Q7PE - College Physics (Urone)

Question

Integrated Concepts 

A lightning bolt strikes a tree, moving 20.0 C of charge through a potential difference of 1.00 x 10² MV. 

(a) What energy was dissipated? 

(b) What mass of water could be raised from $\mathbf{15}\mathbf{°}\mathbf{ }\mathbf{C}$ to the boiling point and then boiled by this energy? 

(c) Discuss the damage that could be caused to the tree by the expansion of the boiling steam.

Step-by-Step Solution

Answer

(a) $∆\mathrm{E}=2.00\times {10}^9 \mathrm{J}$ energy was dissipated.

(b) The mass of water that could be raised is m = 7.65 x 10² kg.

(c) The trees suffer internal tissue damage by expansion of boiling steam.

Step 1: Given data

Charge transferred is - 20.0 C.

Potential difference is- 1.00 x 10² MV.

Initial temperature of water is- $15° \mathrm{C}$.

Step 2: Concept Introduction

Electric Potential Energy: If a particle with charge q is placed at a point where the electric potential of a charged object is V , the electric potential energy of the particle-object system is –

$\mathbf{PE}\mathbf{=}\mathbf{qV}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{(1)}$

The energy Q required to change the temperature of a mass m of a substance by an amount $\mathbf{∆}\mathbf{T}$ is –

$\mathbf{Q}\mathbf{=}\mathbf{mc\Delta T}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{(2)}$

Where c is the specific heat of the substance.

Energy to boil or condense: The energy in joules needed to vaporize a mass m of a liquid at its boiling temperature, or the energy released when a mass m of a gas condensed at that same temperature, is –

$\mathbf{Q}\mathbf{=}\mathbf{\pm }{\mathbf{mL}}_{\mathbf{v}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{(3)}$

Where L_v is the latent heat of vaporization of the substance. The plus sign is used when the substance vaporizes and the minus sign when it condenses.

Step 3: Energy Calculation

(a)

The energy dissipated due to the movement of the charge through the tree is found from Equation ( 1 ) :

 $\mathbf{\Delta E}\mathbf{=}\mathbf{q\Delta V}$

Entering the values for q and $∆\mathrm{V}$, it is obtained:

$\begin{array}{rcl}\mathbf{∆}\mathbf{E}\mathbf{ }& \mathbf{=}& \mathbf{ }\left(20.0 C\right)\left(1.00\times {10}^2 \mathrm{MV}\right)\left(\frac{{10}^6 V}{1 \mathrm{MV}}\right)\\ & \mathbf{=}& \mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}\mathbf{ }}\mathbf{J}\end{array}$

Step 4: Mass of water

(b)

The energy needed to change the temperature of a mass m of water from

$15° \mathrm{C}$ to $100° \mathrm{C}$ is found from Equation (2):

$\begin{array}{rcl}{\mathbf{Q}}_{\mathbf{1}}\mathbf{ }& \mathbf{=}& \mathbf{ }\mathbf{mc}\left(100° C - 15° C\right)\\ & \mathbf{=}& \mathbf{mc}\left(85° C\right)\end{array}$

The energy needed to vaporize the same mass m of water at $100° \mathrm{C}$ is found

from Equation (3):

${\mathbf{Q}}_{\mathbf{2}}\mathbf{=}{\mathbf{mL}}_{\mathbf{v}}$

Therefore, the total energy needed to change the temperature of a mass m 

of water from $15° \mathrm{C}$ to vaporization is:

$\begin{array}{rcl}\mathbf{∆}\mathbf{E}& \mathbf{=}& {\mathbf{Q}}_{\mathbf{1}}\mathbf{+}{\mathbf{Q}}_{\mathbf{2}}\\ & \mathbf{=}& \mathbf{mc}\left(85° C\right)\mathbf{+}{\mathbf{mL}}_{\mathbf{V}}\\ & \mathbf{=}& \mathbf{m}\left[\left(85° C\right)c+L_v\right]\end{array}$

Solve for m:

$\mathbf{m}\mathbf{=}\frac{\mathbf{∆}\mathbf{E}}{\left(85° C\right)\mathbf{c}\mathbf{+}{\mathbf{L}}_{\mathbf{v}}}$

Entering the values for $∆E$ from part (a), c and L_v we obtain:

$\begin{array}{rcl}\mathbf{m}& \mathbf{=}& \frac{\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{ }\mathbf{J}}{\mathbf{(}\mathbf{85}\mathbf{°}\mathbf{ }\mathbf{C}\mathbf{)}\mathbf{(}\mathbf{4186}\mathbf{ }\mathbf{J}\mathbf{/}\mathbf{kg}\mathbf{\times }\mathbf{°}\mathbf{C}\mathbf{)}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{26}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{ }\mathbf{J}\mathbf{/}\mathbf{kg}}\\ & \mathbf{=}& \mathbf{7}\mathbf{.}\mathbf{65}\mathbf{\times }{\mathbf{10}}^{\mathbf{2}}\mathbf{ }\mathbf{kg}\end{array}$

Step 5: Damage caused to the tree

(c)

As we see in part (b), the mass of evaporated water due to the electric current through it is huge. This will cause damage to the internal tissues of the tree since it is mostly water.
 

Therefore, the energy and mass of water is $\mathbf{∆}\mathit{E}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{ }\mathit{J}$and $\mathit{m}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{65}\mathbf{\times }{\mathbf{10}}^{\mathbf{2}}\mathbf{ }\mathit{k}\mathit{g}$respectively. The damage caused to the tree is the damage of internal tissue.