91Ó°ÊÓ

Thermodynamics is the study of energy, heat, and work. It helps us understand how heat pumps work to maintain a certain temperature in a dwelling. The key principles in this problem are the laws of thermodynamics, especially the second law, which explains heat transfer and energy conversion. The heat pump uses energy to move heat from a cooler space (outside) to a warmer space (inside). This process is vital for maintaining comfortable indoor temperatures, especially during chilly weather. By understanding thermodynamics, we can determine how efficiently the heat pump operates and calculate related costs.

Heat transfer refers to the movement of heat from one place to another. In the context of a heat pump, heat is transferred from the outdoors to the indoors. There are different modes of heat transfer, such as conduction, convection, and radiation. In our problem, we focus on the transfer through the walls and roof, which happens mainly through conduction. The rate of heat transfer depends on the temperature difference between indoors and outdoors, and the thermal properties of the building materials. The formula used here is: \[ Q = 2000 \frac{kJ}{h \times ^\text{C}} \times (T_i - T_o) \]where \( T_i \) and \( T_o \) represent the inside and outside temperatures, respectively. This helps compute the total energy transferred, which is crucial for determining the operating cost.

The Coefficient of Performance (COP) is a measure of a heat pump's efficiency. It is defined as the ratio of the amount of heat transported to the work input:\[ COP = \frac{T_i}{(T_i - T_o)} \]In this exercise, we find the COP using the indoor temperature (293 K) and the outdoor temperature (278 K). A higher COP means the heat pump is more efficient, requiring less energy to maintain the desired indoor temperature. This efficiency plays a critical role in minimizing the cost of operating the heat pump. By calculating the COP, we gain insights into how much power is needed and, subsequently, the associated costs.

Energy efficiency is about using less energy to perform the same task. In the case of a heat pump, it means using the least amount of electrical power to transfer heat from a colder place to a warmer one. By optimizing energy efficiency, we reduce operating costs and environmental impact. To calculate the energy efficiency of our heat pump, we use the previously derived COP and determine the required power input as follows:\[ P = \frac{Q}{COP} \]where \( Q \) is the heat transfer rate and \( COP \) is the Coefficient of Performance. Better energy efficiency translates to lower energy consumption and, hence, reduced cost. This exercise's solution shows that by understanding and calculating energy efficiency, we can determine how much it costs to run the heat pump on a daily basis, helping us make informed decisions about household energy use.