A frost-free refrigerator uses about 640 kWH of electrical energy per year.
Express this amount of energy in
J:
kJ:
Cal:
http://www.unitconversion.org/power/kilowatts-to-joules-per-hour-conversion.html
To convert the energy from kilowatt-hours (kWh) to joules (J), we need to multiply the energy in kWh by the conversion factor of 3,600,000.
1 kWh = 3,600,000 J
Therefore, to convert 640 kWh to joules:
640 kWh * 3,600,000 J/kWh = 2,304,000,000 J
So, 640 kWh is equal to 2,304,000,000 J.
To convert the energy from kilowatt-hours (kWh) to kilojoules (kJ), we simply multiply the energy in kWh by 1,000.
1 kWh = 1,000 kJ
Therefore, to convert 640 kWh to kilojoules:
640 kWh * 1,000 kJ/kWh = 640,000 kJ
So, 640 kWh is equal to 640,000 kJ.
To convert the energy from kilowatt-hours (kWh) to calories (Cal), we need to multiply the energy in kWh by the conversion factor of 859.8.
1 kWh = 859.8 Cal
Therefore, to convert 640 kWh to calories:
640 kWh * 859.8 Cal/kWh = 550,272 Cal
So, 640 kWh is equal to 550,272 Cal.