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A heat pump that is used to heat a house has a COP of \(2.5 .\) That is, the heat pump delivers \(2.5 \mathrm{kWh}\) of energy to the house for each \(1 \mathrm{kWh}\) of electricity it consumes. Is this a violation of the first law of thermodynamics? Explain.

Short Answer

Expert verified
Answer: No, a heat pump with a COP of 2.5 does not violate the first law of thermodynamics. The heat pump efficiently transfers heat energy from an external source (such as the outside air or ground) to the house using the consumed electrical energy, maintaining the conservation of energy in the process.

Step by step solution

01

Understand COP and the First Law of Thermodynamics

The Coefficient of Performance (COP) of a heat pump measures the efficiency of the system, defined as the ratio of the heat energy output (delivered to the house) to the electrical energy input (consumed by the heat pump). \[ COP =\frac{Energy\; Output}{Energy\; Input} \] The First Law of Thermodynamics states that energy cannot be created or destroyed, only converted from one form to another.
02

Compute the Total Energy Input and Output

The given heat pump has a COP of 2.5, which means that for every 1 kWh of electricity consumed, it delivers 2.5 kWh of energy to the house. So when the heat pump uses 1 kWh of electrical energy, it transfers 2.5 kWh of heat energy to the house.
03

Analyze Energy Conservation

According to the first law of thermodynamics, energy cannot be created or destroyed, but it can be converted. Here, the heat pump is converting electrical energy (input) into heat energy (output). When considering the energy conservation principle in this situation, we must account for the fact that the heat pump is not creating energy but is transferring heat from another source (usually from outside air or ground).
04

Determine If There Is a Violation of the First Law

In the given scenario, we must consider the total energy in the system. The heat pump consumes 1 kWh of electrical energy and delivers 2.5 kWh of heat energy to the house. The additional 1.5 kWh of energy comes from another source (outside air or ground). The total energy input is then 1 kWh (electricity) + 1.5 kWh (from the external source) = 2.5 kWh, equal to the total energy output. Therefore, the conservation of energy is maintained, and the first law of thermodynamics is not violated. In conclusion, the heat pump with a COP of 2.5 is not violating the first law of thermodynamics. The heat pump efficiently transfers heat energy from one source (typically the outside air or ground) to another (inside the house) using the consumed electrical energy.

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Most popular questions from this chapter

A \(2.4-\mathrm{m}\) high \(200-\mathrm{m}^{2}\) house is maintained at \(22^{\circ} \mathrm{C}\) by an air-conditioning system whose COP is \(3.2 .\) It is estimated that the kitchen, bath, and other ventilating fans of the house discharge a houseful of conditioned air once every hour. If the average outdoor temperature is \(32^{\circ} \mathrm{C},\) the density of air is \(1.20 \mathrm{kg} / \mathrm{m}^{3},\) and the unit cost of electricity is \(\$ 0.10 / \mathrm{kWh}\) the amount of money "vented out" by the fans in 10 hours is \((a) \$ 0.50\) \((b) \$ 1.60\) \((c) \$ 5.00\) \((d) \$ 11.00\) \((e) \$ 16.00\)

A heat pump with refrigerant-134a as the working fluid is used to keep a space at \(25^{\circ} \mathrm{C}\) by absorbing heat from geothermal water that enters the evaporator at \(60^{\circ} \mathrm{C}\) at a rate of \(0.065 \mathrm{kg} / \mathrm{s}\) and leaves at \(40^{\circ} \mathrm{C}\). Refrigerant enters the evaporator at \(12^{\circ} \mathrm{C}\) with a quality of 15 percent and leaves at the same pressure as saturated vapor. If the compressor consumes \(1.6 \mathrm{kW}\) of power, determine \((a)\) the mass flow rate of the refrigerant, \((b)\) the rate of heat supply, \((c)\) the \(\mathrm{COP}\), and \((d)\) the minimum power input to the compressor for the same rate of heat supply.

Using EES (or other) software, determine the maximum work that can be extracted from a pond containing \(10^{5} \mathrm{kg}\) of water at \(350 \mathrm{K}\) when the temperature of the surroundings is \(300 \mathrm{K}\). Notice that the temperature of water in the pond will be gradually decreasing as energy is extracted from it; therefore, the efficiency of the engine will be decreasing. Use temperature intervals of \((a) 5 \mathrm{K},(b) 2 \mathrm{K}\) and \((c) 1 \mathrm{K}\) until the pond temperature drops to \(300 \mathrm{K}\). Also solve this problem exactly by integration and compare the results.

A heat pump supplies heat energy to a house at the rate of \(140,000 \mathrm{kJ} / \mathrm{h}\) when the house is maintained at \(25^{\circ} \mathrm{C} .\) Over a period of one month, the heat pump operates for 100 hours to transfer energy from a heat source outside the house to inside the house. Consider a heat pump receiving heat from two different outside energy sources. In one application the heat pump receives heat from the outside air at \(0^{\circ} \mathrm{C} .\) In a second application the heat pump receives heat from a lake having a water temperature of \(10^{\circ} \mathrm{C}\). If electricity costs \(\$ 0.105 / \mathrm{kWh}\), determine the maximum money saved by using the lake water rather than the outside air as the outside energy source.

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