Exploding shoes. The rain-soaked shoes of a person may explode if ground current from nearby lightning vaporizes the water. The sudden conversion of water to water vapor causes a dramatic expansion that can rip apart shoes. Water has density 1000 kg/m3 and requires 2256 kJ/kg to be vaporized. If horizontal current lasts 2.09 ms and encounters water with resistivity 172 Ω·m, length 15.2 cm, and vertical cross-sectional area 16.7 × 10-5 m2, what average current is required to vaporize the water?

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cjmejiab cjmejiab · Answer 1 · 2019-11-03T00:48:53+01:00

Answer:

I=0.41A

Explanation:

Given,

Density:

[tex]\rho_w = 1000Kg/m^3[/tex]

Heat vaporization

[tex]H_v=2256kJ/Kg=2256*10^3J/Kg[/tex]

[tex]A=16.7*10^{-5}m^2[/tex]

[tex]L=15.2cm[/tex]

[tex]\rho=172 \Omega[/tex]

[tex]t=2.09ms[/tex]

Then we can calculate the mass

[tex]m=\rho_w*A*L = (1000)(16.7*10^{-5})(0.152)= 0.02538Kg[/tex]

We know that,

Heat vaporization, [tex]H_v = \frac{\upsilon}{m}[/tex]

[tex]\upsilon=Hv*m[/tex]

[tex]\upsilon = 57257.28J[/tex]

Rate of energy transference,

[tex](1) P=\frac{\upsilon}{t}[/tex]

[tex](2) P=I*v=I^2*R[/tex]

Where [tex]R= \rho\frac{L}{A}[/tex]

Note that [tex]\rho=R*(A/L)[/tex]

So,

[tex]I^2*\rho \frac{L}{A} = \frac{H_v\rho_w A L }{t}[/tex]

[tex]I=\sqrt{\frac{A^2H_v \rho_w}{\rho t}}[/tex]

[tex]I= \frac{\sqrt{(16.7*10^{-5})^2(2256*10^3)(1000)}}{(172)(2.09)}[/tex]

[tex]I=0.41A[/tex]