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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?

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Solution

## (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. $1\mathrm{kWh}=1\mathrm{kW}×1\mathrm{h}=1000\mathrm{W}×1\mathrm{h}\phantom{\rule{0ex}{0ex}}=1000\left(\mathrm{J}}{\mathrm{s}}\right)×3600\mathrm{s}\left(\because 1\mathrm{W}=1\mathrm{J}}{\mathrm{s}}\right)\phantom{\rule{0ex}{0ex}}=3600000\mathrm{J}\phantom{\rule{0ex}{0ex}}1\mathrm{kWh}=3.6×{10}^{6}\mathrm{J}$ (b) Energy consumed by certain household = 650 units = 650 kWh $1\mathrm{kWh}=3.6×{10}^{6}\mathrm{J}$ $650\mathrm{kWh}=650×3.6×{10}^{6}\mathrm{J}=2.34×{10}^{9}\mathrm{J}$

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