. 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 29 A. The defroster directly heats an area of 0.56 m2 of the rear window. What is the maximum thickness of ice above this area that the defroster can melt in 3.8 minutes?

Explanation:

The heat require to melt the ice formed in the rear window is:

The heat transfer rate given by the defroster is:

But:

The maximum thickness of ice that can be melt is:

Max thickness; h = 4.61 x 10^(-4) m

Explanation:

We are given;

Current; I = 29A

Voltage; V = 12 V

Time; t = 3.8 minutes = 228 seconds

Density;ρ = 917 kg/m³

latent heat of fusion of water; L = 3.35 x 10^(5) J/kg

Area; A = 0.56 m²

We know that volume = Area x Height.

Thus, V = Ah and h = V/A

We also know that density is given by;

ρ = mass/volume = m/V

Amd V = m/ρ

Thus, h can be written as;

h = (m/ρ)/A - - - - - (eq1)

Now, we know that;

The specific latent heat (L) of a material is a measure of the heat energy (Q) per mass (m) released or absorbed during a phase change.

It is defined through the formula

Q = mL

Thus, m = Q/L

So, putting Q/L for m in eq 1,we have; h = (Q/Lρ)/A = Q/LρA

Now, power(P) is; Q/t where Q is energy dissipated and t is time.

Thus, P = Q/t and thus, Q = Pt

Thus, h = Pt/LρA - - - - (eq2)

We also know that Power = IV

Thus, power = 29 x 12 = 348 W

Thus, plugging in the relevant values into eq(2),we have;

h = (348 x 228)/(3.35 x 10^(5) x 917 x 0.56)

h = 79344/(1720.292 x 10^(5))

h = 4.61 x 10^(-4) m