Web Analytics
all2aGgeauB
Answered

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

Verified Answer :

MissPhiladelphia

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:

\(E = 2.556 \times 10^9 J\)

\(E = 2.556 \times 10^6 kJ\)

\(E = 6.11 \times 10^8 Cal\)

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

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

\(E = 6.11 \times 10^8 Cal\)

Related Questions