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An electric water heater having a \(200-L\) capacity heats water from 23 to \(55^{\circ} \mathrm{C}\). Heat transfer from the outside of the water heater is negligible, and the states of the electrical heating element and the tank holding the water do not change significantly. Perform a full exergy accounting, in kJ, of the electricity supplied to the water heater. Model the water as incompressible with a specific heat \(c=4.18 \mathrm{~kJ} / \mathrm{kg}+\mathrm{K}\). Let \(T_{0}=23^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The exergy supplied to the water heater is 26752 kJ.

Step by step solution

01

Calculate the Mass of Water

First, determine the mass of the water. Given the volume of the water heater is 200 liters, convert this to kilograms using the density of water, which is approximately 1 kg per liter. Therefore,\[ m = 200 \text{ L} = 200 \text{ kg} where m is the mass of the water. \]
02

Define Temperature Change

Identify the temperature change. The water heats from 23°C to 55°C, so the temperature change is:\[ \ \ \ ΔT = T_{final} - T_{initial}\ ΔT = 55^{\text{ \∘ }} \text{C} - 23^{\text{∘ }} \text{C} = 32^{\text{∘ }} \text{C} \ where \ ΔT \text{is the temperature change.} \]
03

Calculate the Heat Energy (Q) Required

Use the formula for heat energy required (Q) to raise the temperature of the water, \[ Q = m \times c \times ΔT \ where: \ Q = \text{heat energy required (in kJ)}, \ m = \text{mass of the water (in kg)}, \ c = \text{specific heat capacity of water (4.18 kJ/kg K)}, \ ΔT = \text{temperature change (in K)}. \ Therefore, \ Q = 200 \text{ kg} \times 4.18 \text{ kJ/kg⋅K} \times 32 \text{ K} = 26752 \text{ kJ}. \]
04

Assess the Exergy Supplied

Exergy is the usable portion of the energy supplied. Since all the supplied electrical energy is converted to heat with negligible losses, the exergy is approximately equal to the heat energy provided.Thus, \[ \text{Exergy supplied} = Q = 26752 \text{ kJ} \]
05

Summary

Summarize the calculation to obtain the final result confirming the exergy supplied is 26752 kJ.

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Key Concepts

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

thermodynamics
Thermodynamics is the science that deals with heat and energy transfer. In this problem, we examine the heating process of water in an electric water heater by considering the principles of thermodynamics. When you heat water, you transfer energy from the electrical heating element to the water molecules. This process increases the water's internal energy, raising its temperature.
The First Law of Thermodynamics is essential here. This law states that energy cannot be created or destroyed, only transferred or converted from one form to another. In our scenario, electrical energy is converted into thermal energy, heating the water.
We also discuss the concept of exergy, which refers to the useful portion of energy. When analyzing the exergy supplied to the water heater, we assess how much of the input energy becomes usable heat energy. Since there are negligible losses, almost all supplied electrical energy is converted into heat energy used to warm the water.
heat energy calculation
Calculating heat energy (Q) is crucial in solving thermodynamic problems involving temperature changes. The formula used is:
\[ Q = m \times c \times \triangle T \]
Here,
  • Q is the heat energy required (in kJ),
  • m is the mass of the substance (in kg),
  • c is the specific heat capacity (in kJ/kg K), and
  • \triangle T is the temperature change (in K).
In this exercise, we first calculate the mass of the water, which turns out to be 200 kg. The temperature change (\triangle T) is 32°C (or 32 K, since ΔT in Celsius is the same as in Kelvin). The specific heat capacity (c) of water is given as 4.18 kJ/kg K.
Using the formula, we compute the heat energy as:
\[ Q = 200 \text{ kg} \times 4.18 \text{ kJ/kgâ‹…K} \times 32 \text{ K} = 26752 \text{ kJ} \]
This shows the total energy needed to raise the water temperature from 23°C to 55°C.
specific heat capacity
Specific heat capacity (c) is a property of a substance that indicates how much heat energy is required to change the temperature of a unit mass by one degree (either Celsius or Kelvin). In our exercise, the specific heat capacity of water is provided as 4.18 kJ/kg K.
This means to raise 1 kg of water by 1°C, 4.18 kJ of energy is needed. Understanding this concept helps explain how much energy is required for heating processes. The higher the specific heat, the more energy is needed to raise the temperature.
In the water heater problem, knowing the specific heat capacity is essential for calculating the total heat energy needed. As we have shown, using the specific heat capacity, the mass, and the temperature change, we find out the exact amount of energy required to heat the water to the desired temperature.
incompressible fluids
In thermodynamics, incompressible fluids are those whose density does not change with pressure variations. Water is often modeled as an incompressible fluid, especially when dealing with moderate temperature and pressure ranges. This assumption simplifies calculations as we don't need to account for volume changes due to pressure.
In the exercise, modeling water as an incompressible fluid allows us to use a constant density (1 kg/L) to find the mass of water from its volume. This simplification is crucial for accurately performing calculations without introducing complex variables.
By assuming water is incompressible, the solution focuses on the direct relationship between heat energy, mass, and temperature change, making the calculations straightforward and accurate.

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Most popular questions from this chapter

Two solid blocks, each having mass \(m\) and specific heat \(c\), and initially at temperatures \(T_{1}\) and \(T_{2}\), respectively, are brought into contact, insulated on their outer surfaces, and allowed to come into thermal equilibrium. (a) Derive an expression for the exergy destruction in terms of \(m, c, T_{1}, T_{2}\), and the temperature of the environment, \(T_{0}\) (b) Demonstrate that the exergy destruction cannot be negative. (c) What is the source of exergy destruction in this case?

Refrigerant \(134 \mathrm{a}\) at \(100 \mathrm{lb} / \mathrm{in} .^{2}, 200^{\circ} \mathrm{F}\) enters a valve operating at steady state and undergoes a throttling process. (a) Determine the exit temperature, in \({ }^{\circ} F\), and the exergy destruction rate, in Btu per lb of Refrigerant 134a flowing, for an exit pressure of \(50 \mathrm{lbf} / \mathrm{in}^{2}\). \(^{2}\) (b) Plot the exit temperature, in \({ }^{\circ} \mathrm{F}\), and the exergy destruction rate, in Btu per lb of Refrigerant 134a flowing, each versus exit pressure ranging from 50 to \(100 \mathrm{lbf} /\) in. \({ }^{2}\) Let \(T_{0}=70^{\circ} \mathrm{F}, p_{0}=14.7 \mathrm{lbf} / \mathrm{in}^{2}{ }^{2}\)

Argon enters a nozzle operating at steady state at \(1300 \mathrm{~K}\), \(360 \mathrm{kPa}\) with a velocity of \(10 \mathrm{~m} / \mathrm{s}\) and exits the nozzle at \(900 \mathrm{~K}, 130 \mathrm{kPa}\). Stray heat transfer can be ignored. Modeling argon as an ideal gas with \(k=1.67\), determine (a) the velocity at the exit, in \(\mathrm{m} / \mathrm{s}\), and (b) the rate of exergy destruction, in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of argon flowing. Let \(T_{0}=293 \mathrm{~K}, p_{0}=1\) bar.

Water at \(24^{\circ} \mathrm{C}, 1\) bar is drawn from a reservoir \(1.25 \mathrm{~km}\) above a valley and allowed to flow through a hydraulic turbine- generator into a lake on the valley floor. For operation at steady state, determine the maximum theoretical rate at which electricity is generated, in MW, for a mass flow rate of \(110 \mathrm{~kg} / \mathrm{s}\). Let \(T_{0}=24^{\circ} \mathrm{C}, p_{0}=1\) bar and ignore the effects of motion.

Air enters a counterflow heat exchanger operating at steady state at \(27^{\circ} \mathrm{C}, 0.3 \mathrm{MPa}\) and exits at \(12^{\circ} \mathrm{C}\). Refrigerant 134 a enters at \(0.4 \mathrm{MPa}\), a quality of \(0.3\), and a mass flow rate of \(35 \mathrm{~kg} / \mathrm{h}\). Refrigerant exits at \(10^{\circ} \mathrm{C}\). Stray heat transfer is negligible and there is no significant change in pressure for either stream. (a) For the Refrigerant 134a stream, determine the rate of heat transfer, in \(\mathrm{kJ} / \mathrm{h}\). (b) For each of the streams, evaluate the change in flow exergy rate, in \(\mathrm{kJ} / \mathrm{h}\), and interpret its value and sign. Let \(T_{0}=22^{\circ} \mathrm{C}, p_{0}=0.1 \mathrm{MPa}\), and ignore the effects of motion and gravity.

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