Question

Question #8451f

Answer 1

The given electric kettle is rated as `750W-240V`
This means
`"Power of the heater"P=750W`
`"when connected across "V=240V`

If the resistance of this electric kettle be R ohm,
then
`R=V^2/P=240^2/750Omega=76.8Omega`

Initial temperture of the kettle and its content,water is `20^@C`

And their final temperature is boiling point of water `100^@C`

So gain in temperature`=Deltat=80^@C`

Heat gained by kettle
`DeltaH_k="its heat capacity"xxDeltat`
`=400xx80=32000J`

Heat gained by water
`DeltaH_w="its mass"xx"sp.heat"xxDeltat`
`=0.5xx4200xx80=168000J`

So total heat required
`DeltaH=DeltaH_k+DeltaH_w`
`=(32000+16800)=200000J`

Let this heat is produced in t s by connecting the kettle with `V_m=200V` main. So the heat produced in electric kettle is
`=V_m^2/Rxxt=200^2/76.8xxtJ`

So by the condition

`200^2/76.8xxt=200000`

`=>t=(200000xx76.8)/(200xx200)s`

`=>t=384s=6min" "24s`