Q48P - Fundamentals Of Physics

Question

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 $\mathbf{1000}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$ and requires $\mathbf{2256}\mathbf{ }\mathbf{KJ}\mathbf{/}\mathbf{Kg}$ to be vaporized. If horizontal current lasts 2.00 ms and encounters water with resistivity $\mathbf{150}\mathbf{\Omega }\mathbf{.}\mathbf{m}$, length 12.0 cm, and vertical cross-sectional area , what average current is required to vaporize the water?

Step-by-Step Solution

Answer

The average current required to vaporize the water is 13.0 A.

Step 1: Identification of given data

1. The density of the water is $\mathrm{p}=1000 \mathrm{kg}/{\mathrm{m}}^3$
2. The heat of vaporization is ${\mathrm{L}}_{\mathrm{v}}=2256\mathrm{KJ}/\mathrm{kg}$.
3. The time of horizontal current $∆t=2.00 \mathrm{ms} \mathrm{or} 2.00\times {10}^{-3}\mathrm{s}$.
4. The resistivity of the water is ${\mathrm{p}}_{\mathrm{w}}=150\mathrm{\Omega }.\mathrm{m}$.
5. The horizontal length of water L = 12.0 cm or 0.120m.
6. The vertical cross-sectional area, $A=15\times {10}^{-5}{\mathrm{m}}^2$.

Step 2: Understanding the concept of the flow of current

The current is the flow of charges per unit time through the conductor. The power or rate of energy dissipation, in an electrical device across which a potential difference is equal to the product of current and potential difference. It can also be defined as energy transferred per unit time.

 

Here, we need to use the concept of heat energy required to vaporize the liquid and the concept of heat energy generated due to the current flowing through the resistive material.

 

Formulae:

The mass m of the water over the length L is given by, m = pAL                               …(i)

Here, p is the density, A is the area of cross-section.

 

The energy as heat Q requires vaporizing the water is given by,$Q=L_{v}m$                 …(ii)

Here, $L_v$is the latent heat of vaporization

The thermal energy supplied by Joule heating of the reservoir, $Q=P∆t$                  …(iii)

Here, Q is the heat energy requiredto vaporize the water, 

The power dissipated due to energy transfer,$P=i^{2}R$                                               …(iv)

Here, P is the power, i is current, R is resistance.

The resistance over length L of the water, $R=p_w\frac{L}{A}$                                                 …(v)

Here, R is resistance, L is the length of the wire, $p_w$is the resistivity, A is area of cross-section.

Step 3: Determining the average current

The mass of the water over the length is given using the data in equation (i) as follows:

( p is the density, A is the area of cross-section, and L is the length)

$$\begin{gathered} m=1000\mathrm{kg}/{\mathrm{m}}^3\times 15\times {10}^{-5}{\mathrm{m}}^2\times 0.12\mathrm{m} \\ =0.018 \mathrm{kg} \end{gathered}$$

And the energy as heat Q required to vaporize the water is given by using the given data in equation (ii) as follows:

( $L_v$ is the heat of vaporization)

$$\begin{gathered} Q=2256\mathrm{KJ}/\mathrm{kg}\times 0.018\mathrm{kg} \\ =4.06\times {10}^4\mathrm{J} \end{gathered}$$

The thermal energy supplied by joule heating of the reservoir can be calculated using the above power value and equations (iii), and (iv) as follows:

( P is the power dissipated and $∆t$ is the change in the time)

$Q=i^{2}R∆t$                                                                                                                    …(vi)

The resistance value using the given data and equation (v) can be calculated as follows:

$$\begin{gathered} R=150\mathrm{\Omega }.\mathrm{m}\times \frac{0.120\mathrm{m}}{15\times {10}^{-5}{\mathrm{m}}^2} \\ =1.2\times {10}^5\mathrm{\Omega } \end{gathered}$$

From equation (vi), the average current required to vaporize the water is given by using the given values as follows:

$$\begin{gathered} i=\sqrt{\frac{Q}{R∆t}} \\ =\sqrt{\frac{4.06\times {10}^{4}J}{1.2\times {10}^5\Omega \times 2.0\times {10}^{-3}s}} \\ =13.0A \end{gathered}$$

Hence, the value of the average current in the wire is 13.0 A.