Question

"I want only the procedure" .The answer is 2.16 * 10^7 joules

An electric iron of power 2 kW is used for 3 h per day. How much energy does it consume in 2 days? Express in joule.

If the procedure is helpful I'll mark you the brainliest !
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Answer 1

Explanation:

we know 1kWh = 3.6×10^6J

power of the electric iron p = 2kW

duration of use per day . t = 3h

So, the iron is used in 2 days for , t = 3×2h

= 6h

So, energy consumed in 2days = 2kW × 6h

= 12kWh

= 12× 1kWh

= 12×3.6×10^6joules

= 43.2×10^6J

= 4.32×10^7J