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7 May, 02:46

A 2-kW resistance heater in a water heater runs for 3 hours to raise the water temperature to the desired level. Determine the amount of electric energy used in both kWh and kJ.

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  1. 7 May, 05:58
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    Power (p) = 2kW

    Time (t) = 3 h

    Work done (w) = ?

    p = w / t

    w = p * t

    w = 2kW * 3 h

    w = 6kWh

    Converting 6kWh to Joules,

    6kWh

    6 * (10^3) h (k = 10^3)

    6*10^3 * (60*60) (1 hour = 60 * 60 sec)

    2.2*10^7 J in 2 significant figures
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