The rear window of a van is coated with a layer of ice at 0°C. The density of ic

Question

The rear window of a van is coated with a layer of ice at 0°C. The density of ice is 917 kg/m3, and the latent heat of fusion of water is 3.35 x 105 J/kg. The driver of the van turns on the rear-window defroster, which operates at 12 V and 26 A. The defroster directly heats an area of 0.49 m2 of the rear window. What is the maximum thickness of ice above this area that the defroster can melt in 5.0 minutes? The rear window of a van is coated with a layer of ice at 0°C. The density of ice is 917 kg/m3, and the latent heat of fusion of water is 3.35 x 105 J/kg. The driver of the van turns on the rear-window defroster, which operates at 12 V and 26 A. The defroster directly heats an area of 0.49 m2 of the rear window. What is the maximum thickness of ice above this area that the defroster can melt in 5.0 minutes?

Explanation / Answer

Power is given by:

P = V*i = 26*12

Energy is given by:

E = P*t

t = 5 min = 300 sec

E = 26*12*300 = 93600 J

Mass of ice melt by this energy will be

E = m*Lf

m = E/Lf = 93600/(3.35*10^5) = 0.279 kg

Volume of ice will be

V = m/d = 0.279 kg/(917 kg m^3)

V = 3.04*10^-4 m^3

Now area of window = 0.49 m^2

So thickness of ice above this area will be

Volume = Area*thickness

thickness = Volume/Area

thickness = 3.04*10^-4/0.49 = 6.2*10^-4 m^3 = 0.62 mm

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