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 = 3.0 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 0.70 L of liquid water at a final temperature of 67 oC.

How much longer would it take to warm up the water if it starts as 1/4 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.

Explanation / Answer

heat lost through resistor per unit time , P = V^2/R = 12^2/3 = 48 W

Energy required to heat the water,

E = m*C*(Tf - Ti)

where m =mass of water = 1000*0.7*10^-3 = 0.7 kg

C= specific heat of water = 4186 J/kg.C

Tf = final temperature = 67 deg C

Ti = initital temperaure = 0 deg C

P = 0.7*4186*(67-0)/t = 48

So, t = time taken = 4090 s

Now, for 1/4 th ice and 3/4th water ,

((1/4)*m*L + m*C*(67))/t' = 48

where L = latent heat of ice= 334000 J/kg

t' = time taken in this case

So, (0.7*334000/4 + 0.7*4186*67)/t' = 48

So, t' = 5308 s

So, time difference = 5308 - 4090 = 1218 s = 1218/60 = 20.3 min

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