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 exergy destroyed during this process. The melting temperature and the heat of fusion of ice at atmospheric pressure are \(0^{\circ} \mathrm{C}\) and \(333.7 \mathrm{kJ} / \mathrm{kg}\), respectively. Take \(T_{0}=20^{\circ} \mathrm{C}\).

Short answer

Question: Find the final equilibrium temperature and the exergy destroyed after 80 kg of ice is added to 1 ton of water at 20 °C in an insulated tank. Answer: To find the final equilibrium temperature and exergy destroyed, follow these steps: 1. Calculate the initial energy of the system. 2. Calculate the energy needed to melt the ice. 3. Apply the energy conservation principle and determine the final equilibrium temperature. 4. Calculate the change in entropy for both water and ice. 5. Calculate the total change in entropy of the system and exergy destroyed. By applying these steps, you will find the final equilibrium temperature and exergy destroyed.

Step-by-step solution

Calculate the initial energy of the system

We need to find the total initial energy of water and the ice added to it. We have: 1. Mass of water (\(m_w\)) = 1 ton = 1000 kg 2. Initial temperature of water (\(T_w\)) = 20 °C 3. Mass of ice (\(m_i\)) = 80 kg 4. Initial temperature of ice (\(T_i\)) = -5 °C 5. Melting temperature of ice = 0 °C 6. Heat of fusion of ice (\(L_f\)) = 333.7 kJ/kg 7. Reference temperature (\(T_0\)) = 20 °C Initial energy of water: \(E_{w, initial} = m_w \cdot c_w \cdot (T_w - T_0)\) Initial energy of ice: \(E_{i, initial} = m_i \cdot c_i \cdot (T_i - T_0)\) Total initial energy of the system: \(E_{initial} = E_{w, initial} + E_{i, initial}\) where \(c_w\) and \(c_i\) are the specific heat capacities of water and ice, respectively.

Calculate the energy needed to melt the ice

To find the energy needed to melt the 80 kg of ice we can use the formula: Energy to melt ice: \(E_{melt} = m_i \cdot L_f\)

Apply energy conservation principle and determine the final equilibrium temperature

According to the energy conservation principle, the final energy of the system should be equal to the initial energy: \(E_{final} = E_{initial} - E_{melt}\) Since there is no heat transfer with the surroundings: \(E_{final} = m_w \cdot c_w \cdot (T_f - T_0)\) Where \(T_f\) is the final equilibrium temperature. We can solve this equation to find \(T_f\).

Calculate the change in entropy for both water and ice

Determine the change in entropy for water and ice using the following formulas: \(\Delta S_w = m_w \cdot c_w \cdot \ln \frac{T_f}{T_w}\) \(\Delta S_i = m_i \cdot c_i \cdot \ln \frac{T_f}{T_i} + m_i \cdot \frac{L_f}{T_m}\) where \(T_m\) is the melting temperature (0 °C in Kelvin)

Calculate the total change in entropy of the system and exergy destroyed

Add the change in entropies to find the total change in entropy of the system: \(\Delta S_{total} = \Delta S_w + \Delta S_i\) Now, find the exergy destroyed using the formula: Exergy destroyed: \(X_{destroyed} = T_0 \cdot \Delta S_{total}\) Following these steps, you will successfully find the final equilibrium temperature and the exergy destroyed during this process.

Key concepts

Energy Conservation Principle

The energy conservation principle, often termed the first law of thermodynamics, is a fundamental concept essential for analyzing systems like the one in our example involving water and ice. The principle states that the total energy of an isolated system remains constant over time. This law is based on the idea that energy cannot be created or destroyed; it can only be transformed from one form to another.

When dealing with thermodynamics problems, like the cooling of water by adding ice, it is important to account for all forms of energy entering and exiting the system to predict the system's final state accurately. In this exercise, we consider the internal energy stored in the water and the ice and the energy associated with the phase change of ice as it melts. The primary goal is to find an equilibrium temperature where the energy lost by the water is gained by the ice until it melts completely and both reach the same temperature.

To facilitate problem-solving, it is helpful to use energy balance equations which equate the sum of energies before and after an event, ensuring no energy is unaccounted for.

Equilibrium Temperature Calculation

The equilibrium temperature calculation is integral to solving thermodynamics problems where two substances of different temperatures are mixed, or in our case, where ice is added to water. The objective is to determine the final common temperature at which no net heat exchange occurs between the components.

To find this temperature, one must set up an energy balance equation that equates the total heat lost or gained by each substance. In the given exercise, for example, the heat lost by the water as it cools must equal the heat gained by the ice as it warms up, possibly changes phase and then warms further to the equilibrium temperature.

To improve students' understanding of this concept, emphasizing the role of specific heat capacities, the latent heat of fusion, and the mass of substances involved is vital. Explaining how these factors influence the heat exchange can help students visualize and calculate the final equilibrium temperature more effectively.

Entropy Change

Entropy change is a measure of the disorder or randomness in a system and is a central idea in the second law of thermodynamics. This principle indicates that for any spontaneous process, the entropy of an isolated system will either increase or remain unchanged; it helps us understand the directionality of natural processes.

In the context of our water and ice problem, entropy change is involved during the melting of ice and warming or cooling of water to reach equilibrium. The change in entropy for the water and the ice can be calculated individually using their respective masses, specific heat capacities, the final equilibrium temperature, and, for ice, the heat of fusion as well.

An increased clarity can be achieved by explaining that entropy should always be calculated in reference to a specific temperature - often in Kelvin for absolute values. For students, visualizing entropy as 'energy dispersal' might help comprehend why melting ice, which increases randomness, would lead to an increase in the system's overall entropy.

Exergy Destruction

Exergy destruction is associated with the concept of irreversibility in a system. It quantifies the useful energy that is lost due to the inefficiency of a process. According to the second law of thermodynamics, every real process results in some exergy being destroyed, which represents the lost opportunity to do work.

In our exercise, where water is being cooled by the addition of ice, exergy destruction occurs due to the equilibrium process of reaching the final temperature. It is a form of energy degradation, primarily through the heat exchange and melting of ice, which cannot be used to perform work. The exergy destruction is calculated using the change in total entropy and the reference temperature, demonstrating the loss of potential work as the system moves to a more probable, less ordered state.

Emphasizing that exergy destruction is not actual physical destruction but a measure of energy becoming less available to do work, could help students understand the limitations imposed by natural laws on energy transformations and the importance of designing efficient systems to minimize exergy loss.