Question

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.

Answer 1

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