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Chapter 20: Problem 64

Water in an electric teakettle is boiling. The power absorbed by the water is \(1.00 \mathrm{kW}\). Assuming that the pressure of vapor in the kettle equals atmospheric pressure, determine the speed of effusion of vapor from the kettle's spout, if the spout has a cross-sectional area of \(2.00 \mathrm{cm}^{2}.\)

Short Answer

Expert verified
The speed of effusion of water vapor from the kettle's spout is approximately \(32.659 m/s.\)

Step by step solution

01

Analyze the Power Equation

Power can be written in terms of energy per unit time. In this case, it鈥檚 thermal energy turning into kinetic energy, we can write it as: \(Power =\frac{Energy}{Time}=\frac{Kinetic Energy}{Time}=\frac{1}{2}mV^2/t\). The kinetic energy used here is the energy of the water vapor effusing out of the kettle spout. Given that the power is 1.00kW, we can rewrite the above equation to solve for the velocity as: \(V=\sqrt{\frac{2 \times Power \times t}{m}}\).
02

Calculate the Mass of Water Vapor

Now, let's solve for mass. The mass of the water vapor that exits from the spout can be represented as \(m=蟻AVt\), where \(蟻\) is the density of the water vapor, \(A\) is the cross-sectional area of the spout, and \(t\) is time. Substituting this into our velocity equation we get: \(V=\sqrt{\frac{2 \times Power \times t}{蟻AVt}}\). By simplifying, we get \(V=\sqrt{\frac{2 \times Power}{蟻A}}\).
03

Insert Given Values

Now replace Power with 1.00 kW = 1000 W; the cross-sectional area \(A=2.00 \mathrm{cm}^{2}=2.00 \times 10^{-4} m^2\); and the density of the water vapor under atmospheric pressure \(蟻=0.60 \mathrm{kg/m}^{3}\). As we have all the values, substitute these values into the speed equation, we get: \(V=\sqrt{\frac{2 \times 1000}{0.60 \times 2.00 \times 10^{-4}}}\).
04

Solve for the Speed of Effusion

After calculating, we find the speed \(V \approx 32.659 m/s\). This is the speed at which the water vapor exits the kettle's spout.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermal Energy to Kinetic Energy Conversion
Understanding the relationship between thermal energy and kinetic energy is crucial for a variety of physical phenomena, including the effusion of gases. When we heat water in an electric kettle, the electrical energy is converted into thermal energy. This thermal energy is then used to increase the kinetic energy of water molecules.

During the boiling process, water molecules gain enough energy to overcome intermolecular forces and then escape into the air as vapor. The speed at which these molecules move can be related to their kinetic energy. By applying the equation for kinetic energy, \( \frac{1}{2}mv^2 \), we can relate the absorbed power (rate of energy conversion) to the kinetic energy of the effusing water vapor. Here, \( m \) represents the mass of the water vapor, and \( v \) is the velocity or speed of effusion.
Mass of Water Vapor
The mass of water vapor effusing from a kettle can be determined by considering the density of vapor, the cross-sectional area of the spout, and the time during which the vapor escapes.

In the equation \( m = \rho AVt \), \( \rho \) represents the density of the water vapor, which depends on factors such as temperature and atmospheric pressure. Meanwhile, \( A \) is the cross-sectional area through which the vapor passes, and \( t \) is the time. The product of these three factors gives us the mass of the effusing vapor. Since the water vapor is at atmospheric pressure, we'll use the density corresponding to that specific condition.
Atmospheric Pressure
Atmospheric pressure plays a significant role in the behavior of gases and vapors. It is the force exerted by the weight of the air above us, at sea level, it averages about \( 101.3 \mathrm{kPa} \).

In the context of this problem, the vapor pressure inside the kettle is assumed equal to the atmospheric pressure. This is essential for calculating the density of the water vapor, which is needed to determine the mass of water vapor exiting the kettle. As the atmospheric pressure remains generally constant at a given altitude, it ensures that the density value used in calculations reflects the typical behavior of water vapor under average earthly conditions.

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Most popular questions from this chapter

The surface of the Sun has a temperature of about \(5800 \mathrm{K}\). The radius of the Sun is \(6.96 \times 10^{8} \mathrm{m} .\) Calculate the total energy radiated by the Sun each second. Assume that the emissivity of the Sun is 0.965.

A student obtains the following data in a calorimetry experiment designed to measure the specific heat of aluminum: Initial temperature of water and calorimeter: \(70^{\circ} \mathrm{C}\) Mass of water: \(0.400 \mathrm{kg}\) Mass of calorimeter: \(0.040 \mathrm{kg}\) Specific heat of calorimeter: \(0.63 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\) Initial temperature of aluminum: \(27^{\circ} \mathrm{C}\) Mass of aluminum: \(0.200 \mathrm{kg}\) Final temperature of mixture: \(66.3^{\circ} \mathrm{C}\) Use these data to determine the specific heat of aluminum. Your result should be within \(15 \%\) of the value listed in Table 20.1.

How much energy is required to change a \(40.0-\mathrm{g}\) ice cube from ice at \(-10.0^{\circ} \mathrm{C}\) to steam at \(110^{\circ} \mathrm{C} ?\)

A 670-kg meteorite happens to be composed of aluminum. When it is far from the Earth, its temperature is \(-15^{\circ} \mathrm{C}\) and it moves with a speed of \(14.0 \mathrm{km} / \mathrm{s}\) relative to the Earth. As it crashes into the planet, assume that the resulting additional internal energy is shared equally between the meteor and the planet, and that all of the material of the meteor rises momentarily to the same final temperature. Find this temperature. Assume that the specific heat of liquid and of gaseous aluminum is \(1170 \mathrm{J} / \mathrm{kg} \cdot^{\circ} \mathrm{C}.\)

(a) In air at \(0^{\circ} \mathrm{C},\) a \(1.60-\mathrm{kg}\) copper block at \(0^{\circ} \mathrm{C}\) is set sliding at \(2.50 \mathrm{m} / \mathrm{s}\) over a sheet of ice at \(0^{\circ} \mathrm{C} .\) Friction brings the block to rest. Find the mass of the ice that melts. To describe the process of slowing down, identify the energy input \(Q,\) the work input \(W,\) the change in internal energy \(\Delta E_{\text {int }},\) and the change in mechanical energy \(\Delta K\) for the block and also for the ice. (b) A \(1.60-\mathrm{kg}\) block of ice at \(0^{\circ} \mathrm{C}\) is set sliding at \(2.50 \mathrm{m} / \mathrm{s}\) over a sheet of copper at \(0^{\circ} \mathrm{C}\) Friction brings the block to rest. Find the mass of the ice that melts. Identify \(Q, W, \Delta E_{\text {int }},\) and \(\Delta K\) for the block and for the metal sheet during the process. (c) A thin 1.60-kg slab of copper at \(20^{\circ} \mathrm{C}\) is set sliding at \(2.50 \mathrm{m} / \mathrm{s}\) over an identical stationary slab at the same temperature. Friction quickly stops the motion. If no energy is lost to the environment by heat, find the change in temperature of both objects. Identify \(Q, W, \Delta E_{\text {int }},\) and \(\Delta K\) for each object during the process.
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