Question

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 2.14 ms and encounters water with resistivity 176 Ω·m, length 13.2 cm, and vertical cross-sectional area 17.8 × 10-5 m2, what average current is required to vaporize the water?

Answer 1

Answer: 137.76 A

Explanation: Heat is generated in a material when current is passed through the conducting material, this process is known as Joule heating or Ohmic heating or resistive heating.

Joule's first law, also known as Joule-Lenz law states that the heating power generated by an electrical conductor is directly proportional to the resistance of the conductor and square of the current flowing through it.

Mathematically, we have:

`Q= i^(2) R t`...............................(1)

where:

• Q= heat generated

• i= current flowing through the conductor

• R= resistance of the conductor

• t= time for which the current flows in the conductor

GIven data:

• heat required for the vapourization of water into steam, L 2256
  `J.kg^(-1)`

• heat of vaporization per kg of water,
  `c=2256 kJ.kg^(-1)`

• density of water,
  `d= 1000 kg.m^(-3)`

• time for which the current flows= 2.14 ms =
  `2.14* 10^(-3)` s

• resistivity of water,
  `\rho= 176 \Omega .m`

• lenght of water along which the current flows,
  `l=13.2 cm`

• vertical cross-section area which is normal to the direction of current flow,
  `A=17.8 * 10^(-5) m^(2)`

To find:

• The current required to vaporize the given water, i = ?

Firstly, we calculate the amount of energy required by the given amount of water to vaporize.

We have the density and heat of vaporization per kg of water.

volume of water=
`13.2* 10^(-2)m * 17.8 * 10^(-5)  m^(2)`

`Volume= 2.3496 * 10^(-5) m^(2)`

Now, Mass of water
`= density * volume\\=1000 * 2.3496 * 10^(-5)`

`Mass= 2.3496 * 10^(-2) kg`

∴Total amount of heat energy required,

`Q= m * L\\Q= 2.3496 * 10^(-2)* 2256\\Q= 53.007 J`

we know, resistance
`R= (\rho.l)/(A)`

Putting the given values

`R=(176* 0.132)/(17.8* 10^(-5) ) \\\\R= 130516.85 \Omega`

Using eq. (1)

`Q= i^(2) R t`

`53.007 = i^(2) * 130516.85 * 2.14* 10^(-3)\\ \\i^(2) = (53.007)/(130516.85 * 2.14* 10^(-3)) \\\\i^(2) =0.189781 \\i=0.4356 A`