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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

Systematic use of solar energy can yield a large saving in the cost of winter space heating for a typical house in the north central United States. If the house has good insulation, you may model it as losing energy by heat steadily at the rate \(6000 \mathrm{W}\) on a day in April when the average exterior temperature is \(4^{\circ} \mathrm{C},\) and when the conventional heating system is not used at all. The passive solar energy collector can consist simply of very large windows in a room facing south. Sunlight shining in during the daytime is absorbed by the floor, interior walls, and objects in the room, raising their temperature to \(38^{\circ} \mathrm{C} .\) As the sun goes down, insulating draperies or shutters are closed over the windows. During the period between 5: 00 P.M. and 7: 00 A.M. the temperature of the house will drop, and a sufficiently large "thermal mass" is required to keep it from dropping too far. The thermal mass can be a large quantity of stone (with specific heat \(850 \mathrm{J} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\) ) in the floor and the interior walls exposed to sunlight. What mass of stone is required if the temperature is not to drop below \(18^{\circ} \mathrm{C}\) overnight?

One mole of an ideal gas is contained in a cylinder with a movable piston. The initial pressure, volume, and temperature are \(P_{i}, V_{i},\) and \(T_{i},\) respectively. Find the work done on the gas for the following processes and show each process on a PV diagram: (a) An isobaric compression in which the final volume is half the initial volume. (b) An isothermal compression in which the final pressure is four times the initial pressure. (c) An isovolumetric process in which the final pressure is three times the initial pressure.

A gas is compressed at a constant pressure of 0.800 atm from \(9.00 \mathrm{L}\) to \(2.00 \mathrm{L}\). In the process, \(400 \mathrm{J}\) of energy leaves the gas by heat. (a) What is the work done on the gas? (b) What is the change in its internal energy?

A solar cooker consists of a curved reflecting surface that concentrates sunlight onto the object to be warmed (Fig. P20.63). The solar power per unit area reaching the Earth's surface at the location is \(600 \mathrm{W} / \mathrm{m}^{2} .\) The cooker faces the Sun and has a diameter of \(0.600 \mathrm{m} .\) Assume that \(40.0 \%\) of the incident energy is transferred to \(0.500 \mathrm{L}\) of water in an open container, initially at \(20.0^{\circ} \mathrm{C} .\) How long does it take to completely boil away the water? (Ignore the heat capacity of the container.)

An ideal gas initially at 300 K undergoes an isobaric expansion at \(2.50 \mathrm{kPa}\). If the volume increases from \(1.00 \mathrm{m}^{3}\) to \(3.00 \mathrm{m}^{3}\) and \(12.5 \mathrm{kJ}\) is transferred to the gas by heat, what are (a) the change in its internal energy and (b) its final temperature?
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