Question

One ton of liquid water at \(80^{\circ} \mathrm{C}\) is brought into a well- insulated and well-sealed \(4-\mathrm{m} \times 5-\mathrm{m} \times 7-\mathrm{m}\) room initially at \(22^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\). Assuming constant specific heats for both air and water at room temperature, determine (a) the final equilibrium temperature in the room and \((b)\) the total entropy change during this process, in \(\mathrm{kJ} / \mathrm{K}\).

Short answer

Answer: The final equilibrium temperature in the room is 62.38°C, and the total entropy change during this process is -9.43 kJ/K.

Step-by-step solution

Identify the given values

Here are the given values from the exercise: Water mass: \(m_{w} = 1 \thinspace ton = 1000 \thinspace kg\) (converted from tons to kg) Initial water temperature: \(T_{w,1} = 80^{\circ} \mathrm{C}\) Room dimensions: \(4 \times 5 \times 7 \thinspace m^3\) Initial room temperature: \(T_{1} = 22^{\circ} \mathrm{C}\) Initial room pressure: \(P_{1} = 100 \thinspace kPa\)

Calculate the mass of air in the room

First, we need to find the mass of the air in the room. To do this, we can use the ideal gas law formula: \(PV = mRT\), where P is the pressure in pascal (Pa), V is the volume in cubic meters (m^3), m is the mass of air in kilograms (kg), R is the specific gas constant for air, and T is the temperature in Kelvin (K). First, convert the room temperature to Kelvin: \(T_{1} = 22^{\circ} \mathrm{C} + 273.15 = 295.15 \thinspace K\) Next, convert the pressure to Pa: \(P_{1} = 100 \thinspace kPa \times 1000 = 100,000 \thinspace Pa\) Now, find the volume of the room: \(V_{room} = 4 \cdot 5 \cdot 7 = 140 \thinspace m^3\) Using the specific gas constant for air, which is: \(R_{air} = 0.287 \thinspace kJ/kg \cdot K = 287 \thinspace J/kg \cdot K\) (Notice that we convert it from \(kJ/kg\cdot K\) to \(J/kg\cdot K\)) Now, we can solve for the mass of the air (m): \(m_{air} = \frac{PV_{room}}{R_{air}T_{1}} = \frac{100,000 \cdot 140}{287 \cdot 295.15} = 166.71 \thinspace kg\)

Calculate the final equilibrium temperature

We can determine the final equilibrium temperature using the conservation of energy principle. The heat lost by water (Q_water) equals the heat gained by air (Q_air). \(m_{w}C_{pw}(T_f - T_{w,1}) = m_{air}C_{p,air}(T_f - T_{1})\) The specific heat of water and air at room temperature is approximately: \(C_{p,w} = 4.18 \thinspace kJ/kg \cdot K\) \(C_{p,air} = 1.006 \thinspace kJ/kg \cdot K\) Now, solve for the final equilibrium temperature (T_f): \(T_f = \frac{m_{w}C_{pw}T_{w,1} + m_{air}C_{p,air}T_{1}}{m_{w}C_{pw} + m_{air}C_{p,air}} = \frac{1000 \cdot 4.18 \cdot 80 + 166.71 \cdot 1.006 \cdot 22}{1000 \cdot 4.18 + 166.71 \cdot 1.006} = 62.38^{\circ} \mathrm{C}\)

Calculate the total entropy change

To find the total entropy change, compute the entropy change separately for both water and air, then sum them up: \(\Delta S_{total} = \Delta S_{w} + \Delta S_{air}\) The entropy change formulas for water and air are: \(\Delta S_{w} = m_{w}C_{pw} \ln (\frac{T_{f}}{T_{w,1}})\) \(\Delta S_{air} = m_{air}C_{p,air} \ln (\frac{T_{f}}{T_{1}})\) Convert the final equilibrium temperature to Kelvin: \(T_f = 62.38^{\circ} \mathrm{C} + 273.15 = 335.53 \thinspace K\) Calculate the entropy change for water and air: \(\Delta S_{w} = 1000 \cdot 4.18 \cdot 10^3 \cdot \ln (\frac{335.53}{353.15}) = -21494.94 \thinspace J/K\) \(\Delta S_{air} = 166.71 \cdot 1.006 \cdot 10^3 \cdot \ln (\frac{335.53}{295.15}) = 12064.87 \thinspace J/K\) Now, calculate the total entropy change: \(\Delta S_{total} = \Delta S_{w} + \Delta S_{air} = -21494.94 + 12064.87 = -9430.07 \thinspace J/K = -9.43 \thinspace kJ/K\) So, the final equilibrium temperature in the room is \(62.38^{\circ} \mathrm{C}\), and the total entropy change during this process is \(-9.43 \thinspace kJ/K\).

Key concepts

Entropy Change Calculation

Entropy—a concept often introduced in thermodynamics as a measure of disorder in a system—plays a vital role in understanding how energy transformations impact the disorder within a closed system.

The second law of thermodynamics states that in an isolated system, the entropy will either increase or remain constant. During an energy exchange, such as the one between air and water in our textbook problem, the entropy of each substance changes, and calculating this change is pivotal for understanding the whole process.

For a given substance, the entropy change \( \Delta S \) can be calculated using the formula:
\[ \Delta S = m \cdot C_p \cdot \ln\left(\frac{T_f}{T_i}\right) \]
where

• \( m \) is the mass,
• \( C_p \) is the specific heat at constant pressure,
• \( T_f \) is the final temperature in Kelvin, and
• \( T_i \) is the initial temperature in Kelvin.

This calculation involves the natural logarithm (ln) of the ratio of the final and initial temperatures. In the exercise, the entropy change for water and air were calculated separately and then summed to find the total change for the system.

While entropy can sometimes be difficult to conceptualize, analyzing it through such calculations helps students grasp the subtle yet profound implications of thermodynamics in practical scenarios.

Specific Heats

The specific heat capacity, or simply 'specific heat', is a property that determines how much heat energy is needed to raise the temperature of a substance by a given amount. Specific heat is crucial in calculations involving heat transfer because it represents the substance's ability to absorb or release heat.

In our exercise, the specific heats \( C_p \) of both water and air are given constant values at room temperature, which simplifies the calculation of heat transfer. For water, the value is \( C_{p,w} = 4.18 \thinspace kJ/kg \cdot K \) and for air, it is \( C_{p,air} = 1.006 \thinspace kJ/kg \cdot K \). The higher specific heat of water indicates that water can absorb more heat before its temperature increases compared to air.

Applications and Importance

Understanding specific heat is not only fundamental in thermodynamics but also in various fields such as meteorology, cooking, and engineering. For example, in climate science, the specific heat of ocean water affects how temperature changes in different layers of the ocean can drive weather patterns. In designing heating and cooling systems, engineers rely on specific heat values to predict how substances will react to heat flow, ensuring the systems operate efficiently.

Conservation of Energy Principle

One of the most fundamental concepts in physics is the principle of conservation of energy. This principle posits that energy cannot be created or destroyed; it can only be transformed from one form to another or transferred from one object to another.

In thermodynamics, this principle is applied to deduce the final equilibrium temperature of a system, such as in our example where we measure the heat exchange between air and water within a closed room. The formulas for calculating the heat lost by water and the heat gained by air reflect this fundamental principle:

\[ m_w C_{pw} (T_f - T_{w,i}) = m_{air} C_{p,air} (T_f - T_{i}) \]
Here, the amount of heat lost by the water (\( Q_{water} \)) must equal the heat gained by the air (\( Q_{air} \)), because the energy within the system is conserved.

Energy conservation is a cornerstone in various engineering problems, where balancing heat, work, and internal energy is necessary to design efficient systems. From simple tasks like calculating the energy needs for heating a home to complex industrial processes such as power generation, the conservation of energy principle is integral to environmental management, sustainable development, and technological innovation.