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 0.75 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 1/2 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

volume of ice = volume of water = 0.75/2 = 0.375 L

mass of water m1 = 0.375*10^-3*1000 = 0.375 kg

mass of ice m2 = 0.375*10^-3*917 = 0.344 kg

heat supplied = Q = (m1*Cw*dt) + (m2*Lf) + (m2*cw*dt)

Q = (0.375*4190*69) + (0.344*334000)+(0.334*4190*69)

Q = 319874.9 J

power = V^2/R

P = Q/t

319874.9/t = 12^2/1.5

t = 3332.03 s

t = 55.53 min

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