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)?

1 Answer
Oct 6, 2016

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#