Question

An electric kettle is rated 500W,220V ,it is used to heat 400g of water for 30 sec. assuming the voltage to be 220V,calculate the rise in the temp. of the water,specific heat capacity of water=4200J/KGDEGREE CELCIUS

Answer 1

Power of cattle  p= 500w 
working v=200v
heat developed in 30s
h=pt, 500*30=15000j
let delta t be rise in tamp: of water 
therefor 
h=m*s*delta t
15000=0.4*4200*delta t
delta t = 8.93 deg centigrade.

Answer 2

The rise in temperature of water is 8.93°C

Given data:

P of an electric kettle = 500W

V for an electric kettle = 220 V

Mass of water = 400 g

Time = 30 sec

Specific heat of water = 4200 J/kg °C

Solution:

$Energy\quad =\quad Power\quad \times \quad Time$

$({ E })\quad =\quad 500{ W }\quad \times \quad 30{ sec }\quad =\quad 15000\quad { J }$

Since, $E\quad =\quad mass\quad \times \quad Specific\quad Heat\quad \times \quad \Delta T$

$15000\quad =\quad 0.4\quad \times \quad 4200\quad \times \quad \Delta T$

Hence, $\Delta T\quad =\quad \frac { 15000 }{ 0.4\quad \times \quad 4200 } \quad=\quad 8.93\quad Degree\quad Celsius$