Question

1000 W to 250 V of electric kettle is used to bring water at 20°c to its boiling point.If kettle is used for 11 mins and 12 secs, calculate:(1) resistance (2) current and (3) mass of water in the kettle ? (PLS ANS URGENTLY I Request you)?

Answer 1

The electric kettle is rated 1000W 250V.This means if it is connected across 250V supply voltage, it will spend electrical energy @ 1000J/s.

So `P->"Power"=1000"J/s"`

`V->"Applied Voltage"=250V`

(1) If its resistance be `ROmega` then
`P=V^2/R=>R=V^2/P=250^2/1000=62.5Omega`

(2) `"Current "(I)=V/R=250/62.5=4A`

(3) Let the mass of water be mkg.The heat energy required to raise its temperature from `20^@C " to " 100^@C` is given by `H=mkgxx4200J/(kg^@C)xx(100-20)^@C`
(where `4200J/(kg^@C)` is the sp.heat capacity of water.)

`H=4200xx80mJ`

To raise the temperature Power is supplied for t = 11 min 12s or 672s.

So energy supplied
`=Pxxt=1000xx672=672000J`

If total energy goes only to increase the temperature of water ,then

`4200xx80m=672000`

`=>m=672000/(4200xx80)=2kg`