Question

A two-phase liquid-vapor mixture of \(\mathrm{H}_{2} \mathrm{O}\), initially at 1.0 MPa with a quality of \(90 \%\), is contained in a rigid, wellinsulated tank. The mass of \(\mathrm{H}_{2} \mathrm{O}\) is \(2 \mathrm{~kg}\). An electric resistance heater in the tank transfers energy to the water at a constant rate of \(60 \mathrm{~W}\) for \(1.95 \mathrm{~h}\). Determine the final temperature of the water in the tank, in \({ }^{\circ} \mathrm{C}\).

Short answer

After heating, the final temperature of the water is calculated using steam tables and energy balance. This temperature must be checked with steam tables, considering constant volume and quality transitions.

Step-by-step solution

Identify given data and initial conditions

The initial pressure of the \(\text{H}_2\text{O}\) is 1.0 MPa, the initial quality (x) is 90%, and the mass is 2 kg. The energy rate of the heater is 60 W, and it operates for 1.95 h.

Calculate initial specific volumes

Use steam tables to find the specific volumes for saturated liquid (\(v_f\)) and saturated vapor (\(v_g\)) at 1.0 MPa. Then, calculate the initial specific volume using \( v = x \cdot v_g + (1 - x) \cdot v_f \).

Determine the total energy added to the system

The energy added \(Q\) by the heater is given by \( Q = \text{power} \times \text{time} \), where power is 60 W and time is 1.95 hours (convert hours to seconds). Thus, \( Q = 60 \text{W} \times 1.95 \times 3600 \ \text{seconds} \).

Find the initial internal energy of the mixture

Using the steam tables, find the specific internal energies for saturated liquid (\(u_f\)) and saturated vapor (\(u_g\)) at 1.0 MPa. The initial specific internal energy is \( u_i = x \cdot u_g + (1 - x) \cdot u_f \). Then, calculate the total initial internal energy \( U_i = m \cdot u_i \).

Calculate the total internal energy after heating

The final internal energy \( U_f \) after heating is \( U_f = U_i + Q \).

Determine the final temperature of the water

Assuming the process is at constant volume, use steam tables to find the final temperature corresponding to the final internal energy and specific volume. If needed, interpolate between table values.

Key concepts

thermodynamics

Thermodynamics is a field of physics that explores energy transformations and the properties of matter. The exercise demonstrates a process involving a two-phase liquid-vapor mixture of water. It's governed by the first law of thermodynamics, which suggests that the energy added to the system equals the change in its internal energy. This principle is used when considering how energy provided by an electric heater affects the internal energy and temperature of water in a tank. The system's isolation, ensured by the tank's insulation, simplifies the process as there is no heat transfer with the surroundings.

internal energy

Internal energy is the energy possessed by a system due to the kinetic and potential energies of its molecules. For a two-phase liquid-vapor mixture, it’s crucial to use steam tables to find specific internal energies for both phases. In this exercise, you calculate the mixture’s initial internal energy using quality (x), which represents the vapor mass fraction. The formula is:
\[ u_i = x \times u_g + (1 - x) \times u_f \]
Here, \( u_f \) and \( u_g \) are specific internal energies of the saturated liquid and vapor, respectively. The initial and final internal energies help determine how much energy the system has after heating, essential for finding the final temperature.

specific volume

Specific volume is the volume per unit mass of a substance. For a two-phase mixture, it's calculated by:
\[ v = x \times v_g + (1 - x) \times v_f \]
here, \( v_f \) and \( v_g \) are specific volumes of the saturated liquid and vapor. The initial conditions give us the specific volume, which remains constant in this rigid tank scenario. During heating, the mixture’s temperature needs to be found at constant volume. Steam tables provide data needed to match the final internal energy level at the given specific volume.

electric resistance heating

Electric resistance heating involves converting electrical energy into heat. The heater in the exercise supplies energy to the water. We calculate the energy provided by multiplying the heater's power by time:
\[ Q = \text{power} \times \text{time} \]
Where power is given in watts (W) and time should be converted into seconds. Specifically, for a power of 60 W over 1.95 hours, we get:
\[ Q = 60 \text{W} \times 1.95 \times 3600 \text{seconds} \]
This energy directly increases the system's internal energy since the tank is well-insulated and energy does not escape.

steam tables

Steam tables are invaluable tools in thermodynamics for determining properties of water and steam under various conditions. They provide specific volumes, internal energies, enthalpies, and entropies for different temperatures and pressures. In this exercise, steam tables are used to find:

• Specific volumes of saturated liquid (\(v_f\)) and vapor (\(v_g\))
• Specific internal energies (\(u_f\)) and (\(u_g\))
• Temperature corresponding to final internal energy and specific volume

By interpolating between table values, precise data relevant to a mixture’s current state or projected state can be determined, aiding in accurate thermodynamic analysis.