Question

A closed, rigid tank having a volume of \(1 \mathrm{~m}^{3}\) contains a mixture of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) and water vapor at \(75^{\circ} \mathrm{C}\). The respective masses are \(12.3 \mathrm{~kg}\) of carbon dioxide and \(0.5 \mathrm{~kg}\) of water vapor. If the tank contents are cooled to \(20^{\circ} \mathrm{C}\), determine the heat transfer, in \(\mathrm{kJ}\), assuming ideal gas behavior.

Short answer

-484.44 kJ

Step-by-step solution

Understand the Parameters

Identify and list the given parameters. The volume of the tank is 1 m³. The masses of carbon dioxide and water vapor are 12.3 kg and 0.5 kg, respectively. The initial and final temperatures are 75°C and 20°C. We are to find the heat transfer in kJ.

Convert Temperatures to Kelvin

Convert the temperatures from Celsius to Kelvin. Initial temperature: \(T_1 = 75^\circ\text{C} + 273.15 = 348.15 \text{ K}\)Final temperature: \(T_2 = 20^\circ\text{C} + 273.15 = 293.15 \text{ K}\)

Use Ideal Gas Law to Find Initial Pressures

Use the ideal gas law to find the initial pressures of CO₂ and water vapor. \(PV = nRT\)For CO₂: \(P_{CO2} V = n_{CO2} R_{CO2} T_1\)Where \(R_{CO2} = 0.1889 \frac{kJ}{kg.K}\) and \(n_{CO2} = 12.3 \text{ kg}\)Calculate pressure: \(P_{CO2} = \frac{n_{CO2} R_{CO2} T_1}{V} = \frac{12.3 \times 0.1889 \times 348.15}{1} = 812.64 \text{ kPa}\)For water vapor: \(P_{H2O} V = n_{H2O} R_{H2O} T_1\)Where \(R_{H2O} = 0.4615 \frac{kJ}{kg.K}\) and \(n_{H2O} = 0.5 \text{ kg}\)Calculate pressure: \(P_{H2O} = \frac{n_{H2O} R_{H2O} T_1}{V} = \frac{0.5 \times 0.4615 \times 348.15}{1} = 80.25 \text{ kPa}\)

Use Ideal Gas Law to Find Final Pressures

Calculate the final pressures using the ideal gas law at 20°C. For CO₂: \(P_{CO2} V = n_{CO2} R_{CO2} T_2\)Calculate pressure: \(P_{CO2} = \frac{n_{CO2} R_{CO2} T_2}{V} = \frac{12.3 \times 0.1889 \times 293.15}{1} = 684.92 \text{ kPa}\)For water vapor: \(P_{H2O} = \frac{n_{H2O} R_{H2O} T_2}{V} = \frac{0.5 \times 0.4615 \times 293.15}{1} = 67.69 \text{ kPa}\)

Calculate Internal Energy Change

Determine the change in internal energy. Use the specific heat capacities at constant volume \(C_v\): \(\Delta U = m_{CO2} C_{v, CO2} \Delta T + m_{H2O} C_{v, H2O} \Delta T\)Where \(C_{v, CO2} = 0.657 \frac{kJ}{kg.K}\) and \(C_{v, H2O} = 1.421 \frac{kJ}{kg.K}\), and \(\Delta T = T_2 - T_1 = 293.15 - 348.15 = -55 \text{ K}\)Calculate the total change in internal energy: \(\Delta U = (12.3 \cdot 0.657 \cdot -55) + (0.5 \cdot 1.421 \cdot -55) = -444.85 - 39.59 = -484.44 \text{ kJ}\)

Determine the Heat Transfer

Since the tank is rigid and closed, the work done is zero, and the heat transfer is equal to the change in internal energy. Therefore, \(Q = \Delta U = -484.44 \text{ kJ}\). The negative sign indicates that heat is released.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in thermodynamics. It relates the pressure (P), volume (V), temperature (T), and amount of gas (n) in a system. The equation is given by: \( PV = nRT \), where R is the specific gas constant. In simpler terms, this law helps us understand how gases behave under different conditions of temperature and pressure.
For instance, in the problem, we used the Ideal Gas Law to calculate the pressures of CO₂ and water vapor at different temperatures. By applying the formula and knowing the initial conditions (volume, temperature, and mass), we could find out how the pressure changed as the gas cooled down.

Heat Transfer

Heat transfer is the process by which thermal energy moves from one object or substance to another due to a temperature difference. There are three primary modes of heat transfer: conduction, convection, and radiation.
In the given problem, we considered heat transfer in a closed, rigid tank cooling down. When the substance within the tank loses heat, its temperature and internal energy decrease. Since the tank is rigid, there is no change in volume, and the process only involves the heat leaving the gas mixture, which is why there's a negative sign in the final heat transfer value.

Internal Energy Change

Internal energy refers to the total energy contained within a system due to the microscopic motion and interactions of molecules. When the temperature of a system changes, its internal energy also changes.
In the problem's solution, we calculated this change in internal energy (\( \Delta U \)) as the tank's temperature dropped from 75°C to 20°C. The formula used is \( \Delta U = m_{CO2} C_{v, CO2} \Delta T + m_{H2O} C_{v, H2O} \Delta T \), which combines the specific heat capacities and the mass of each substance along with the change in temperature.

Specific Heat Capacity

Specific heat capacity is a property of matter that measures the amount of heat required to change the temperature of a unit mass of a substance by one degree Kelvin. It plays a crucial role in calculating the heat transfer and internal energy changes in thermodynamic systems.
In the exercise, we used specific heat capacities at constant volume for CO₂ and water vapor—denoted as \( C_v \). The value for CO₂ is \( 0.657 \frac{kJ}{kg.K} \) and for water vapor is \( 1.421 \frac{kJ}{kg.K} \). These values are essential for determining how much energy is stored or released when the gas mixture cools down.