Published on *EGEE 102: Energy Conservation and Environmental Protection* (https://www.e-education.psu.edu/egee102)

What is the energy consumption of a refrigerator with a wattage rating of 700 Watts when it is operated for 24 hours a day?

To solve, use the following formula:

$$\text{EnergyConsumption=PowerConsumption}\times \text{NumberofHoursOperated}$$Where:

Energy Consumption= Watt/Hours (Wh) or KiloWatt/Hours (kWh)

Power Consumption= Watts (W) or kW (KiloWatts)Number of Hours Operated= Hours (h)For the example above:

Energy Consumption = 700 W x 24 h

Energy Consumption = 16800 W/h

To convert from W/h to kWh/h, remember that 1kWh/h = 1000 W/h

To solve, set up as a ratio and use linear algebra to solve for ?.

$$\begin{array}{l}\begin{array}{ccc}\frac{1\text{}kWh}{1000\text{}Wh}& =& \frac{?kWh}{16800\text{}Wh}\\ \frac{16,800\text{}Wh\left(1\text{}kWh\right)}{1000\text{}Wh}& =& ?KWh\\ 16.8\text{}KWh& =& ?KWh\end{array}\\ \end{array}$$Use the following link to generate a random practice problem [1].

**Links**

[1] https://courseware.e-education.psu.edu/courses/egee102/L01Instruction/activity/check012701.html