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 0.65 L of liquid water at a final temperature of 74 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)

Explanation / Answer

density of water = 1 g/mL
specific heat capacity of water, C = 4.186 J/goC
Latent heat of fusion of ice, Lf = 334 J/g
mass = volume * density = 650 mL*1 g/mL = 650 g

when whole substance is water,
Energy required= m*C*delta T
V^2/R * t = m*C*delta T
12^2/1.5 * t = 650* 4.186*74
t= 2097.4 s
= 35 min

when there is water and ice:
mass of ice, m1 = (1/4)*650 = 162.5 g
mass of water, m2 = (3/4)*650 = 487.5 g

Energy required= energy to melt ice + energy to raise temperature of water and melted ice
V^2/R * t = m1*Lf + (m1+m2)*C*delta T
12^2/1.5 * t = 162.5*334 + (162.5+ 487.5)* 4.186*74
12^2/1.5 * t = 255621.6
t = 2662.7 s
= 44.4 min

Extra time taken = 44.4 - 35 = 9.4 min

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