Question

In order to cool 1 -ton of water at \(20^{\circ} \mathrm{C}\) in an insulated tank, a person pours \(80 \mathrm{kg}\) of ice at \(-5^{\circ} \mathrm{C}\) into the water. Determine ( \(a\) ) the final equilibrium temperature in the \(\operatorname{tank}\) and \((b)\) the entropy generation during this process. The melting temperature and the heat of fusion of ice at atmospheric pressure are \(0^{\circ} \mathrm{C}\) and \(333.7 \mathrm{kJ} / \mathrm{kg}\).

Short answer

Answer: The final equilibrium temperature is 2.34°C, and the entropy generation during the cooling process is -194.87 J/K.

Step-by-step solution

Calculate the heat given to the ice to raise it from -5°C to 0°C

First, we need to find the amount of heat required to raise the temperature of the ice from -5°C to its melting point at 0°C. We can use the equation: \(Q_1 = m_ic_i \Delta T_{ice}\) Where: \(Q_1\) is the heat given to the ice, \(m_i = 80 \mathrm{kg}\) is the mass of the ice, \(c_i = 2.1 \mathrm{kJ/(kg \cdot K)}\) is the specific heat of ice, \(\Delta T_{ice} = 5 \mathrm{K}\) is the temperature difference. \(Q_1 = 80 \mathrm{kg} \cdot 2.1 \mathrm{kJ/(kg \cdot K)} \cdot 5 \mathrm{K} = 840 \mathrm{kJ}\).

Calculate the heat given to the ice to melt it completely into water at 0°C

Next, we need to find the amount of heat required to melt the ice completely. We can use the equation: \(Q_2 = m_i L_f\) Where: \(Q_2\) is the heat given to the ice to melt it, \(L_f = 333.7 \mathrm{kJ/kg}\) is the heat of fusion of ice. \(Q_2 = 80 \mathrm{kg} \cdot 333.7 \mathrm{kJ/kg} = 26,696 \mathrm{kJ}\).

Calculate the heat given to the water to raise it from 20°C to the final equilibrium temperature

Now, we need to find the heat given to the water to raise its temperature from 20°C to the final equilibrium temperature. We can use the equation: \(Q_3 = m_w c_w \Delta T_{water}\) Where: \(Q_3\) is the heat given to the water, \(m_w = 1000 \mathrm{kg}\) is the mass of the water, \(c_w = 4.18 \mathrm{kJ/(kg \cdot K)}\) is the specific heat of water, \(\Delta T_{water} = T_{f} - 20^{\circ} \mathrm{C}\) is the temperature difference, with \(T_f\) being the unknown final temperature.

Determine the final equilibrium temperature

The total energy in the mixture of water and ice is conserved. The sum of the heats given to the ice and the heat given to the water is zero: \(Q_1 + Q_2 + Q_3 = 0\) Substitute expressions for \(Q_1\), \(Q_2\), and \(Q_3\): \(840 \mathrm{kJ} + 26,696 \mathrm{kJ} + 1000 \mathrm{kg} \cdot 4.18 \mathrm{kJ/(kg \cdot K)} (T_f - 20^{\circ} \mathrm{C}) = 0\) Solve for \(T_f\): \(T_f = \frac{840 \mathrm{kJ} + 26,696 \mathrm{kJ}}{{1000 \mathrm{kg} \cdot 4.18 \mathrm{kJ/(kg \cdot K)}}} + 20^{\circ} \mathrm{C} \approx 2.34^{\circ} \mathrm{C}\) So the final equilibrium temperature is \(2.34^{\circ} \mathrm{C}\).

Calculate the entropy generation during this process

To find the entropy generation, we'll calculate the entropy change for each subprocess and sum them up: 1. Entropy change in heating the ice from -5°C to 0°C: \(\Delta S_{ice} = m_i c_i \int_{-5}^0 \frac{dT_{ice}}{T_{ice}} = m_i c_i \ln{\frac{T_2}{T_1}}\) \(\Delta S_{ice} = 80 \mathrm{kg} \cdot 2.1 \mathrm{kJ/(kg \cdot K)} \cdot \ln{\frac{273.15 \mathrm{K}}{268.15 \mathrm{K}}} \approx 52.27 \mathrm{J/K}\) 2. Entropy change in melting the ice: \(\Delta S_{melt} = \frac{Q_2}{T_m}\) \(\Delta S_{melt} = \frac{26,696 \mathrm{kJ}}{273.15 \mathrm K} \approx 97.73 \mathrm{J/K}\) 3. Entropy change in heating the water from 20°C to 2.34°C: \(\Delta S_{water} = m_w c_w \int_{20}^{2.34} \frac{dT_{water}}{T_{water}} = m_w c_w \ln{\frac{T_f}{T_w}}\) \(\Delta S_{water} = 1000 \mathrm{kg} \cdot 4.18 \mathrm{kJ/(kg \cdot K)} \cdot \ln{\frac{275.49 \mathrm{K}}{293.15 \mathrm{K}}} \approx -344.87 \mathrm{J/K}\) Total entropy generation: \(\Delta S_{total} = \Delta S_{ice} + \Delta S_{melt} + \Delta S_{water} = 52.27 \mathrm{J/K} + 97.73 \mathrm{J/K} - 344.87 \mathrm{J/K} = -194.87 \mathrm{J/K}\) The entropy generation during this process is \(-194.87 \mathrm{J/K}\).

Key concepts

Equilibrium Temperature Calculation

Understanding the equilibrium temperature in a thermodynamic system such as the one in our exercise, where ice is added to water, is fundamental. The equilibrium temperature represents the condition where both substances—the ice and the water—reach a state of thermal balance, with no net heat flow between them.

Mathematically, this is found by setting the heat lost by one part of the system equal to the heat gained by the other. In our example, heat flows from the warmer water to the colder ice until both reach the same temperature. The solution involves using specific heat capacities and the heat of fusion, ensuring energy conservation principles are adhered to. The step-by-step provided in the exercise leads you to the correct equilibrium temperature using these concepts.

For greater clarity, you can visualize this scenario as a seesaw where both sides must be balanced. The mass and specific heat of the water and the ice act like the weights on either side, and our task is to find the point where those weights are balanced in terms of thermal energy.

Entropy Generation

The concept of entropy generation revolves around the Second Law of Thermodynamics, which states that in any energy exchange, if no energy enters or leaves the system, the potential energy of the state will always be less than that of the initial state—this is known as entropy. In our exercise, entropy generation signifies the measure of disorder or randomness introduced by the mixing of water and ice.

To compute the entropy generation, we consider the entropy changes in heating the ice, melting the ice, and finally adjusting the water temperature to the new equilibrium state. Entropy increases when ice melts and water cools, but it decreases when we warm up the water slightly to reach equilibrium. It's like mixing two different colors of clay—once you mix them, you cannot fully separate them again, signifying an increase in disorder. The calculation steps provided guide you through this comparison quantitatively, giving you the total entropy generated in the process.

Heat Transfer

Heat transfer is a key aspect in thermodynamics, and it refers to the movement of thermal energy between physical systems depending on the temperature difference. In the exercise, heat transfer occurs in three main stages: heating the ice to its melting point, the phase change from ice to water, and finally adjusting the water's temperature up to equilibrium.

Heat moves from the warmer water to the cooler ice, demonstrated by the heat equation used in each step. The specific heat and heat of fusion characteristics dictate how much energy transfers in these processes. Understanding heat transfer can be equated to understanding how a hot coffee cools down in a cold room; energy transfers from the coffee (higher temperature) to the room (lower temperature) until equilibrium is reached. The equations provided in the steps effectively showcase the energy balance throughout this heat exchange.