A particular frost-free refrigerator uses about 705 kWh of electrical energy per year. Express this amount of energy in each unit. d. Cal

Verified step by step guidance

1. Identify the conversion factor between kilowatt-hours (kWh) and calories (Cal). Note that 1 kWh = 3.6 x 10^6 J and 1 Cal = 4184 J.
2. Convert the energy from kilowatt-hours to joules using the conversion factor: Energy (J) = Energy (kWh) x 3.6 x 10^6 J/kWh.
3. Use the conversion factor between joules and calories to convert the energy from joules to calories: Energy (Cal) = Energy (J) / 4184 J/Cal.
4. Perform the calculation to find the energy in calories.
5. Ensure the units are consistent and the conversion is correct by checking each step.

Key Concepts

Energy Units

Energy can be measured in various units, including kilowatt-hours (kWh) and calories (Cal). A kilowatt-hour is the amount of energy consumed when a device uses one kilowatt of power for one hour. Understanding how to convert between these units is essential for solving energy-related problems.

Conversion Factors

To convert energy from one unit to another, specific conversion factors are used. For instance, 1 kWh is equivalent to approximately 860 Cal. Knowing these conversion factors allows for accurate calculations when expressing energy in different units.

Thermodynamics

Thermodynamics is the branch of physics that deals with heat and energy transfer. In the context of refrigerators, it is important to understand how energy is consumed and transformed, as this relates to the efficiency and operation of the appliance. This knowledge aids in comprehending the significance of energy consumption figures.