You have purchased a portable resistive water heater, consisting of a resistor R

Question

You have purchased a portable resistive water heater, consisting of a resistor R = 1.5 connected to a 12 V outlet in your car. On a cold-weather camp-out you want to warm up some water that has been left out over night for your morning tea. Assume the water is in an insulated cup so no heat is lost to the surroundings.

You end up with 0.50 L of liquid water at a final temperature of 69 oC.

How much longer would it take to warm up the water if it starts as 1/3 ice and the rest water rather than it all initially being liquid at 0 oC?

Give your answer in minutes (there 60 seconds in a minute) to at least three significant figures.

Do not include units in your answer.

Explanation / Answer

pwoer provided by the outlet=voltage^2/resistance=12^2/1.5=96 W

if time taken to heat the water is t seconds, then energy provided=power*time=96*t J

now latent heat of

fusion of ice=334 kJ/kg

specific heat of water=4187 J/(kg.K)

density of water=1 kg/ltr

total mass of water=0.5 L=0.5 kg

mass of ice=(0.5/3)=(1/6) kg

energy required to melt this much ice=mass*latent heat of fusion of ice

=(1/6)*334*1000=5.5667*10^4 J

now , total water is liquid form at 0 degree celcius

to heat it up till 69 degree celcius , energy required=mass*specific heat of water*temperature difference

=0.5*4187*(69-0)=1.4445*10^5 J

then total energy required=5.6667*10^4+1.4445*10^5= 201117 J

hence 96*t= 201117

==>t=2095 seconds

as 1 minute=60 seconds,

t=34.916 minutes

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