91Ó°ÊÓ

When it comes to household refrigerators, the coefficient of performance (COP) is a crucial metric. It's a measure of a refrigerator's efficiency. In simple terms, COP is the ratio of useful heating or cooling provided to the work required. For our refrigerator exercise, the COP is given as 2.4. This means the refrigerator can remove 2.4 units of heat for every unit of electrical energy it consumes. The formula to determine COP is: \(COP = \frac{Q_L}{W}\). Here, \(Q_L\) represents the heat removed (in watts), and \(W\) is the work input (also in watts). Knowing the COP helps us calculate how much work the refrigerator needs to do its job. In our example, we rearranged the formula to solve for \(W\): \(W = \frac{Q_L}{COP} = \frac{200 \text{ W}}{2.4} = 83.33 \text{ W}\). The result tells us the refrigerator requires 83.33 watts of power to remove 200 watts of heat.

Calculating energy consumption is essential in understanding how much power an appliance uses over time. Once we know the power input (83.33 W from our previous calculation), the next step is to determine how much energy the refrigerator consumes. We use the formula: \(\text{Energy} = W \times t\). Here, \(W\) is the power input, and \(t\) is the operating time in hours. For our scenario, the refrigerator operates for 360 hours in a month. So, \(\text{Energy} = 83.33 \text{ W} \times 360 \text{ h} \approx 30000 \text{ Wh}\). Since energy is often billed in kilowatt-hours (kWh), we convert watt-hours to kilowatt-hours: \(30000 \text{ Wh} = 30 \text{ kWh}\). This tells us the refrigerator uses 30 kWh in a month.

Evaluating electricity costs involves understanding how much energy our appliance uses and the cost per unit of energy. Electricity is commonly billed in kilowatt-hours (kWh). In our case, the energy consumption is 30 kWh per month. The cost of electricity is given as \(0.08 per kWh. To find the total monthly cost of running the refrigerator, we multiply the energy consumed by the cost per kWh: \(\text{Cost} = 30 \text{ kWh} \times 0.08 \$ / \text{kWh} = 2.4 \$\). Thus, the monthly cost of operating the refrigerator is \)2.40. This simple multiplication helps you understand how efficiency and power use affect your electricity bills.