Question

An electric kettle rated 1000w ,220v is used to bring water at 20 degrees to its boiling point . if the kettle is switched on for 10minutes .calculate :(1)the Resistance of the element kettle (2)current flowing through the element (3)mass of water in the kettle.

Answer 1

I=P/V
=1000/220
=4.54A

R=V/I
=220/4.54
48.4

Sorry but i don't know the mass.
Hope it helps.

Answer 2

Answer:

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=1000J/s

V→Applied Voltage=250V

(1) If its resistance be  RΩ

then P=V2R

⇒R=V2P

=2502

        1000

=62.5Ω

(2)  Current  (I)=VR=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=mkg×4200Jkg∘C×(100−20)∘C

(where  4200Jkg∘Cis the sp.heat capacity of water.)

H=4200×80mJ

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

So energy supplied=P×t

=1000×672

=672000J

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

4200×80m

=672000

⇒m=672000

4200×80=

2kg

Explanation:

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