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 23
a. the defroster directly heats an area of 0.65 m2 of the rear window. what is the maximum thickness of ice above this area that the defroster can melt in 4.3 minutes?

Answer 1

Let h =  the thickness of ice, m.

Because the ice covers an area of 0.65 m², the volume of ice is
 V = 0.65h m³

The density of ice is 917 kg/m³, therefore the mass of ice is
m = (917 kg/m³)*(0.65h m³) = 596.05h kg

The latent heat of fusion of ice is 3.35 x 10⁵ J/kg, therefore the heat required to melt the ice is
Q₁ = (596.05h kg)*(3.35 x 10⁵ J/kg) = 2h x 10⁸ J

The electric heater supplies heat at 12 V, 23 A for 4.3 minutes to melt the ice.
Therefore the heat provided is
Q₂ = (12 V)*(23 A)*(4.3*60 s) =  71208 J

Because Q₁ = Q₂, obtain
2h x 10⁸ = 71208
h = 356 x 10⁻⁶ m = 356 x 10⁻³ mm = 0.356 mm

Answer: 0.356 mm