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#