Power
Energy and Thermal Physics

Power and domestic appliances

Physics Narrative for 11-14 Supporting Physics Teaching

Home appliances working

Let's think about making a pot of tea. Boiling enough water for a pot of tea takes 180 second with my kettle. The kettle is marked 3.0 kilowatt, which means that it costs me 3000 joule every second to run the kettle. To boil the kettle, I must therefore pay the electricity board for 3000 joule / second  ×  180 second which can be worked out to be 540,000 joule. Other domestic appliances cost me different numbers of joules, as they work at different rates for different lengths of time. Some are high power, but work only for a short time (cooker, kettle). Others are lower power, but work more or less continuously (refrigerator) or for long periods of time (lighting).

Here are some typical annual costs:

applianceenergy / megajoule
freezer2380
cooking2380
dishwasher1700
lighting1300
refrigerator1080
tumble dryer1010
kettle900
television792
washing machine84
iron270
vacuum cleaner90

To find out how much energy each appliance uses, simply keep a log of how long you run it for (time in second – the duration), then multiply this quantity by the power of the appliance (power in watt).

energy = power × duration

energyjoule = powerwatt × durationsecond

Or

energykilojoule = powerkilowatt × durationsecond

Of course the averages given above vary with lifestyle and occupancy. Here are some estimates of how the annual total energy per household might vary:

householdenergy / megajoule
working couple14 820
single person1100
family with two children (parents working, children at school)19 730
Power
appears in the relation P=VI P=I^2R P=V^2/R ΔQ=PΔt
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