Question

The rear window of a van is coated with a layer of ice at 0°C.
The density of ice is 917 kg/m^{3}, and the latent heat of
fusion of water is 3.35 x 10^{5} 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 m^{2} 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^{o}C

ice density = 917 kg/m^{3}

Latent heat of fusion of water = 3.35 x 10^{5} J/Kg

defroster V=12 V and I= 18 A

Area = 0.69 m^{2}

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 m ^{3}**

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

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

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