Question

A well-insulated \(3-m \times 4-m \times 6-m\) room initially at \(7^{\circ} \mathrm{C}\) is heated by the radiator of a steam heating system. The radiator has a volume of \(15 \mathrm{L}\) and is filled with super-heated vapor at \(200 \mathrm{kPa}\) and \(200^{\circ} \mathrm{C}\). At this moment both the inlet and the exit valves to the radiator are closed. A 120 -W fan is used to distribute the air in the room. The pressure of the steam is observed to drop to \(100 \mathrm{kPa}\) after \(45 \mathrm{min}\) as a result of heat transfer to the room. Assuming constant specific heats for air at room temperature, determine the average temperature of air in 45 min. Assume the air pressure in the room remains constant at \(100 \mathrm{kPa}\).

Short answer

Answer: The average temperature of air in the room after 45 minutes is approximately 7.04°C.

Step-by-step solution

Calculate the mass of air in the room

First, we need to find the volume of the room and the mass of the air in the room. The volume of the room, V_room, can be found using the given dimensions: V_room = 3m × 4m × 6m = 72 m³ Now, we need to find the mass of the air. We know that the air pressure in the room is 100 kPa and the initial temperature is 7°C. We'll use the ideal gas law to find the mass of air: PV = mRT where P is the pressure, V is the volume, m is the mass, R is the specific gas constant for air, and T is the temperature. The specific gas constant for air, R_air, is 0.287 kJ/(kg*K). We can rearrange the formula to find the mass of air: m_air = PV / (R_air * T_air_initial) Don't forget to convert the initial temperature of air to Kelvin: T_air_initial = 7 + 273.15 = 280.15 K Now we can plug in the values: m_air = (100 kPa * 72 m³) / (0.287 kJ/(kg*K) * 280.15 K) ≈ 891.2 kg

Find the energy transferred from the radiator to the room

We are given the initial pressure and temperature of the steam as well as the volume of the radiator. First, we need to find the mass of steam in the radiator. We can use the specific volume of steam at the given initial pressure and temperature as follows: V_radiator = 15 L = 0.015 m³ v_steam_initial = 0.39256 m³/kg (from steam tables at 200 kPa and 200°C) m_steam = V_radiator / v_steam_initial ≈ 0.0382 kg Now we can find the specific enthalpies of steam at the initial and final pressures: h_initial = 2852.8 kJ/kg (at 200 kPa, 200°C from steam tables) h_final = 2686.9 kJ/kg (at 100 kPa from steam tables) The energy transferred from the radiator to the room, Q_room, can be calculated as: Q_room = m_steam * (h_initial - h_final) ≈ 6.344 kJ

Calculate the final temperature of the air

Now we can calculate the energy gained by the air in the room using the specific heat of air at constant pressure, c_p_air, which is 1.005 kJ/(kg*K): ΔQ_air = m_air * c_p_air * (T_air_final - T_air_initial) Since the energy transferred to the room is equal to the energy gained by the air, we can write: Q_room = ΔQ_air We can rearrange the formula to find the final temperature of the air: T_air_final = (Q_room / (m_air * c_p_air)) + T_air_initial ≈ 280.22 K Now we can convert the final temperature back to Celsius: T_air_final = 280.22 K - 273.15 = 7.07 °C Finally, the average temperature of the air in the room during the 45 minutes can be calculated as the average of the initial and final temperatures: T_air_average = (T_air_initial + T_air_final) / 2 ≈ 7.04 °C So, the average temperature of air in the room after 45 minutes is approximately 7.04°C.

Key concepts

Understanding the Ideal Gas Law

The ideal gas law is a crucial equation in thermodynamics that connects the pressure, volume, temperature, and number of moles of an ideal gas. It is represented by the equation
\(PV = nRT\)
where \(P\) stands for pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin. In practice, engineers often use a form of the ideal gas law that includes the mass (\(m\)) of the gas and the specific gas constant (\(R_{specific}\)), leading to:
\(PV = mR_{specific}T\).
The ideal gas law assumes that the gas particles do not attract or repel each other and that they occupy no volume. Although no real gas behaves ideally, this assumption is reasonably accurate for gases under normal conditions.
During the process of solving problems like heating a room with a radiator, as in the exercise, we use this law to calculate the mass of the gas in the room, assuming air behaves like an ideal gas. This initial step is fundamental to solve subsequent parts of the problem, such as calculating heat transfer and temperature changes.
It is essential to remember that temperature must always be in Kelvin for the ideal gas law to be valid, as it is an absolute scale. Any slight misunderstanding of this concept can result in incorrect solutions.

Specific Heat Capacity Explained

Specific heat capacity, often referred to simply as specific heat, is a substance's capacity to absorb heat for a given increase in temperature. It is defined as the amount of heat needed to raise the temperature of one kilogram of the substance by one degree Celsius (or Kelvin, as the size of the degree is the same on both scales). The formula to calculate the heat absorbed or released is
\(Q = mc\Delta T\),
where \(Q\) is the quantity of heat, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature.
In the context of the given problem, the specific heat capacity of air is used to determine how much energy the air in the room will absorb as its temperature increases. It's a key value in calculating the final temperature of the room after a specified amount of heat is added or removed. Knowledge of this concept is essential for designing heating and cooling systems and for understanding heat transfer in various engineering applications.

Phase Change Enthalpy and its Role

Phase change enthalpy refers to the heat energy required to change the phase of a substance without changing its temperature. Each phase change, be it melting, vaporization, or condensation, has an associated enthalpy value known as latent heat.

Enthalpy of Vaporization and Condensation

For example, when you're dealing with steam in a heating system, as in our problem, you must consider the enthalpy of vaporization if the steam is turning into water, or the enthalpy of condensation if the process is the reverse.
A critical application of phase change enthalpy is in calculating the energy exchange during these phase transitions. In the given problem, the enthalpy values for steam at different pressures are taken from steam tables, which are essential resources in thermodynamics. These values allow us to calculate the heat transferred to the room due to the condensation of steam within the radiator, as steam loses energy and transfers it to the surrounding air.
Understanding the phase change enthalpy is vital for engineers engaged in designing systems that rely on phase changes, such as heating systems, refrigeration, and power generation. It's a central concept in explaining why phase changes are such effective methods of energy transfer in various engineering applications.