/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 A heart defibrillator passes \(1... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 20

A heart defibrillator passes \(10.0\) A through a patient's torso for \(5.00 \mathrm{~ms}\) in an attempt to restore normal beating. (a) How much charge passed? (b) What voltage was applied if \(500 \mathrm{~J}\) of energy was dissipated? (c) What was the path's resistance? (d) Find the temperature increase caused in the \(8.00 \mathrm{~kg}\) of affected tissue.

Short Answer

Expert verified
(a) 0.050 C; (b) 10000 V; (c) 1000 惟; (d) 0.015 掳C.

Step by step solution

01

Calculate the Charge Passed

To find the charge passed, we use the formula for charge, which is given by \( Q = I \times t \), where \( I = 10.0 \mathrm{~A} \) is the current and \( t = 5.00 \times 10^{-3} \mathrm{~s} \) is the time.\[Q = 10.0 \times (5.00 \times 10^{-3}) = 0.050 \mathrm{~C}\]Thus, the charge passed is \( 0.050 \mathrm{~C} \).
02

Calculate the Voltage Applied

The energy dissipated can be calculated using the formula \( E = V \times Q \), where \( E = 500 \mathrm{~J} \) and \( Q = 0.050 \mathrm{~C} \). We can solve for \( V \).\[V = \frac{E}{Q} = \frac{500}{0.050} = 10000 \mathrm{~V}\]Therefore, the voltage applied is \( 10000 \mathrm{~V} \).
03

Calculate the Path's Resistance

The power can also be expressed in terms of resistance as \( P = I^2 R \), where \( P = \frac{E}{t} \). Solving for \( R \) using \( I = 10.0 \mathrm{~A} \):\[R = \frac{E}{I^2 t} = \frac{500}{10.0^2 \times 5.00 \times 10^{-3}}\]\[R = 1000 \mathrm{~\Omega}\]Hence, the path's resistance is \( 1000 \mathrm{~\Omega} \).
04

Determine the Temperature Increase

To determine the temperature increase, we use the formula relating heat energy and temperature change: \( Q = mc\Delta T \), where \( m = 8.00 \mathrm{~kg} \) is the mass, \( c = 4186 \mathrm{~J/kg \cdot 掳C} \) (specific heat capacity of water, assuming tissue is similar), and \( Q = 500 \mathrm{~J} \) is the energy dissipated.\[\Delta T = \frac{Q}{mc} = \frac{500}{8.00 \times 4186}\]\[\Delta T \approx 0.015 \mathrm{~掳C}\]Thus, the temperature increase in the tissue is approximately \( 0.015 \mathrm{~掳C} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Current
Current is the flow of electric charge in a circuit. It helps define how much charge moves in a time period. In our heart defibrillator example, we had a current of 10.0 A moving through the patient's torso. But what exactly is an ampere (A)? It signifies one coulomb of charge passing through a point per second. Thus, in this situation, 10 coulombs of charge pass each second. Using the formula \( Q = I \times t \), we calculated the charge to be 0.050 C by multiplying the current by time. Current is a fundamental quantity that drives most electrical phenomena, including the defibrillator's ability to save lives.
Voltage
Voltage, often referred to as electric potential difference, is the push that makes electric charges move. Consider it like the pressure that pushes water through a hose. In the defibrillator scenario, we found the voltage to be a whopping 10000 V. But how did we reach this number? By using the energy relationship \( E = V \times Q \). We divided the known energy, 500 J, by the calculated charge, 0.050 C. This shows that voltage is crucial for energy transfer in a system, determining how efficiently electrical energy is applied to achieve desired results, such as restoring a heartbeat.
Resistance
Resistance is like the friction that opposes the flow of electric charge. It determines how much current will flow for a given voltage. Resistors and components in a circuit present obstacles for the current. We calculated a resistance of 1000 \( \Omega \) for the defibrillator's path. By using the formula \( R = \frac{E}{I^2 \cdot t} \), it showed how challenging it is for the current to move through the patient's chest. Resistance plays a critical role in controlling currents to ensure that devices work safely and effectively, as excessively low resistance could mean overheating.
Charge
Electric charge is a property that allows objects to exert forces on each other without touching. In the exercise, we saw that the defibrillator passed a charge of 0.050 C through a person's body. This charge results from the flow of electrons pushed by the defibrillator's voltage. Charge is measured in coulombs (C) and forms the basis of electricity as it builds up and directs currents in circuits. Understanding charge is key to grasping how various electrical phenomena, from lighting a bulb to pacing a heart, operate efficiently.
Energy Dissipation
Energy dissipation happens when energy is transformed from one form to another, often losing some energy as heat. We calculated that the defibrillator dissipated 500 J of energy. By utilizing the formula \( E = V \times Q \), we connected energy dissipation with both charge and voltage. This process is critical since the lost energy as heat requires careful consideration, especially in medical equipment where excessive energy might damage tissues. Proper energy management through monitoring dissipation ensures that devices like defibrillators work effectively without unintended consequences.
Temperature Change
Temperature change refers to the variation in thermal energy within a system. In our scenario, the defibrillator caused a temperature increase of roughly 0.015 掳C in human tissue. This was determined using the formula \( \Delta T = \frac{Q}{mc} \), where \( m \) is mass and \( c \) is specific heat capacity. Here, we assume human tissue behaves similarly to water, which has a specific heat of 4186 J/kg掳C. Minimal temperature rise ensures safe operation without harming tissues. This understanding of heat and temperature illustrates how energy transformations affect both living organisms and mechanical systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An electric water heater consumes \(5.00 \mathrm{~kW}\) for \(2.00 \mathrm{~h}\) per day. What is the cost of running it for one year if electricity costs \(12.0\) cents \(/ \mathrm{kW} \cdot \mathrm{h}\) ? See Figure

If aluminum and copper wires of the same length have the same resistance, which has the larger diameter? Why?

(a) An immersion heater utilizing \(120 \mathrm{~V}\) can raise the temperature of a \(1.00 \times 10^{2}-\mathrm{g}\) aluminum cup containing \(350 \mathrm{~g}\) of water from \(20.0^{\circ} \mathrm{C}\) to \(95.0^{\circ} \mathrm{C}\) in \(2.00 \mathrm{~min}\). Find its resistance, assuming it is constant during the process. (b) A lower resistance would shorten the heating time. Discuss the practical limits to speeding the heating by lowering the resistance.

The average television is said to be on 6 hours per day. Estimate the yearly cost of electricity to operate 100 million TVs, assuming their power consumption averages \(150 \mathrm{~W}\) and the cost of electricity averages \(12.0\) cents \(/ \mathrm{kW} \cdot \mathrm{h}\).

Why do incandescent lightbulbs grow dim late in their lives, particularly just before their filaments break?
See all solutions

Recommended explanations on Physics Textbooks

Kinematics Physics

Read Explanation

Fluids

Read Explanation

Work Energy and Power

Read Explanation

Magnetism

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Geometrical and Physical Optics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.