/*! 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 62 A rigid, well-insulated tank of ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 62

A rigid, well-insulated tank of volume \(0.5 \mathrm{~m}^{3}\) is initially evacuated. At time \(t=0\), air from the surroundings at 1 bar, \(21^{\circ} \mathrm{C}\) begins to flow into the tank. An electric resistor transfers energy to the air in the tank at a constant rate of \(100 \mathrm{~W}\) for \(500 \mathrm{~s}\), after which time the pressure in the tank is 1 bar. What is the temperature of the air in the tank, in \({ }^{\circ} \mathrm{C}\), at the final time?

Short Answer

Expert verified
The final temperature of the air in the tank is approximately 137.05掳C.

Step by step solution

01

Define the known quantities

Identify all the given data in the problem. Initial conditions: - Volume, V = 0.5 m鲁 - Initial pressure, P鈧� = 1 bar - Initial temperature, T鈧� = 21掳C or 294 K - Power of electric resistor, P = 100 W - Time duration, t = 500 s - Final pressure, P鈧� = 1 bar
02

Determine the energy added by the resistor

Calculate the total energy added by the resistor using the formula: \( Q = P \times t \) Substituting the values: \( Q = 100 \text{ W} \times 500 \text{ s} = 50000 \text{ J} \)
03

Use the ideal gas law

Since we know the initial and final pressures and volume of the tank, use the ideal gas law to determine the initial and final states. The ideal gas law is: \( PV = nRT \)
04

Determine the mass of the air

Calculate the amount of air (in moles) using the initial conditions. Convert the pressure from bar to Pascals (1 bar = 100000 Pa): \( P_0 = 1 \text{ bar} = 100000 \text{ Pa} \) Using the ideal gas law: \( n = \dfrac{PV}{RT} \) Substituting the values for initial conditions: \( n = \dfrac{100000 \text{ Pa} \times 0.5 \text{ m}^3}{8.314 \times 294 \text{ K}} \) \( n \approx 20.46 \text{ moles} \)
05

Use the first law of thermodynamics

Apply the first law of thermodynamics for the process to find the final temperature: \( Q = nC_v螖T \) where \( C_v \) for air (assuming it as an ideal gas) is approximately 718 J/(kg路K) and mass \( m \) can be calculated as \( m = nM \) with the molar mass of air (\( M \)) around 29 g/mol or 0.029 kg/mol. \( n \approx 20.46 \times 0.029 \approx 0.5933 \text {kg} \) Rearranging the equation: \( 螖T = \dfrac{Q}{nC_v} \) Substituting the values: \( 螖T = \dfrac{50000 \text{ J}}{0.5933 \text{ kg} \times 718 \text{ J/(kg路K)}} \) \( 螖T \approx 116.2 \text{ K} \)
06

Calculate the final temperature

Add the temperature rise to the initial temperature to obtain the final temperature: \( T_f = T_0 + 螖T \) \( T_f = 294 \text{ K} + 116.2 \text{ K} \) \( T_f \approx 410.2 \text{ K} \) Converting to 掳C: \( T_f = 410.2 \text{ K} - 273.15 \approx 137.05掳 \text{C} \)

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 fundamental in understanding how gases behave under different conditions of temperature and pressure. The law is expressed as: \( PV = nRT \). Here:

鈥� **P** is the pressure of the gas
鈥� **V** is the volume of the gas
鈥� **n** is the number of moles
鈥� **R** is the universal gas constant, approximately 8.314 J/(mol路K)
鈥� **T** is the absolute temperature in Kelvin

In the given problem, we use this law to find out the initial and final states of the gas in the tank. By rearranging the equation, we can solve for the number of moles of gas: \( n = \frac{PV}{RT} \). This helps to know how much air enters the tank initially. Understanding the relationship between these variables becomes essential as we use it for further calculations in solving the problem.
First Law of Thermodynamics
The First Law of Thermodynamics, also known as the Law of Energy Conservation, states that energy cannot be created or destroyed in an isolated system. The law is represented as: \( Q = nC_v螖T \), where:

鈥� **Q** is the heat added to the system
鈥� **n** is the number of moles of the gas
鈥� **C_v** is the specific heat at constant volume
鈥� **螖T** is the change in temperature

In the problem, we calculate the energy added by an electric resistor using the equation: \( Q = P \times t \). The resistor transfers energy to the gas, increasing its temperature. By applying this law, we rearrange to find the change in temperature: \( 螖T = \frac{Q}{nC_v} \). This approach allows us to determine how much the temperature of the air inside the tank increases due to the energy supplied.
Energy Transfer
Energy transfer in this problem happens through the electric resistor. It continuously supplies energy to the air inside the tank at a rate of 100 W for 500 seconds. This transfer of energy can be calculated using the formula: \( Q = P \times t \). Substituting the values gives: \( Q = 100 \text{ W} \times 500 \text{ s} = 50000 \text{ J} \). Understanding this transfer of energy is crucial because it directly affects the temperature rise in the air within the tank. The concept of energy transfer helps us in determining the amount of heat introduced into the system, leading to a detailed understanding of the entire thermodynamic process.
Heat Capacity
Heat capacity is a property that describes how much heat a substance can store. For gases, we typically refer to two specific heat capacities: \( C_v \) (at constant volume) and \( C_p \) (at constant pressure). In this problem, since the volume of the tank remains constant, we use the specific heat at constant volume, \( C_v \), which for air is approximately 718 J/(kg路K).

This is important as it tells us how much heat energy is needed to raise the temperature of a kilogram of air by one Kelvin. The relationship is given by: \( Q = nC_v螖T \). Rearranging this, we get: \( 螖T = \frac{Q}{nC_v} \). This aids in determining the final temperature of the in-tank air after the energy has been transferred from the resistor.

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 feedwater heater operates at steady state with liquid water entering at inlet 1 at 7 bar, \(42^{\circ} \mathrm{C}\), and a mass flow rate of \(70 \mathrm{~kg} / \mathrm{s}\). A separate stream of water enters at inlet 2 as a two-phase liquid-vapor mixture at 7 bar with a quality of \(98 \%\). Saturated liquid at 7 bar exits the feedwater heater at 3 . Ignoring heat transfer with the surroundings and neglecting kinetic and potential energy effects, determine the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\), at inlet 2 .

A tiny hole develops in the wall of a rigid tank whose volume is \(0.75 \mathrm{~m}^{3}\), and air from the surroundings at 1 bar, \(25^{\circ} \mathrm{C}\) leaks in. Eventually, the pressure in the tank reaches 1 bar. The process occurs slowly enough that heat transfer between the tank and the surroundings keeps the temperature of the air inside the tank constant at \(25^{\circ} \mathrm{C}\). Determine the amount of heat transfer, in \(\mathrm{kJ}\), if initially the tank (a) is evacuated. (b) contains air at \(0.7\) bar, \(25^{\circ} \mathrm{C}\).

A well-insulated rigid tank of volume \(10 \mathrm{~m}^{3}\) is connected to a large steam line through which steam flows at 15 bar and \(280^{\circ} \mathrm{C}\). The tank is initially evacuated. Steam is allowed to flow into the tank until the pressure inside is \(p\). (a) Determine the amount of mass in the tank, in \(\mathrm{kg}\), and the temperature in the tank, in \({ }^{\circ} \mathrm{C}\), when \(p=15\) bar. (b) Plot the quantities of part (a) versus \(p\) ranging from \(0.1\) to 15 bar.

Steam at 160 bar, \(480^{\circ} \mathrm{C}\), enters a turbine operating at steady state with a volumetric flow rate of \(800 \mathrm{~m}^{3} / \mathrm{min}\). Eighteen percent of the entering mass flow exits at 5 bar, \(240^{\circ} \mathrm{C}\), with a velocity of \(25 \mathrm{~m} / \mathrm{s}\). The rest exits at another location with a pressure of \(0.06\) bar, a quality of \(94 \%\), and a velocity of \(400 \mathrm{~m} / \mathrm{s}\). Determine the diameters of each exit duct, in \(\mathrm{m}\).

Propane vapor enters a valve at \(1.6 \mathrm{MPa}, 70^{\circ} \mathrm{C}\), and leaves at \(0.5 \mathrm{MPa}\). If the propane undergoes a throttling process, what is the temperature of the propane leaving the valve, in \({ }^{\circ} \mathrm{C}\) ?
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Engineering Physics

Read Explanation

Particle Model of Matter

Read Explanation

Rotational Dynamics

Read Explanation

Thermodynamics

Read Explanation

Electromagnetism

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.