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?

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

1 Answer
Aug 25, 2017

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