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An electric heater supplies 1.8 kW of power in the form of heat to a tank of water. How long will it take to heat the 200 kg of water in the tank from `10^@` to `70^@`C ? Assume heat losses to the surroundings to be negligible. |
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Answer» The heat added to tank `DeltaQ=` power`xx`time `DeltaQ=1.8xx10^3xxt(J)` .(i) The heat absorbed in water `DeltaQ=mcDeltaQ=mcDeltaT=200xx10^3xx1xx60=12xx10^6cal` `=12xx10^66xx4.2J` Hence `t=(12xx10^6xx4.2)/(1.8xx10^3)=2.78xx10^4s=7.75h` |
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