/*! 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 27 What must be the efficiency of a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 27

What must be the efficiency of an electric kettle marked \(500 \mathrm{~W}, 230 \mathrm{~V}\), if it was found to bring \(1 \mathrm{~kg}\) of water at \(15^{\circ} \mathrm{C}\) to boiling point in 15 min? (Given specific heat capacity of water \(=4200 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}\) ) (a) \(79 \%\) (b) \(81 \%\) (c) \(72 \%\) (d) \(69 \%\)

Short Answer

Expert verified
The efficiency of the kettle is (a) 79%.

Step by step solution

01

Determine the energy supplied by the kettle

The power of the kettle is given as \(500 \mathrm{~W}\). To find the energy supplied, use the formula: \( \text{Energy supplied} = \text{Power} \times \text{Time} \). Here, the time is 15 minutes, which needs to be converted into seconds: \(15 \times 60 = 900\) seconds. Thus, \( \text{Energy supplied} = 500 \times 900 = 450,000 \mathrm{~J}\).
02

Calculate the energy required to heat the water

First, determine the temperature change \( \Delta T \) which is from \(15^\circ \mathrm{C}\) to \(100^\circ \mathrm{C}\), so \( \Delta T = 100 - 15 = 85^\circ \mathrm{C}\). The mass of the water is \(1 \mathrm{~kg}\), and the specific heat capacity \( c \) is given as \(4200 \mathrm{~J/kg}^\circ \mathrm{C}\). Using the formula: \[ \text{Energy required} = \text{mass} \times \text{specific heat capacity} \times \Delta T \]substitute the given values: \[ \text{Energy required} = 1 \times 4200 \times 85 = 357,000 \mathrm{~J} \].
03

Calculate the efficiency of the kettle

Efficiency is given by the formula: \[ \text{Efficiency} = \left( \frac{\text{Energy required}}{\text{Energy supplied}} \right) \times 100\% \] Substitute the values from the previous steps: \[ \text{Efficiency} = \left( \frac{357,000}{450,000} \right) \times 100\% = 79.33\% \] Therefore, the efficiency of the kettle is approximately \(79\%\).
04

Choose the closest efficiency

The calculated efficiency is approximately \(79\%\), which matches option (a) \(79\%\). Therefore, the correct answer is (a) \(79\%\).

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.

Specific Heat Capacity
Specific heat capacity is a crucial concept in thermodynamics, helping us understand how different materials react to heating. It tells us the amount of heat required to raise the temperature of 1 kilogram of a substance by 1 degree Celsius.
The specific heat capacity of water, in this case, is given as \(4200 \mathrm{~J/kg} \degree \mathrm{C}\). This number indicates that for every kilogram of water, we need 4200 Joules of energy to increase the temperature by one degree Celsius.
This is especially important when heating large volumes, as the energy required can quickly add up. Knowing the specific heat capacity allows us to calculate how much energy is needed to achieve a desired temperature change. It also informs us about the efficiency of the heating process.
Energy Calculation
Energy calculation is essential when determining how much power or work is needed in a process. In the example of an electric kettle, energy calculation shows how much electricity is used to heat water.
To calculate energy supplied, you multiply power by time. Here, power is 500 watts, and time is converted from 15 minutes to seconds (15 minutes equals 900 seconds). Therefore, the energy supplied by the kettle is \(500 \times 900 = 450,000 \mathrm{~J}\).
Understanding energy calculations helps determine the efficiency of appliances. By comparing the energy supplied to the energy used for heating, you can assess if an appliance is performing optimally. For effective energy use, it is crucial to ensure that most of the supplied energy accomplishes the needed task without excessive waste.
Temperature Change
Temperature change is the difference in temperature that results from heating or cooling a substance. In the process of heating water with a kettle, the temperature change is calculated by subtracting the initial temperature from the final temperature.
In the exercise, water starts at \(15^{\circ} \mathrm{C}\) and reaches boiling point at \(100^{\circ} \mathrm{C}\). The temperature change \(\Delta T\) is therefore \(100 - 15 = 85^{\circ} \mathrm{C}\).
A correct temperature change calculation is crucial for determining how much energy is required based on the specific heat capacity of the substance in question. It directly influences how efficiently devices like electric kettles can convert electrical energy into heat energy to reach the desired temperature.

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

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

A metallic resistor is connected across a battery. If the number of collisions of the free electrons with the lattice is some how decreased in the resistor (for example by cooling it), the current will (a) remains constant (b) increase (c) decrease (d) become zero

A wire of length \(l\) is drawn such that its diameter is reduced to half of its original diameter. If initial resistance of the wire were \(10 \Omega\), its resistance would be (a) \(160 \Omega\) (b) \(120 \Omega\) (c) \(140 \Omega\) (d) \(100 \Omega\)

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

Assertion The temperature depending on resistance is usually given as \(R=R_{0}(1+\alpha \Delta t)\). The resistance of a wire changes from \(100 \Omega\) to \(150 \Omega\). When its temperature is increased from \(27^{\circ} \mathrm{C}\) to \(227^{\circ} \mathrm{C}\). This implies that \(\alpha=2.5 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\). Reason \(R=R_{0}(1+\alpha \Delta t)\) is valid when then change in the temperature \(\Delta T\) is small and \(\Delta R=\left(R-R_{0}\right)<
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Atoms and Radioactivity

Read Explanation

Quantum Physics

Read Explanation

Classical Mechanics

Read Explanation

Thermodynamics

Read Explanation

Translational Dynamics

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.