Question

A heat pump is used to maintain a house at \(25^{\circ} \mathrm{C}\) by extracting heat from the outside air on a day when the outside air temperature is \(4^{\circ} \mathrm{C}\). The house is estimated to lose heat at a rate of \(110,000 \mathrm{kJ} / \mathrm{h},\) and the heat pump consumes \(4.75 \mathrm{kW}\) of electric power when running. Is this heat pump powerful enough to do the job?

Short answer

Answer: No, the 4.75 kW heat pump is not powerful enough to maintain the house at \(25^{\circ} \mathrm{C}\) temperature, as it provides only 17100 kJ/h of heat energy, which is lesser than the rate of heat energy loss of the house (110000 kJ/h).

Step-by-step solution

Calculate the heat energy provided by the heat pump

To calculate the rate of heat energy provided by the heat pump, let's first convert the electrical power consumed (4.75 kW) into kilojoules per hour (kJ/h). We know that 1 kW = 1000 J/s, and there are 3600 seconds in an hour. Therefore, Power = 4.75 kW \(\times\) 1000 J/s \(\times\) 3600 s/h = 17100 kJ/h.

Compare the heat energy provided with the heat energy loss of the house

We now have the rate of heat energy provided by the heat pump, 17100 kJ/h. The house is estimated to lose heat at a rate of 110000 kJ/h. Therefore, we need to compare these two values to determine if the heat pump is powerful enough.

Determine the sufficiency of the heat pump

Since the rate of heat energy provided by the heat pump (17100 kJ/h) is lesser than the rate of heat energy loss of the house (110000 kJ/h), one can conclude that this heat pump is not powerful enough to maintain the house at \(25^{\circ} \mathrm{C}\) temperature.

Key concepts

Thermodynamic Cycles

Understanding thermodynamic cycles is crucial when analyzing the efficiency of a heat pump. A heat pump operates on the basic principle of a thermodynamic cycle, where it moves heat from a cooler place to a warmer place. This seemingly simple process involves complex interactions described by laws of thermodynamics.

The efficiency of this cycle depends on the temperature difference between the heat source and the heat sink. In other words, it's easier for the heat pump to maintain the inside temperature when the outside temperature isn't too low. However, as the outside temperature drops, more work - and hence, more electrical energy - is required to transfer the same amount of heat. This correlation is essential to determine if the device can perform adequately under specified conditions. In this case, we're considering a house that needs to be kept at a cozy 25°C - a task for which the existing heat pump might be insufficient.

Heat Transfer

Heat transfer is another fundamental concept involved in heat pump operation. It describes how thermal energy moves from one place to another. There are three main modes of heat transfer: conduction, convection, and radiation. In the scenario of heat pumps, convection plays a major role as it moves heat through a fluid - such as a refrigerant - that absorbs heat at the evaporator coil and releases it at the condenser coil.

For a heat pump to be powerful enough, it must transfer heat at a rate that compensates for the heat lost from the house. The heat loss rate of 110,000 kJ/h refers to the amount of energy per unit time that the house loses to its colder surroundings. So, the pump's heat transfer rate must be at least equal to this loss for the desired temperature to be maintained.

Energy Conservation in Systems

In the context of heat pumps, energy conservation relates to how effectively the system can convert one form of energy into another without excessive loss. The Law of Conservation of Energy states that energy cannot be created or destroyed, only transformed from one form to another.

When assessing a heat pump, the energy conversion efficiency is described by the coefficient of performance (COP). It is the ratio of heat energy delivered to the electrical energy consumed. Essentially, for a heat pump to be considered efficient, it must transfer more energy as heat than the electrical energy it uses. However, with the current heat pump consuming 4.75 kW but only providing 17,100 kJ/h, when compared to a heat loss of 110,000 kJ/h, the system is evidently not conserving energy efficiently enough to maintain the desired indoor temperature.

Electrical to Thermal Energy Conversion

The conversion from electrical to thermal energy is a defining factor in the operation of a heat pump. In this process, the heat pump uses electrical energy to drive a compressor, which in turn increases the temperature and pressure of the refrigerant. This heated refrigerant then releases thermal energy into the house, warming it up.

A heat pump's performance is severely influenced by its ability to convert electricity into a sufficient amount of heat. Given the electrical consumption of 4.75 kW and the provided heat energy, the system's capability can be quantified. By analyzing the ratio of electrical input to thermal output, we see that the current setup does not meet the necessary thermal output to counteract the heat loss of the house, thus the heat pump is inadequate for the given conditions.