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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: \( T_f = 410.2 \text{ K} - 273.15 \approx 137.05掳 \text{C} \)

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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.

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Most popular questions from this chapter

An insulated rigid tank is initially evacuated. A valve is opened and atmospheric air at \(20^{\circ} \mathrm{C}, 1 \mathrm{~atm}\) enters until the pressure in the tank becomes 1 bar, at which time the valve is closed. Is the final temperature of the air in the tank equal to, greater than, or less than \(20^{\circ} \mathrm{C}\) ?

Refrigerant \(134 \mathrm{a}\) enters an air conditioner compressor at \(3.2\) bar, \(10^{\circ} \mathrm{C}\), and is compressed at steady state to \(10 \mathrm{bar}, 70^{\circ} \mathrm{C}\). The volumetric flow rate of refrigerant entering is \(3.0 \mathrm{~m}^{3} / \mathrm{min}\). The power input to the compressor is \(55.2 \mathrm{~kJ}\) per \(\mathrm{kg}\) of refrigerant flowing. Neglecting kinetic and potential energy effects, determine the heat transfer rate, in \(\mathrm{kW}\).

Sensibly Built Homes Cost No More (see box Sec. 4.3). In energy-efficient homes, indoor air quality can be a concern. Research the issue of carbon monoxide and radon buildup in tightly sealed houses. Write a report including at least three references.

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 control volume has one inlet and one exit. The mass flow rates in and out are, respectively, \(\dot{m}_{i}=1.5\) and \(\dot{m}_{\mathrm{e}}=\) \(1.5\left(1-e^{-0.002}\right.\) ), where \(t\) is in seconds and \(\dot{m}\) is in \(\mathrm{kg} / \mathrm{s}\). Plot the time rate of change of mass, in \(\mathrm{kg} / \mathrm{s}\), and the net change in the amount of mass, in \(\mathrm{kg}\), in the control volume versus time, in s, ranging from 0 to \(3600 \mathrm{~s}\).
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