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A closed, rigid tank contains \(5 \mathrm{~kg}\) of air initially at \(300 \mathrm{~K}\), 1 bar. As illustrated in Fig. P6.71, the tank is in contact with a thermal reservoir at \(600 \mathrm{~K}\) and heat transfer occurs at the boundary where the temperature is \(600 \mathrm{~K}\). A stirring rod transfers \(600 \mathrm{~kJ}\) of energy to the air. The final temperature is \(600 \mathrm{~K}\). The air can be modeled as an ideal gas with \(c_{v}=\) \(0.733 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\) and kinetic and potential energy effects are negligible. Determine the amount of entropy transferred into the air and the amount of entropy produced, each in \(\mathrm{kJ} / \mathrm{K}\).

Step by step solution

01

- Identify the given values

List the given values: \(m = 5 \text{ kg}\)\(T_1 = 300 \text{ K}\)\(P_1 = 1 \text{ bar}\)\(T_{\text{reservoir}} = 600 \text{ K}\)\(Q = 600 \text{ kJ}\)\(T_2 = 600 \text{ K}\)\(c_v = 0.733 \text{ kJ}/(\text{kg} \cdot \text{K})\)

02

- Calculate the initial internal energy

Using the formula for internal energy for an ideal gas for the initial state, \(U_1 = m \cdot c_v \cdot T_1\)\[ U_1 = 5 \text{ kg} \cdot 0.733 \text{ kJ}/(\text{kg} \cdot \text{K}) \cdot 300 \text{ K} = 1099.5 \text{ kJ} \]

03

- Calculate the final internal energy

Using the formula for internal energy for an ideal gas for the final state, \(U_2 = m \cdot c_v \cdot T_2\)\[ U_2 = 5 \text{ kg} \cdot 0.733 \text{ kJ}/(\text{kg} \cdot \text{K}) \cdot 600 \text{ K} = 2199 \text{ kJ} \]

04

- Apply the first law of thermodynamics

Using the first law to relate heat transfer, energy transferred by stirring, and the change in internal energy,\(Q + W = \Delta U\)\[ 600 \text{ kJ} + W = U_2 - U_1\]\[ 600 \text{ kJ} + W = 2199 \text{ kJ} - 1099.5 \text{ kJ}\]\[ W = 1099.5 \text{ kJ} - 600 \text{ kJ} = 499.5 \text{ kJ}\]

05

- Calculate entropy transferred into the air

The entropy transferred due to the reservoir temperature is given by,\(\Delta S_{\text{transferred}} = \frac{Q}{T_{\text{reservoir}}}\)\[ \Delta S_{\text{transferred}} = \frac{600 \text{ kJ}}{600 \text{ K}} = 1 \text{ kJ}/\text{K}\]

06

- Calculate the change in entropy of the air

Using the formula for change in entropy for an ideal gas,\(\Delta S_{\text{air}} = m \cdot c_v \cdot \ln \left( \frac{T_2}{T_1} \right)\)\[ \Delta S_{\text{air}} = 5 \text{ kg} \cdot 0.733 \text{ kJ}/(\text{kg} \cdot \text{K}) \cdot \ln \frac{600 \text{ K}}{300 \text{ K}}\]\[ \Delta S_{\text{air}} = 5 \text{ kg} \cdot 0.733 \text{ kJ}/(\text{kg} \cdot \text{K}) \cdot \ln 2\]\[ \Delta S_{\text{air}} = 5 \cdot 0.733 \cdot 0.693 = 2.54 \text{ kJ}/\text{K}\]

07

- Calculate the entropy produced

Entropy production is the total entropy change minus the entropy transferred,\(\sigma = \Delta S_{\text{air}} - \Delta S_{\text{transferred}}\)\[ \sigma = 2.54 \text{ kJ}/\text{K} - 1 \text{ kJ}/\text{K} = 1.54 \text{ kJ}/\text{K}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Law of Thermodynamics

The First Law of Thermodynamics is a fundamental principle in thermodynamics. It states that energy cannot be created or destroyed, only transformed. This law can be expressed mathematically as \(Q + W = \Delta U\), where \(Q\) is the heat added to the system, \(W\) is the work done on the system, and \(\Delta U\) is the change in internal energy. In the context of the exercise, heat transfer (\(Q = 600 \text{ kJ}\)) and work done by stirring (\(W\)) contribute to the change in internal energy of the air inside the tank. The First Law helps determine how energy inputs alter the system's state.

Ideal Gas Law

The Ideal Gas Law characterizes the behavior of an idealized gas using the equation \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is temperature. In this exercise, since the air is modeled as an ideal gas, we use its specific heat at constant volume (\(c_v\)) to calculate changes in properties like internal energy and entropy. This law simplifies the calculations by providing straightforward relationships between pressure, volume, and temperature for an ideal gas.

Entropy Change

Entropy is a measure of disorder or randomness in a system. The change in entropy (\(\Delta S\)) is significant in understanding how energy distribution within the system changes. For an ideal gas, the change in entropy can be calculated using \(\Delta S = m \cdot c_v \cdot \ln \( \frac{T_2}{T_1}\)\). In the given exercise, this equation helps calculate the change in the air's entropy as it heats up from \(300 \text{ K}\) to \(600 \text{ K}\). The positive entropy change reflects the increased disorder as the air's temperature rises.

Internal Energy

Internal energy (\(U\)) refers to the total energy contained within a system due to both kinetic and potential energy of its molecules. For an ideal gas, the internal energy depends solely on temperature and can be calculated using \(\text{U} = m \cdot c_v \cdot T\). In the provided exercise, the initial and final internal energies are calculated to determine the energy change caused by heating and stirring. With \(c_v = 0.733 \text{ kJ}/(\text{kg} \cdot \text{K})\), the air’s internal energy increases significantly as it reaches the final temperature of \(600 \text{ K}\).

Entropy Production

Entropy production (\(\sigma\)) quantifies the irreversibility of a process within a thermodynamic system. It is calculated as the total entropy change minus the entropy transferred to or from the surroundings. For the given exercise, entropy production is determined by the formula \(\sigma = \Delta S_{\text{air}} - \Delta S_{\text{transferred}}\). This value indicates that the actual process of heating and stirring incurs an inherent generation of entropy, highlighting the natural tendency towards increased disorder in real thermodynamic processes.