. When the ice absorbs heat and begins to melt, the arrangement of its molecules becomes more disordered, increasing its entropy. This is calculated by considering the heat exchange and the average temperature during the transition processes.
- **Ice Warming to Melt:** The entropy change for the warming of ice is computed using the formula: \[ \Delta S_1 = \frac{q_1}{T_{avg}} \] Here, \( q_1 \) is the heat absorbed by the ice and \( T_{avg} \) is the average temperature in Kelvin during warming.- **Ice Melting:** The entropy change for the phase change: \[ \Delta S_2 = \frac{q_2}{273.15} \] Where \( q_2 \) is the latent heat required for the phase change at the melting point temperature.- **Water Cooling:** The entropy change for the water is calculated as it cools and provides the heat absorbed by the ice.
Overall, the entropy changes reflect the balance of disorder added and retained within the system as energy and heat are transferred.

Specific Heat Capacity

Specific heat capacity is a critical concept in thermodynamics. It describes the amount of heat needed to change the temperature of a given substance by one degree Celsius per unit mass. It's denoted by \( c \). This property varies across different materials and plays a key role in energy transfer processes, such as heating or cooling.
In the problem, both ice and water have specific heat capacities that determine how much they heat up or cool down, respectively. The calculations involved:- For ice, the specific heat capacity is \( c_{ice} = 2.09 \, \text{J/g°C} \) , which tells us how much heat is required to raise the temperature of 1 gram of ice by 1°C.- For water, the specific heat capacity is \( c_{water} = 4.18 \, \text{J/g°C} \) , which is higher than that of ice, reflecting water's ability to absorb and retain more heat.
Calculating the heat (\( q \) ) necessary to change the temperature relies on the specific heat formula: \[ q = m \cdot c \cdot \Delta T \] where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the temperature change.

Latent Heat

Latent heat is the amount of heat energy required to change the phase of a substance without changing its temperature. It is crucial to understanding phase changes such as melting, freezing, and vaporization, where energy is absorbed or released to break the molecular bonds without altering the substance's temperature.
- **Latent Heat of Fusion:** For ice, the latent heat of fusion is \( 334 \, \text{J/g} \). This means that 334 joules of energy are needed to convert 1 gram of ice at 0°C into water without any change in temperature.- **Energy Calculations:** In the exercise, the ice requires this latent heat energy to transition completely into the liquid phase. The calculation used is \[ q_2 = m \cdot L_f \] where \( m \) is the mass of the ice and \( L_f \) is the latent heat of fusion.
Latent heat is thus pivotal in phase change processes, influencing how substances absorb and release energy while their states change.

Phase Change

Phase change refers to the transformation of a substance from one state of matter to another, such as from solid to liquid, liquid to gas, or vice versa. These changes occur at specific temperatures and involve energy absorption or release.
In the context of the exercise, the phase change discussed is the melting of ice into water: - **Melting:** This occurs when the ice, initially a solid, absorbs enough heat to break the molecular bonds holding the ice structure together, changing into liquid water. - **Energy Roles:** The energy required for this phase change is given by the latent heat, specifically the latent heat of fusion for ice. During melting, the temperature of the substance remains constant at the melting point even as the ice transitions to the liquid state.
Understanding phase change helps in quantifying the energy involved and the behavior of substances as they transition between different states.