Question

please help me......

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Answer 1

Answer:

The specific heat of water is 4.184 J/g*°C.

mass of given water = 0.5 Kg = 500g.

temperature to be raised =(100°C - 50°C)= 50°C

Heat energy required for this process =m*c*t.

so heat required is = 500*4.184*50 104600 J =104.6 KJoules.

electric kettle is operating at 220V.

it is drawing current of 3A.

so its power is = P=V*I

P=220*3= 660 Watts.

1 Watt = 1 Joule/sec.

so to provide 104.6 KJ of energy to water it will take

Energy = Power/Time

660 = 104600 /t

t= 158.48 sec

= 2.64 mins

Answer 2

$\huge\bf\green{Answer}$ = 2.64 min ...