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An electric heater of resistance 10Omega connected to 220 V power supply is immersed in the water of 1 kg. How long the electrical heater has to be switched on to increase its temperature from 30^(@)C to 60^(@)C. (The specific heat of water is s=4200J kg^(-1)/C) |
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Answer» Solution :According to Joule.s heating law `H=I^(2)Rt` The current PASSED through the ELECTRICAL heater `=(220V)/(10Omega)=22A` The heat produced in one second by the electrical heater `=(220V)/(10Omega)=22A` The heat produced in one second by the electrical heater`H=I^(2)R` The heat produced inone second `H=(22)^(2)xx10=4840J=4.84kJ`. In FACT the power rating of this electrical heater is 4.84 k W. The amount of energy to increase the temperatureof 1 kg water from `30^(@)C` to `60^(@)C` is `Q=ms DeltaT` Here `m=1kg, s=4200J kg^(-1), DeltaT=30` so `Q=1xx4200xx30=126kJ` The time required to producethis heat energy `t=(Q)/(I^(2)R)=(126xx10^(3))/(4840)=26.03s` |
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