Question

The heater coil of an electric kettle is rated as 2000W at 200V. How much time will it take to heat one litre of water from 20 degree C 100 degree C, assuming that entire electric energy liberated from heater coil is utilized for heating water?
Density of water = 1 g cm^-3

solve that !! ​

Answer 1

Given,

Potential of an electric kettle = V = 200V

Power of an electric kettle= p = 2000 W

As we know,

current = I = $\frac{P}{V}$

                                   = $\frac{2000}{200}$

                                   = 10A

Heat produced in time T =  $\frac{VIt}{J}$

                                            = $\frac{200*10*t}{4.2}cal$

                                             

Now,

Heat gain by water = mc∅                                      [∅ = thetha]

                          = 1000* 1*80 cal

                           

We all know,

Heat lost by electric kettle = heat gained by water

On putting the values,

$\frac{2000t}{4.2} = 1000*80$ = 1000*80

By solving we get,

t = 168 sec.

Answer 2

Answer:

given

p=2000w

v=200v

row/H2O=1g/cm^3

1cm^3=0.001

density of water =10^3g/L

volume of water =1liter

so mass of water=row*v=10^3g=1kg

we know that

since

p=i^2R

and q=i^2*R*t

so

q=p*t

q=2000t_________(1

change in temprature =80 °C

change in temprature =m*c*q

q=m*c/80____________(2

from 1 and 2

t=mc/160000

where value of c =1

so t=1/160000