/*! 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 51 An electric heater of \(1.08 \ma... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 51

An electric heater of \(1.08 \mathrm{~kW}\) is immersed in water. After the water has reached a temperature of \(100^{\circ} \mathrm{C}\), how much time will be required to produce \(100 \mathrm{~g}\) of steam? (Latent heat of steam \(=540 \mathrm{calg}^{-1}\) )

Short Answer

Expert verified
209.21 seconds are needed to produce 100 g of steam with the electric heater.

Step by step solution

01

Understanding the Problem

We need to find out how long it will take a 1.08 kW electric heater to convert 100 g of water into steam at 100掳C. The latent heat required for this phase change is 540 cal/g.
02

Converting Power Units

An electric heater's power of 1.08 kW needs to be converted from kilowatts to calories per second. We know that 1 watt is equivalent to 0.239005736 calories/second. Therefore, 1.08 kW is equivalent to \(1.08 \times 1000 \times 0.239\) cal/s. This results in approximately 258.12 cal/s.
03

Calculating Total Energy Required

We must calculate the total energy required to convert 100 g of water at 100掳C into steam. Using the latent heat of vaporization, we find the total heat required as \( 100 \times 540 \) cal, which equals 54,000 cal.
04

Calculating Time Required

Now, divide the total energy required by the power converted into calories per second: \( \frac{54000 \, \text{cal}}{258.12 \, \text{cal/s}} \approx 209.21 \, \text{seconds}. \) Therefore, it will take approximately 209.21 seconds to produce 100 g of steam.

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.

Phase Change
When water transitions from liquid to gas, it undergoes a process known as a phase change, specifically vaporization. During a phase change, a substance absorbs energy without an increase in temperature. In the problem you encountered, water at boiling point needs extra heat to transform into steam. This is different from simply heating water, as the temperature remains constant at 100掳C during this phase change.
A phase change like boiling involves latent heat, essential for breaking the molecular bonds in the liquid. This means you supply energy to change the state, not the temperature. Here, the latent heat of vaporization tells us how much energy each gram of water needs to convert to steam without changing temperature.
Energy Conversion
Energy conversion is a crucial aspect of working with an electric heater to produce steam. It involves translating electrical energy, measured in kilowatts, into heat energy, measured in calories. The electric heater in the problem converts power to heat, which then helps transition water into steam.
Understanding this conversion helps determine how efficiently an electric device can deliver energy to its surroundings. Here鈥檚 a breakdown:
  • The heater's power rating means it delivers energy at a rate of 1.08 kW.
  • This energy needs conversion from kilowatts to the calorie unit, a more convenient measure for thermal energy calculations.
  • Once in calories, this energy can quantify how fast it can cause the phase change, leading to steam production.
Calorie to Joule Conversion
Calories and joules are units used to measure energy, with many scientific calculations requiring conversions between these units. In this context, a calorie is the energy needed to raise the temperature of 1 gram of water by 1掳C, but in many scenarios, conversions to joules are necessary since it is the standard metric unit of energy.
To convert calories to joules, remember that 1 calorie equals approximately 4.184 joules. This conversion is essential when comparing energy quantities in different systems or scenarios.
In our problem, although we're primarily dealing in calories, you might need to understand conversions when presented with results in different units or for further scientific analysis.

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

In a neon gas discharge tube \(\mathrm{Ne}^{+}\)ions moving through a cross- section of the tube each second to the right is \(2.9 \times 10^{18}\), while \(1.2 \times 10^{18}\) electrons move towards left in the same time. The electronic charge being \(1.6 \times 10^{-19} \mathrm{C}\), the net electric current is (a) \(0.27\) A to the right (b) \(0.66\) A to the right (c) \(0.66 \mathrm{~A}\) to the left (d) zero

A capacitor of \(10 \mu \mathrm{F}\) has a potential difference of \(40 \mathrm{~V}\) across it. If it is discharged in \(0.2 \mathrm{~s}\), the average current during discharge is (a) \(2 \mathrm{~mA}\) (b) \(4 \mathrm{~mA}\) (c) \(1 \mathrm{~mA}\) (d) \(0.5 \mathrm{~mA}\)

If two identical heaters each rated as (1000 W-220 V) are connected in parallel to \(220 \mathrm{~V}\), then the total power consumed is (a) \(200 \mathrm{~W}\) (b) \(2500 \mathrm{~W}\) (c) \(250 \mathrm{~W}\) (d) \(2000 \mathrm{~W}\)

A battery of emf \(E\) and internal resistance \(r\) is connected with an external voltage source (generator) through a resistance \(R\) as shown in figure. Choose the correct statements. (a) In order to charge the battery, the output voltage \(V\) of the generator must be greater than \(E\) (b) In order to charge the battery, the output voltage \(V\) of the generator must be at least twice of \(E\) (c) The charging current \(i\) through the circuit is given by \(i=\frac{V-E}{(R+r)}\) (d) The charging current \(i\) through the circuit is given by \(i=\frac{V}{(R+r)}\)

The measurement of an unknown resistance \(R\) is to be carried out using Wheatstones bridge. Two students perform an experiment in two ways. The first students takes \(R_{2}=10 \Omega\) and \(R_{1}=5 \Omega\). The other student takes \(R_{2}=1000 \Omega\) and \(R_{1}=500 \Omega\). In the standard arm, both take \(R_{3}=5 \Omega\). Both find \(\begin{aligned} R &=\frac{R_{2}}{R_{1}} R_{3}=10 \Omega \text { within errors. } \\ \text { [NCERT Exemplar] } \end{aligned}\) (a) The errors of measurement of the two students are the same (b) Errors of measurement do depend on the accuracy with which \(R_{2}\) and \(R_{1}\) can be measured (c) If the student uses large values of \(R_{2}\) and \(R_{1}\), the currents through the arms will be feeble. This will make determination of null point accurately more difficult (d) Wheatstone bridge is a very accurate instrument and has no errors of measurement
See all solutions

Recommended explanations on Physics Textbooks

Electricity and Magnetism

Read Explanation

Geometrical and Physical Optics

Read Explanation

Force

Read Explanation

Fluids

Read Explanation

Thermodynamics

Read Explanation

Quantum Physics

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.