Question

A particular frost-free refrigerator uses about 710kWh of electrical energy per year. Express this amount of energy in J, kJ, & Cal.

Answer 1

A particular frost-free refrigerator uses about 710kWh of electrical energy per year. You are to express this amount of energy in J, kJ, & Calories. 

1 year (365 days / 1 year)(24 hours / 1 day)(3600s / 1h) = 31,536,000s

710 kWh/yr (1 yr) = 710 kWh
710 x 10^3 Wh = 710 x 10^3(J/s)(31,536,000s) = 2.24 x 10^13 J
2.24 x 10^13 J = 2.24 x 10^10 kJ = 5.35 x 10^12 cal

Answer 2

Answer:

$E = 2.556 \times 10^9 J$

$E = 2.556 \times 10^6 kJ$

$E = 6.11 \times 10^8 Cal$

Explanation:

As we know that

$1 kWh = 10^3 \times 3600 h J$

so here we have

$1 kWh = 3.6 \times 10^6 J$

now we have

$E = 710 kWh$

$E = 710 \times 3.6 \times 10^6 J$

$E = 2.556 \times 10^9 J$

now we know that

$1 kJ = 10^3 J$

so we have

$E = 2.556 \times 10^6 kJ$

also we know that

$1 cal = 4.18 J$

so we have

$E = 2.556 \times 10^9 \times \frac{1}{4.18} cal$

$E = 6.11 \times 10^8 Cal$