Answer to Question #148293 in General Chemistry for liam donohue

General Chemistry

Question #148293

A Christmas LED bulb burned for 24 hours and consumed 6532KJ during this period.

What is the wattage or watt (W) rating of this lead?

Expert's answer

Power(kW)= Energy(kJ)/Time(s)

24hours= 86400s

∴ Power = 6532/86400

= 0.0756kW

= 75.6W

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