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

`E = 2.556 * 10^9 J`

`E = 2.556 * 10^6 kJ`

`E = 6.11 * 10^8 Cal`

Step-by-step explanation:

As we know that

`1 kWh = 10^3 * 3600 h J`

so here we have

`1 kWh = 3.6 * 10^6 J`

now we have

`E = 710 kWh`

`E = 710 * 3.6 * 10^6 J`

`E = 2.556 * 10^9 J`

now we know that

`1 kJ = 10^3 J`

so we have

`E = 2.556 * 10^6 kJ`

also we know that

`1 cal = 4.18 J`

so we have

`E = 2.556 * 10^9 * (1)/(4.18) cal`

`E = 6.11 * 10^8 Cal`