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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掳C to 100.0掳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 \(100.0^{\circ}\mathrm{C}\).

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Most popular questions from this chapter

If the total power per unit area from the Sun incident on a horizontal leaf is \(9.00 \times 10^{2} \mathrm{W} / \mathrm{m}^{2},\) and we assume that \(70.0 \%\) of this energy goes into heating the leaf, what would be the rate of temperature rise of the leaf? The specific heat of the leaf is $3.70 \mathrm{kJ} /\left(\mathrm{kg} \cdot^{\circ} \mathrm{C}\right),$ the leaf's area is \(5.00 \times 10^{-3} \mathrm{m}^{2},\) and its mass is $0.500 \mathrm{g}$.
A blacksmith heats a 0.38 -kg piece of iron to \(498^{\circ} \mathrm{C}\) in his forge. After shaping it into a decorative design, he places it into a bucket of water to cool. If the available water is at \(20.0^{\circ} \mathrm{C},\) what minimum amount of water must be in the bucket to cool the iron to \(23.0^{\circ} \mathrm{C} ?\) The water in the bucket should remain in the liquid phase.
A hotel room is in thermal equilibrium with the rooms on either side and with the hallway on a third side. The room loses heat primarily through a 1.30 -cm- thick glass window that has a height of \(76.2 \mathrm{cm}\) and a width of $156 \mathrm{cm} .\( If the temperature inside the room is \)75^{\circ} \mathrm{F}$ and the temperature outside is \(32^{\circ} \mathrm{F}\), what is the approximate rate (in \(\mathrm{kJ} / \mathrm{s}\) ) at which heat must be added to the room to maintain a constant temperature of \(75^{\circ} \mathrm{F} ?\) Ignore the stagnant air layers on either side of the glass.
If \(125.6 \mathrm{kJ}\) of heat are supplied to \(5.00 \times 10^{2} \mathrm{g}\) of water at \(22^{\circ} \mathrm{C},\) what is the final temperature of the water?
A 10.0 -g iron bullet with a speed of $4.00 \times 10^{2} \mathrm{m} / \mathrm{s}\( and a temperature of \)20.0^{\circ} \mathrm{C}$ is stopped in a \(0.500-\mathrm{kg}\) block of wood, also at \(20.0^{\circ} \mathrm{C} .\) (a) At first all of the bullet's kinetic energy goes into the internal energy of the bullet. Calculate the temperature increase of the bullet. (b) After a short time the bullet and the block come to the same temperature \(T\). Calculate \(T\), assuming no heat is lost to the environment.
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