# Question #8451f

##### 1 Answer
Aug 30, 2016

The given electric kettle is rated as $750 W - 240 V$
This means
$\text{Power of the heater} P = 750 W$
$\text{when connected across } V = 240 V$

If the resistance of this electric kettle be R ohm,
then
$R = {V}^{2} / P = {240}^{2} / 750 \Omega = 76.8 \Omega$

Initial temperture of the kettle and its content,water is ${20}^{\circ} C$

And their final temperature is boiling point of water ${100}^{\circ} C$

So gain in temperature$= \Delta t = {80}^{\circ} C$

Heat gained by kettle
$\Delta {H}_{k} = \text{its heat capacity} \times \Delta t$
$= 400 \times 80 = 32000 J$

Heat gained by water
$\Delta {H}_{w} = \text{its mass"xx"sp.heat} \times \Delta t$
$= 0.5 \times 4200 \times 80 = 168000 J$

So total heat required
$\Delta H = \Delta {H}_{k} + \Delta {H}_{w}$
$= \left(32000 + 16800\right) = 200000 J$

Let this heat is produced in t s by connecting the kettle with ${V}_{m} = 200 V$ main. So the heat produced in electric kettle is
$= {V}_{m}^{2} / R \times t = {200}^{2} / 76.8 \times t J$

So by the condition

${200}^{2} / 76.8 \times t = 200000$

$\implies t = \frac{200000 \times 76.8}{200 \times 200} s$

$\implies t = 384 s = 6 \min \text{ } 24 s$