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

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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:

[tex]E = 2.556 \times 10^9 J[/tex]

[tex]E = 2.556 \times 10^6 kJ[/tex]

[tex]E = 6.11 \times 10^8 Cal[/tex]

Explanation:

As we know that

[tex]1 kWh = 10^3 \times 3600 h J[/tex]

so here we have

[tex]1 kWh = 3.6 \times 10^6 J[/tex]

now we have

[tex]E = 710 kWh[/tex]

[tex]E = 710 \times 3.6 \times 10^6 J[/tex]

[tex]E = 2.556 \times 10^9 J[/tex]

now we know that

[tex]1 kJ = 10^3 J[/tex]

so we have

[tex]E = 2.556 \times 10^6 kJ[/tex]

also we know that

[tex]1 cal = 4.18 J[/tex]

so we have

[tex]E = 2.556 \times 10^9 \times \frac{1}{4.18} cal[/tex]

[tex]E = 6.11 \times 10^8 Cal[/tex]

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