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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 28 A. The defroster directly heats an area of 0.44 m2 of the rear window. What is the maximum thickness of ice above this area that the defroster can melt in 4.5 minutes?

Respuesta :

Answer:

thickness of ice melted = 1.242*10^{-3} m

Explanation:

power delivered = V*I = 12 * 28 =    336 Watts

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heat delivered = power * time = 336 W * 300 seconds =   168000 Joules

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mass of ice melted [tex]= \frac{heat}{latent heat dffusion}[/tex]

mass of ice melted = 97200 J / 335000 J/kg = 0.5014 kg

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volume of ice melted = mass / density =  0.5014 / 917 = 5.46*10{-4} m3

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thickness of ice melted = volume / area = 5.46*10{-4} / 0.44   =  1.242*10^{-3} m

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