A 240 V electric kettle has a heating element with a resistance of 20 ohms. The kettle boils 750 ml water in 1.5 minutes. Assuming 100% efficiency, what was the initial temperature of the water just before the kettle was switched on?

Question

a) 17
b) 18
c) 25
d) 38

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Answer 1 · 2017-08-25T09:43:05

The mass of 750mL water $m=0.75kg$

Specific heat of water $s=4200Jkg^(-1)""^@C^(-1)$

If initial temperature of water be $t^@C$ , then heat required to raise the temperature to boilinh point $100^@C$ will be given by
$H=mxxsxx(100-t)$

The voltage applied to the kettle $V=240$ Volt

Resistance of the kettle $R=20Omega$

Time to raise the temperature of water to boiling point is $T=1.5min=1.5xx60=90sec$

Assuming 100% efficiency of the kettle we can write

$H=mxxsxx(100-t)=V^2/RxxT$

$=>(100-t)=(V^2T)/(Rxxmxxs)$

$=>(100-t)=(240^2xx90)/(20xx0.75xx4200)~~82$

$=>t=18^@C$