Consider the piston— cylinder apparatus shown in Figure P20.81. The bottom of the cylinder contains 2.00 kg of water at just under 100.0ºc. The cylinder has a radius of r = 7.50 cm. The piston of mass m = 3.00 kg sits on the surface of the water. An electric heater in the cylinder base transfers energy into the water at a rate of 100 W. Assume the cylinder is much taller than shown in the figure, so we don’t need to be concerned about the piston reaching the top of the cylinder. (a) Once the water begins boiling, how fast is the piston rising? Model the steam as an ideal gas. (b) After the water has completely turned to steam and the heater continues to transfer energy to the steam at the same rate, how fast is the piston rising?
(a)
The speed of the rise of the piston.
Answer to Problem 20.81CP
The speed of the rise of the piston is
Explanation of Solution
Given info: The mass of the water is
Write the expression for the volume of the piston.
Here,
Write the expression for the density of the piston.
Here,
Substitute
Write the expression for the rate of energy required during a phase change.
Here,
Substitute
Substitute
Conclusion:
Therefore, the rise of the piston is
(b)
The rise of the piston.
Answer to Problem 20.81CP
The rise of the piston is
Explanation of Solution
Given info: The mass of the water is
Write the expression for the area of the piston.
Write the expression for the force applied.
Here,
Write the expression for the pressure due to the weight of the piston.
Substitute
Write the expression for the total pressure.
Here,
Substitute
Write the expression for the ideal gas equation.
Here,
Write the expression for the energy required to increase the temperature.
Here,
Differentiate above equation with respect to time.
Substitute
Substitute
Substitute
Conclusion:
Therefore, the rise of the piston is
Want to see more full solutions like this?
Chapter 20 Solutions
EBK PHYSICS:F/SCI.+ENGRS.,TECH.UPDATED
- A vertical cylinder of cross-sectional area A is fitted with a tight-fitting, frictionless piston of mass m (Fig. P18.40). The piston is not restricted in its motion in any way and is supported by the gas at pressure P below it. Atmospheric pressure is P0. We wish to find the height h in Figure P18.40. (a) What analysis model is appropriate to describe the piston? (b) Write an appropriate force equation for the piston from this analysis model in terms of P, P0, m, A, and g. (c) Suppose n moles of an ideal gas are in the cylinder at a temperature of T. Substitute for P in your answer to part (b) to find the height h of the piston above the bottom of the cylinder. Figure P18.40arrow_forwardA cylinder that has a 40.0-cm radius and is 50.0 cm deep is filled with air at 20.0C and 1.00 atm (Fig. P10.74a). A 20.0-kg piston is now lowered into the cylinder, compressing the air trapped inside as it takes equilibrium height hi (Fig. P16.74b). Finally, a 25.0-kg dog stands on the piston, further compressing the air, which remains at 20C (Fig. P16.74c). (a) How far down (h) does the piston move when the dog steps onto it? (b) To what temperature should the gas be warmed to raise the piston and dog back to hi?arrow_forwardA vertical cylinder of cross-sectional area A is fitted with a tight-fitting, frictionless piston of mass m (Fig. P16.56). The piston is not restricted in its motion in any way and is supported by the gas at pressure P below it. Atmospheric pressure is P0. We wish to find die height h in Figure P16.56. (a) What analysis model is appropriate to describe the piston? (b) Write an appropriate force equation for the piston from this analysis model in terms of P, P0, m, A, and g. (c) Suppose n moles of an ideal gas are in the cylinder at a temperature of T. Substitute for P in your answer to part (b) to find the height h of the piston above the bottom of the cylinder.arrow_forward
- A gas is in a container of volume V0 at pressure P0. It is being pumped out of the container by a piston pump. Each stroke of the piston removes a volume Vs through valve A and then pushes the air out through valve B as shown in Figure P19.74. Derive an expression that relates the pressure Pn of the remaining gas to the number of strokes n that have been applied to the container. FIGURE P19.74arrow_forwardA high-pressure gas cylinder contains 60.0 L of toxic gas at a pressure of 1.43 ✕ 107 N/m2 and a temperature of 22.0°C. Its valve leaks after the cylinder is dropped. The cylinder is cooled to dry ice temperature (−78.5°C) to reduce the leak rate and pressure so that it can be safely repaired. 1.What is the final pressure in the tank in pascals, assuming a negligible amount of gas leaks while being cooled and that there is no phase change? 2.What is the final pressure in pascals if one-tenth of the gas escapes?arrow_forwardA pressure versus volume (pv) diagram for a system is shown in the figure. The arrows of the curve indicate the direction of the process, and the points of interest are labeled. The values for the points in the diagram are shown in the table. Volume (m3) Pressure (Pa) v0=27.4 p0=1.00×104 v1=19.3 p1=1.00×104 v2=16.0 p2=4.92×103 v3=13.3 p3=4.92×103 v4=13.3 p4=3.20×103 v5=7.51 p5=1.00×103 Calculate the amount of work done on the system from 0–2 (W02) and then for the entire curve from 0–5 (W05).arrow_forward
- In a 30.0-s interval, 500 hailstones strike a glass window of area 0.600 m2 at an angle of 45.0° to the window surface. Each hailstone has a mass of 5.00 g and a speed of 8.00 m/s. Assuming the collisions are elastic, find (a) the average force and (b) the average pressure on the window during this interval.arrow_forwardI've posted the below question earlier, and I received the answer. This is the question: A J-shaped tube is filled with air at 760 Torr and 22 °C. The long arm is closed off at the top and is 100.0 cm long; the short arm is 40.00 cm high. Mercury is poured through a funnel into the open end. When the mercury spills over the top of the short arm, what is the pressure on the trapped air? Let h be the length of mercury in the long arm. I need further explanation of one step: - when P2 is obtained in the short arm, the equation is 40-h+P1 1. How did we get this equation? 2. the short arm does not contain any air, why do we create an equation of P2 in the short arm? Please clarify these concerns. I have received the answer shown in the attached picture but it is not correct.arrow_forward
- Physics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
- Physics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning