Question

a particular frost - free refrigerator uses about 775 kwh of electrical energy per year.
part a
express this amount of energy in j.
express your answer in joules to three significant figures.
part b
express this amount of energy in kj.
express your answer in kilojoules to three significant figures.
part c
express this amount of energy in cal.
express your answer in calories to three significant figures.

Explanation:

Step1: Recall energy - unit conversion factors

1 kWh = 1000 W×3600 s = 3.6×10^6 J, 1 kJ = 1000 J, 1 Cal = 4184 J.

Step2: Convert kWh to J for Part A

Given 775 kWh. Multiply by the conversion factor: 775 kWh×3.6×10^6 J/kWh = 2.79×10^9 J.

Step3: Convert J to kJ for Part B

Since 1 kJ = 1000 J, then 2.79×10^9 J÷1000 = 2.79×10^6 kJ.

Step4: Convert J to Cal for Part C

Since 1 Cal = 4184 J, then 2.79×10^9 J÷4184 J/Cal≈6.67×10^5 Cal.

Answer:

Part A: 2.79×10^9 J
Part B: 2.79×10^6 kJ
Part C: 6.67×10^5 Cal