Question

(a) Derive the relation between commercial unit of energy (kWh) and SI unit of energy (joule).
(b) A certain household consumes 650 units of electricity in a month. How much is this electricity in joules?

Answer 1

(a) Commercial unit of energy in 1 kWh. One kilowatt-hour is the amount of energy consumed in one hour at a constant rate of one kilowatt. SI unit of energy is joule.

$$\begin{gathered} 1\mathrm{kWh}=1\mathrm{kW}\times 1\mathrm{h}=1000\mathrm{W}\times 1\mathrm{h} \\ =1000\left(\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.\right)\times 3600\mathrm{s}(\because 1\mathrm{W}=1\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.) \\ =3600000\mathrm{J} \\ 1\mathrm{kWh}=3.6\times {10}^6\mathrm{J} \end{gathered}$$

(b) Energy consumed by certain household = 650 units = 650 kWh

$1\mathrm{kWh}=3.6\times {10}^6\mathrm{J}$

$650\mathrm{kWh}=650\times 3.6\times {10}^6\mathrm{J}=2.34\times {10}^9\mathrm{J}$