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)?

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 $P \to \text{Power"=1000"J/s}$

$V \to \text{Applied Voltage} = 250 V$

(1) If its resistance be $R \Omega$ then
$P = {V}^{2} / R \implies R = {V}^{2} / P = {250}^{2} / 1000 = 62.5 \Omega$

(2) $\text{Current } \left(I\right) = \frac{V}{R} = \frac{250}{62.5} = 4 A$

(3) Let the mass of water be mkg.The heat energy required to raise its temperature from ${20}^{\circ} C \text{ to } {100}^{\circ} C$ is given by $H = m k g \times 4200 \frac{J}{k {g}^{\circ} C} \times {\left(100 - 20\right)}^{\circ} C$
(where $4200 \frac{J}{k {g}^{\circ} C}$ is the sp.heat capacity of water.)

$H = 4200 \times 80 m J$

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

So energy supplied
$= P \times t = 1000 \times 672 = 672000 J$

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

$4200 \times 80 m = 672000$

$\implies m = \frac{672000}{4200 \times 80} = 2 k g$