/*! 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 83 An electric immersion heater nor... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 26: Problem 83

An electric immersion heater normally takes 100 min to bring cold water in a well-insulated container to a certain temperature, after which a thermostat switches the heater off. One day the line voltage is reduced by \(6.00 \%\) because of a laboratory overload. How long does heating the water now take? Assume that the resistance of the heating element does not change.

Short Answer

Expert verified
The heating now takes approximately 113.17 minutes.

Step by step solution

01

Understanding the Problem

We are given that an immersion heater normally takes 100 minutes to heat water. One day, the voltage is reduced by \(6.00\%\). We need to calculate the new time it will take to heat the water given that the resistance of the heater remains constant.
02

Identify Relationship Between Power, Voltage, and Time

Power \(P\) is related to energy \(E\) and time \(t\) by the formula \(P = \frac{E}{t}\). The power consumed by a circuit is also determined by the voltage \(V\) and resistance \(R\) as \(P = \frac{V^2}{R}\). Since \(R\) is constant, changes in \(V\) will affect \(P\) and, consequently, \(t\).
03

Calculate Original Power

The original power is expressed as \( P_{ ext{original}} = \frac{V^2}{R} \). The time taken for heating is \( t_{ ext{original}} = 100 \) minutes, so \( E = P_{ ext{original}} \times t_{ ext{original}} \).
04

Calculate New Voltage and Power

If voltage is reduced by \(6.00\%\), the new voltage \( V_{ ext{new}} = 0.94V \). The new power \( P_{ ext{new}} = \frac{(0.94V)^2}{R} = 0.8836 \times \frac{V^2}{R} = 0.8836P_{ ext{original}} \).
05

Relate Power and Time After Voltage Change

The energy required to heat the water remains the same as before, so we can set \( E = P_{ ext{new}} \times t_{ ext{new}} \). Therefore, \( t_{ ext{new}} = \frac{E}{P_{ ext{new}}} = \frac{E}{0.8836P_{ ext{original}}} \). Simplifying gives \( t_{ ext{new}} = \frac{t_{ ext{original}}}{0.8836} \).
06

Compute the New Heating Time

Substitute \( t_{ ext{original}} = 100 \) minutes: \( t_{ ext{new}} = \frac{100}{0.8836} \approx 113.17 \) minutes.

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.

Power Calculations
Power calculations are an essential part of understanding how electric circuits operate, especially when working with devices like heating elements. The power in a circuit is the rate at which energy is used. It's calculated using the formula \( P = \frac{E}{t} \), where \( P \) represents power, \( E \) is energy, and \( t \) is time. For electric devices, power can also be expressed as \( P = \frac{V^2}{R} \). Here, \( V \) stands for voltage, \( R \) is resistance.
  • Power allows us to understand how much energy is consumed over time. High power means more energy usage and faster operation, whereas low power means slower operation.
  • In our exercise, the heater requires a certain power to bring water to the desired temperature in a specific amount of time.
  • When voltage changes, it directly impacts power, altering how quickly an electric device can perform its function.
Understanding these calculations is crucial for determining how changes in a circuit, like voltage drop, will affect overall performance.
Voltage Reduction
Voltage reduction in an electric circuit can significantly affect the performance of devices such as heating elements. Voltage is the potential difference that drives current through a circuit. In our scenario, a \(6\%\) reduction in voltage affects how efficiently the heater operates.
  • With lower voltage, the power, calculated as \( P = \frac{V^2}{R} \), decreases. Since power is proportional to the square of the voltage, even a small voltage reduction can lead to a noticeable drop in power.
  • This reduced power output means that the heater takes longer to reach the same level of performance, which in our case, is bringing water to a certain temperature.
  • Voltage reduction can be caused by various factors like lab overloads and can impact not just heating times but also the functionality of other sensitive equipment.
Being aware of how voltage drops affect performance is important for maintaining efficiency and effectiveness in electric circuits.
Heating Elements
Heating elements are devices used to convert electrical energy into heat. They are made from materials with high electrical resistance, allowing them to generate heat efficiently when current passes through.
  • They are used in a wide range of appliances, from water heaters to ovens and even irons.
  • The efficiency of a heating element is affected by the power available, which is dictated by the voltage and the intrinsic resistance of the element itself.
  • In the given exercise, since the resistance remains constant, the heating element's power adapts solely based on changes in the voltage.
When the voltage decreases, as shown in our problem, the time taken to achieve the desired heating increases because the power output of the heating element is reduced. Understanding the interplay between voltage, resistance, and power in heating elements helps in optimizing their use across various applications.

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

A copper wire of cross-sectional area \(2.00 \times 10^{-6} \mathrm{~m}^{2}\) and length \(4.00 \mathrm{~m}\) has a current of 2.00 A uniformly distributed across that area. (a) What is the magnitude of the electric field along the wire? (b) How much electrical energy is transferred to thermal energy in \(30 \mathrm{~min} ?\)

A coil of current-carrying Nichrome wire is immersed in a liquid. (Nichrome is a nickel-chromium-iron alloy commonly used in heating elements.) When the potential difference across the coil is \(12 \mathrm{~V}\) and the current through the coil is \(5.2 \mathrm{~A},\) the liquid evaporates at the steady rate of \(21 \mathrm{mg} / \mathrm{s}\). Calculate the heat of vaporization of the liquid (see Module \(18-4\) ).

Two conductors are made of the same material and have the same length. Conductor \(A\) is a solid wire of diameter \(1.0 \mathrm{~mm} .\) Conductor \(B\) is a hollow tube of outside diameter \(2.0 \mathrm{~mm}\) and inside diameter \(1.0 \mathrm{~mm} .\) What is the resistance ratio \(R_{A} / R_{B}\), measured between their ends?

The magnitude \(J(r)\) of the current density in a certain cylindrical wire is given as a function of radial distance from the center of the wire's cross section as \(J(r)=B r,\) where \(r\) is in meters, \(J\) is in amperes per square meter, and \(B=2.00 \times 10^{5} \mathrm{~A} / \mathrm{m}^{3} .\) This function applies out to the wire's radius of \(2.00 \mathrm{~mm}\). How much current is contained within the width of a thin ring concentric with the wire if the ring has a radial width of \(10.0 \mu \mathrm{m}\) and is at a radial distance of \(1.20 \mathrm{~mm} ?\)

The legend that Benjamin Franklin flew a kite as a storm approached is only a legend-he was neither stupid nor suicidal. Suppose a kite string of radius \(2.00 \mathrm{~mm}\) extends directly upward by \(0.800 \mathrm{~km}\) and is coated with a \(0.500 \mathrm{~mm}\) layer of water having resistivity \(150 \Omega \cdot \mathrm{m} .\) If the potential difference between the two ends of the string is \(160 \mathrm{MV},\) what is the current through the water layer? The danger is not this current but the chance that the string draws a lightning strike, which can have a current as large as 500000 A (way beyond just being lethal).
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Particle Model of Matter

Read Explanation

Geometrical and Physical Optics

Read Explanation

Physical Quantities And Units

Read Explanation

Rotational Dynamics

Read Explanation

Linear Momentum

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.