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 18 A. The defroster directly heats an area of 0.69 m² of the rear window. What is the maximum thickness of ice above this area that the defroster can melt in 4.4 minutes?

Answer 1

Given:

ice temp = 0^oC

ice density = 917 kg/m³

Latent heat of fusion of water = 3.35 x 10⁵ J/Kg

defroster V=12 V and I= 18 A

Area = 0.69 m²

time=264 s

Solution

power delivered P = VI = 12 x 18 =    216 Watts

heat delivered H = power x time = 216 W x 264 seconds =   57024 Joules

mass of ice melted = heat / latent heat offusion = 57024 J / 335000 J/kg =  0.170220896 kg

volume of ice melted = mass / density = 0.170220896 / 917 = 0.000185628 m³

thickness of ice melted = volume / area = 0.000185628 / 0.69 =   0.000269026 meter

<<<<<<<<<<<<<<<<<<<>>>>>>>>>>>>>>>>>>>>>>>>>>>.