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 = 2.0 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 0.75 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

Mass of the water is

M = 0.75*10^-3*1000=0.75 kg =750 g

Mass of the ice is,

m = 0.75/4 =0.1875 kg=187.5 g

The heat energy is,

Q= mL + McdT = 187.5*333.55+750*4.186*67=2.73*10^5 J

Thus, the time is,

t = QR/V^2 = 3790 s = 63.1 min

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.