/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 Air is contained in a vertical p... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 50

Air is contained in a vertical piston-cylinder assembly by a piston of mass \(50 \mathrm{~kg}\) and having a face area of \(0.01 \mathrm{~m}^{2}\). The mass of the air is \(5 \mathrm{~g}\). and initially the air occupies a volume of 5 liters. The atmosphere exerts a pressure of \(100 \mathrm{kPa}\) on the top of the piston. The volume of the air slowly decreases to \(0.002 \mathrm{~m}^{3}\) as the specific internal energy of the air decreases by \(260 \mathrm{~kJ} / \mathrm{kg}\). Neglecting friction between the piston and the cylinder wall, determine the heat transfer to the air, in \(\mathrm{kJ}\).

Short Answer

Expert verified
Heat transfer to the air, Q, is approximately -1.253 kJ.

Step by step solution

01

Calculate Initial Pressure

The initial state includes the atmospheric pressure and the pressure due to the weight of the piston. Convert the mass of the piston to force: \[ F = m \times g = 50 \text{ kg} \times 9.81 \text{ m/s}^2 = 490.5 \text{ N} \]Then, calculate the pressure exerted by the piston: \[ P_{\text{piston}} = \frac{F}{A} = \frac{490.5 \text{ N}}{0.01 \text{ m}^2} = 49050 \text{ Pa} = 49.05 \text{ kPa} \]Finally, add this to the atmospheric pressure to find the total initial pressure: \[ P_1 = P_{\text{atmosphere}} + P_{\text{piston}} = 100 \text{ kPa} + 49.05 \text{ kPa} = 149.05 \text{ kPa} \]
02

Calculate Initial and Final Volumes

The initial volume is given as 5 liters: \[ V_1 = 5 \text{ L} = 0.005 \text{ m}^3 \]The final volume is given as: \[ V_2 = 0.002 \text{ m}^3 \]
03

Use the Ideal Gas Law to Find Initial and Final State Properties

Using the ideal gas law: \[ P_1 V_1 = m R T_1 \]Initially: \[ 149.05 \text{ kPa} \times 0.005 \text{ m}^3 = 0.005 \text{ kg} \times R \times T_1 \]Solve for initial temperature \(T_1\)
04

Apply the Energy Equation for a Closed System

The energy equation can be written as: \[ Q - W = \text{螖U} \]Where \(Q\) is the heat transfer, \(W\) is the work done by the system, and \(\text{螖U}\) is the change in internal energy.
05

Determine the Work Done by the System

Work done by the system is calculated as: \[ W = P \times (V_2 - V_1) \]Given the process is quasi-static and neglecting changes in external pressure: \[ W = 149.05 \text{ kPa} \times (0.002 \text{ m}^3 - 0.005 \text{ m}^3) \]\[ W = 149.05 \text{ kPa} \times -0.003 \text{ m}^3 \]
06

Calculate the Change in Internal Energy

The change in internal energy is: \[ \text{螖U} = m \times \text{螖u} = 0.005 \text{ kg} \times (-260 \text{ kJ/kg}) = -1.3 \text{ kJ} \]
07

Solve for Heat Transfer

Using the energy equation: \[ Q - W = \text{螖U} \]Rearrange to solve for \(Q\): \[ Q = \text{螖U} + W \]Substitute the values found in previous steps to calculate \(Q\):

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

ideal gas law
The ideal gas law is a fundamental equation in thermodynamics. It relates the pressure, volume, temperature, and amount of a gas. The formula is given by:\( PV = nRT \), where
  • \(P\) is the pressure,
  • \(V\) is the volume,
  • \(n\) is the number of moles of gas,
  • \(R\) is the universal gas constant, and
  • \(T\) is the temperature.
In our exercise, we use a modified form to accommodate given mass. For initial calculations:
  • \( P_1V_1 = mRT_1\)
Here, we rearrange to find the initial temperature (\(T_1\)). This foundation helps solve further steps, especially when properties of the system (like volume) change.
internal energy change
The internal energy of an ideal gas depends mainly on its temperature. Any change in temperature results in a change in internal energy. In thermodynamics, the change in internal energy (\(\text{螖U}\)) is critical in determining the system's heat exchange.
  • General formula: \(\text{螖U} = m \times \text{螖u}\)
  • In our exercise, specific internal energy change is given. Multiply by mass (\(m\)) to get the total change in internal energy.
This concept is pivotal when applying the energy equation for closed systems later.
work done by system
In thermodynamics, work done by a system during a volume change can be calculated. For quasi-static processes, the formula is:\( W = P \times (V_2 - V_1)\)
Here, since the process is slow (quasi-static), the initial pressure (\(P_1\)) is constant.
  • Insert known values: \(149.05 \text{ kPa} \times (0.002 \text{ m}^3 - 0.005 \text{ m}^3)\)
  • Calculate work done: Negative value indicates work done on the system.
Understanding work done helps quantify the overall energy change in the system.
heat transfer
Heat transfer (\(Q\)) reflects energy exchanged due to temperature differences. It can be computed using the first law of thermodynamics for closed systems:
  • Equation: \(Q - W = \text{螖U}\)
  • Rearrange to solve for \(Q: Q = \text{螖U} + W\)
    Practically, this helps find the heat transfer required/expelled for specific internal and external work changes.
To solve:
  • Substitute \(W\) (found in 'Work Done by System') and \(\text{螖U}\) (found in 'Internal Energy Change').
This step aggregates various concepts, finalizing our calorimetric analysis.
pressure calculation
Calculating pressure encompasses atmospheric and additional applied forces (like piston weight). This involves:
  • Convert mass (\(m\)) to force (\(F\)):\(F = m \times g\)
  • From force, calculate pressure due to piston: \(P_{\text{piston}} = \frac{F}{A}\)
  • Total Initial Pressure:\(P_1 = P_{\text{atmosphere}} + P_{\text{piston}}\)
These calculations set the stage for applying the ideal gas law and further thermodynamic analyses. Subsequently, ensure correct units (\(kPa\) and \(Pa\)).
volume change
Volume change in gases impacts other thermodynamic quantities like pressure and temperature. For the piston-cylinder assembly:
  • Initial volume (\(V_1\)) and final volume (\(V_2\)) are given.\(V_1 = 0.005 \text{ m}^3\), \(V_2 = 0.002 \text{ m}^3\)
  • A negative volume change indicates compression.\(\text{Volume change} = V_2 - V_1\)
This data aids in further calculations like work done by system and applying ideal gas law for state properties.

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

A vertical cylindrical mass of \(5 \mathrm{~kg}\) undergoes a process during which the velocity decreases from \(30 \mathrm{~m} / \mathrm{s}\) to \(15 \mathrm{~m} / \mathrm{s}\), while the elevation remains unchanged. The initial specific internal energy of the mass is \(1.2 \mathrm{~kJ} / \mathrm{kg}\) and the final specific internal energy is \(1.9 \mathrm{~kJ} / \mathrm{kg}\). During the process, the mass receives \(2 \mathrm{~kJ}\) of energy by heat transfer through its bottom surface and loses \(1 \mathrm{~kJ}\) of energy by heat transfer through its top surface. The lateral surface experiences no heat transfer. For this process, evaluate (a) the change in kinetic energy of the mass in \(\mathrm{kJ}\), and (b) the work in \(\mathrm{kJ}\).

\(.\) A block of mass \(10 \mathrm{~kg}\) moves along a surface inclined \(30^{\circ}\) relative to the horizontal. The center of gravity of the block. is elevated by \(3.0 \mathrm{~m}\) and the kinetic energy of the block decreases by \(50 \mathrm{~J}\). The block is acted upon by a constant force \(\mathbf{R}\) parallel to the incline, and by the force of gravity. Assume frictionless surfaces and let \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\). Determine the magnitude and direction of the constant force \(\mathbf{R}\), in \(\mathrm{N}\).

\(.\) An electric generator supplies electricity to a storage battery at a rate of \(15 \mathrm{~kW}\) for a period of 12 hours. During this \(12-\mathrm{h}\) period there also is heat transfer from the battery to the surroundings at a rate of \(1.5 \mathrm{~kW}\). Then, during the next 12 -h period the battery discharges electricity to an external load at a rate of \(5 \mathrm{~kW}\) while heat transfer from the battery to the surroundings occurs at a rate of \(0.5 \mathrm{~kW}\). (a) For the first 12 -h period, determine, in \(\mathrm{kW}\), the time rate of change of energy stored within the battery. (b) For the second 12 -h period, determine, in \(\mathrm{kW}\), the time rate of change of energy stored in the battery. (c) For the full 24-h period, determine, in \(\mathrm{kJ}\), the overall change in energy stored in the battery.

An air-conditioning unit with a coefficient of performance of \(2.93\) provides \(5,275 \mathrm{~kJ} / \mathrm{h}\) of cooling while operating during the cooling season 8 hours per day for 125 days. If you pay 10 cents per \(\mathrm{kW} \cdot \mathrm{h}\) for electricity, determine the cost, in, dollars, for the cooling season.

A mass of \(5 \mathrm{~kg}\) undergoes a process during which there is heat transfer from the mass at a rate of \(10 \mathrm{~kJ}\) per \(\mathrm{kg}\), an elevation decrease of \(75 \mathrm{~m}\), and an increase in velocity from \(10 \mathrm{~m} / \mathrm{s}\) to \(20 \mathrm{~m} / \mathrm{s}\). The specific internal energy decreases by \(10 \mathrm{~kJ} / \mathrm{kg}\) and the acceleration of gravity is constant at \(9.8 \mathrm{~m} / \mathrm{s}^{2}\). Determine the work for the process, in \(\mathrm{kJ}\).
See all solutions

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Measurements

Read Explanation

Physical Quantities And Units

Read Explanation

Magnetism

Read Explanation

Physics of Motion

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.