Q6PE - College Physics (Urone)

Question

Integrated Concepts 

(a) What is the average power output of a heart defibrillator that dissipates 400J of energy in 10.0 ms? 

(b) Considering the high-power output, why doesn’t the defibrillator produce serious burns?

Step-by-Step Solution

Answer

(a) The heart defibrillator’s mean power output is $\mathrm{P}=4.00\times {10}^4 \mathrm{W}$.

(b) Applying gel on the deliberator pads prevents serious burns despite the high-power output.

Step 1: Concept Introduction

Power: The power of a process is the amount of energy converted into a different type divided by the time interval $\mathbf{∆}\mathbf{t}$ in which the process occurred –

 $\mathbf{P}\mathbf{=}\frac{\mathbf{\Delta E}}{\mathbf{\Delta t}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Step 2: Average power distribution

(a)

The defibrillator’s mean power output is -

 $\mathbf{P}\mathbf{=}\frac{\mathbf{\Delta E}}{\mathbf{\Delta t}}$

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

 $\begin{array}{rcl}\mathbf{P}& \mathbf{=}& \frac{\mathbf{400}\mathbf{ }\mathbf{J}}{\mathbf{0}\mathbf{.}\mathbf{0100}\mathbf{ }\mathbf{s}}\\ & \mathbf{=}& \mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{ }\mathbf{W}\\ & & \end{array}$

Step 3: Reasons for no serious burns

(b)

A gel is used on the pads of the defibrillator before using it. The human body is a conductor of the electric current so it allows the passage to current. The gel applied on the pad does not allow much energy to reach the body, thus protecting form burns.

 

Therefore, the power value obtained is $\mathrm{P}=4.00\times {10}^4 \mathrm{W}$and due to the application of gel on the pad of the deliberator, there are no serious burns.