Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 24

A heating coil inside an electric kettle delivers \(2.1 \mathrm{kW}\) of electric power to the water in the kettle. How long will it take to raise the temperature of \(0.50 \mathrm{kg}\) of water from \(20.0^{\circ} \mathrm{C}\) to \(100.0^{\circ} \mathrm{C} ?\) (tutorial: heating)

Short Answer

Expert verified
Answer: It takes approximately 79.752 seconds for the electric kettle to raise the temperature of 0.50 kg of water from 20.0°C to 100.0°C.

Step by step solution

01

List the given values

We are given the following values: - Power of heating coil: \(P=2.1\,\mathrm{kW} = 2100\,\mathrm{W}\) (converted to watts); - Mass of water: \(m=0.50\,\mathrm{kg}\); - Initial temperature: \(T_{1}=20.0^{\circ}\mathrm{C}\); - Final temperature: \(T_{2}=100.0^{\circ}\mathrm{C}\); Additionally, we know that the specific heat capacity of water is \(c_{water}=4.186\,\mathrm{J/g\cdot K}\).
02

Calculate the temperature change

To calculate the temperature change, subtract the initial temperature from the final temperature: \(\Delta T = T_{2} - T_{1} = 100.0^{\circ}\mathrm{C} -20.0^{\circ}\mathrm{C} = 80.0\,\mathrm{K}\) We can use kelvin or degrees Celsius as the unit here, since we're just interested in the temperature difference.
03

Calculate the energy required to heat the water

We can calculate the energy required to heat the water using the formula: \(Q=m\times c_{water} \times \Delta T\) Plug in the values: \(Q = 0.50\,\mathrm{kg} \times 4.186\,\mathrm{J/g\cdot K} \times 1000\,\mathrm{g/kg} \times 80.0\,\mathrm{K}\) \(Q=167480\,\mathrm{J}\)
04

Determine the time needed to heat the water

To determine the time needed to heat the water, we divide the energy required by the power of the heating coil: \(t=\frac{Q}{P}\) \(t=\frac{167480\,\mathrm{J}}{2100\,\mathrm{W}} = 79.752\,\mathrm{s}\)
05

Write down the final answer

It will take approximately \(79.752\,\mathrm{s}\) for the electric kettle to raise the temperature of \(0.50\,\mathrm{kg}\) of water from \(20.0^{\circ}\mathrm{C}\) to \(100.0^{\circ}\mathrm{C}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 incandescent light bulb radiates at a rate of \(60.0 \mathrm{W}\) when the temperature of its filament is \(2820 \mathrm{K}\). During a brownout (temporary drop in line voltage), the power radiated drops to \(58.0 \mathrm{W} .\) What is the temperature of the filament? Neglect changes in the filament's length and cross-sectional area due to the temperature change. (tutorial: light bulb)
A hiker is wearing wool clothing of \(0.50-\mathrm{cm}\) thickness to keep warm. Her skin temperature is \(35^{\circ} \mathrm{C}\) and the outside temperature is \(4.0^{\circ} \mathrm{C} .\) Her body surface area is \(1.2 \mathrm{m}^{2} .\) (a) If the thermal conductivity of wool is $0.040 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K}),$ what is the rate of heat conduction through her clothing? (b) If the hiker is caught in a rainstorm, the thermal conductivity of the soaked wool increases to \(0.60 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})\) (that of water). Now what is the rate of heat conduction?
Convert \(1.00 \mathrm{kJ}\) to kilowatt-hours \((\mathrm{kWh})\).
A 75 -kg block of ice at \(0.0^{\circ} \mathrm{C}\) breaks off from a glacier, slides along the frictionless ice to the ground from a height of $2.43 \mathrm{m},$ and then slides along a horizontal surface consisting of gravel and dirt. Find how much of the mass of the ice is melted by the friction with the rough surface, assuming \(75 \%\) of the internal energy generated is used to heat the ice.
Five ice cubes, each with a mass of \(22.0 \mathrm{g}\) and at a temperature of \(-50.0^{\circ} \mathrm{C},\) are placed in an insulating container. How much heat will it take to change the ice cubes completely into steam?
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Mechanics and Materials

Read Explanation

Space Physics

Read Explanation

Oscillations

Read Explanation

Electric Charge Field and Potential

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.

Sign-up for free