/*! 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 55 A closed, rigid tank having a vo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A closed, rigid tank having a volume of \(1.8 \mathrm{~m}^{3}\) contains a mixture of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) and water vapor at \(87^{\circ} \mathrm{C}\). The respective masses are \(12.5 \mathrm{~kg}\) of carbon dioxide and \(0.06\) \(\mathrm{kg}\) of water vapor. If the tank contents are cooled to \(25^{\circ} \mathrm{C}\), determine the heat transfer, in \(\mathrm{kJ}\), assuming ideal gas behavior.

Short Answer

Expert verified
Determine the heat transfer by applying Q= −12.5 kg * 0.657 kJ/kgK (360 K - 298 K ) + 0.06 kg * 1.43 kJ/kgK (360 K - 298 K ) which equals approximately -481.65 kJ.

Step by step solution

01

State the given data

Volume (V=1.8 m^3), mass of carbon dioxide ( m_{CO2} = 12.5 kg), mass of water vapor ( m_{H2O} = 0.06 kg), initial temperature (T1 = 87°C (360K) ), final temperature (T2 = 25°C (298 K)). Assume ideal gas behavior.
02

Ideal gas constant values

The molar mass of CO2 is 44.01 g/mol and the universal gas constant (R_u) is 8.314 J/(mol·K). Calculate the gas constant for CO2: R_{CO2} = R_u / 44.01 = 0.1889 J/gK = 0.1889 kJ/kg·K. Do the same for water vapor:
03

Calculate mass fraction

Calculate the total mass of the mixture: m_{total} = m_{CO2} + m_{H2O} = 12.5 kg + 0.06 kg = 12.56 kg.
04

Determine the specific heat capacities

For CO2: C_{v,CO2} = 0.657 kJ/kg·K.). For H2O: C_{v,H2O} = 1.430 kJ/kg·K..
05

Calculate change in internal energy

The change in internal energy ( ∆U ) is given by: ∆U = (mass of CO2 * C_{v,CO2} + mass of H2O * C_{v,H2O} ) * ∆T where ∆T = T1 - T2
06

Calculate the heat transfer

The heat transfer ( Q) is negative of internal energy change: Q = - ∆U

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 91Ó°ÊÓ!

Key Concepts

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

Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. It describes how energy moves and changes form. Understanding thermodynamics is crucial for solving problems related to energy and heat transfer, such as the one in our exercise. The core principles of thermodynamics can be broken down into several key laws:

1. The **First Law of Thermodynamics** states that energy cannot be created or destroyed, only transferred or converted from one form to another.
2. The **Second Law of Thermodynamics** states that entropy, or disorder, tends to increase in an isolated system.

In this exercise, we're primarily concerned with the First Law of Thermodynamics. The energy changes we're studying involve heat transfer due to temperature changes within a closed and rigid tank. This means there’s no change in volume and no work done by the system. We assume that the gases behave ideally, which means we can use simplified equations to relate their pressure, volume, and temperature.
Heat Transfer Calculations
Heat transfer is the movement of thermal energy from one object or substance to another. It can occur in three ways:

1. **Conduction** – direct heat transfer through a material
2. **Convection** – heat transfer through fluid movement
3. **Radiation** – heat transfer through electromagnetic waves

In our exercise, we're looking at heat transfer as the mixture cools down from 87°C to 25°C inside a tank. Using the specific heat capacities (\(C_v\)) of the gases involved, we can calculate how much heat (\(Q\)) energy is needed to achieve this temperature change.

The formula for heat transfer in such a scenario, assuming ideal gas behavior, is:

\[ Q = - \big( m_{CO2} \times C_{v,CO2} + m_{H2O} \times C_{v,H2O} \big) \times (T_1 - T_2)\]

Here, we calculate the negative of the internal energy change (\( \big( m_{CO2} \times C_{v,CO2} + m_{H2O} \times C_{v,H2O} \big) \times (T_1 - T_2) \)) to find the heat transfer, since the system is cooling down.
Specific Heat Capacity
Specific heat capacity (\(C_v\)) is a measure of how much heat energy is required to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). Different substances have different specific heat capacities, which determine how they respond to heat energy. In our exercise, we are dealing with carbon dioxide (\text{CO2}) and water vapor (\text{H2O}).

The values of specific heat capacities are:

\[ C_{v, CO2} = 0.657 \frac{kJ}{kg\bullet K} \]

\[ C_{v, H2O} = 1.430 \frac{kJ}{kg\bullet K} \]

These values indicate that water vapor can absorb more heat per unit mass compared to carbon dioxide for the same temperature change. This information is crucial when calculating the internal energy change and, thus, the heat transfer in the system.

By knowing the specific heat capacities and the masses of the gases, we can determine the total amount of heat energy transferred during the cooling process. This concept helps us understand how different gases contribute to heat exchange in a mixture, leading to the final calculation of heat transfer by application of the formula above.

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 mixture of \(0.6 \mathrm{~kg}\) of carbon dioxide and \(0.4 \mathrm{~kg}\) of nitrogen is compressed from \(p_{1}=1\) bar, \(T_{1}=295 \mathrm{~K}\) to \(p_{2}=3.5\) bar in a polytropic process for which \(n=1.26\). Determine (a) the final temperature, in \(\mathrm{K}\). (b) the work, in \(\mathrm{kJ}\). (c) the heat transfer, in \(\mathrm{kJ}\).

A \(27 \mathrm{~m}^{3}\) tank initially filled with \(\mathrm{N}_{2}\) at \(21^{\circ} \mathrm{C}, 35 \mathrm{kPa}\) is connected by a valve to a large vessel containing \(\mathrm{O}_{2}\) at \(21^{\circ} \mathrm{C}\), \(138 \mathrm{kPa}\). Oxygen is allowed to flow into the tank until the pressure in the tank becomes \(103 \mathrm{kPa}\). If heat transfer with the surroundings maintains the tank contents at a constant temperature, determine (a) the mass of oxygen that enters the tank, in \(\mathrm{kg}\). (b) the heat transfer, in \(\mathrm{kJ}\).

Moist air at \(33^{\circ} \mathrm{C}\) and \(60 \%\) relative humidity enters a dehumidifier operating at steady state with a volumetric flow of rate of \(230 \mathrm{~m}^{3} / \mathrm{min}\). The moist air passes over a cooling coil and water vapor condenses. Condensate exits the dehumidifier saturated at \(12^{\circ} \mathrm{C}\). Saturated moist air exits in a separate stream at the same temperature. There is no significant loss of energy by heat transfer to the surroundings and pressure remains constant at 1 bar. Determine (a) the mass flow rate of the dry air, in \(\mathrm{kg} / \mathrm{min}\). (b) the rate at which water is condensed, in \(\mathrm{kg}\) per \(\mathrm{kg}\) of dry air flowing through the control volume. (c) the required refrigerating capacity, in tons.

Moist air at \(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\), and \(40 \%\) relative humidity enters an evaporative cooling unit operating at steady state consisting of a heating section followed by a soaked pad evaporative cooler operating adiabatically. The air passing through the heating section is heated to \(43^{\circ} \mathrm{C}\). Next, the air passes through a soaked pad exiting with \(40 \%\) relative humidity. Using data from the psychrometric chart, determine (a) the humidity ratio of the entering moist air mixture. (b) the rate of heat transfer to the moist air passing through the heating section, in \(\mathrm{kJ}\) per kg of mixture. (c) the humidity ratio and temperature, in \({ }^{\circ} \mathrm{C}\), at the exit of the evaporative cooling section.

An air-handling system is being designed for a \(12 \mathrm{~m} \times\) \(12 \mathrm{~m} \times 2.5 \mathrm{~m}\) biological research facility that houses 3000 laboratory mice. The indoor conditions must be maintained at \(24^{\circ} \mathrm{C}, 60 \%\) relative humidity when the outdoor air conditions are \(32^{\circ} \mathrm{C}, 70 \%\) relative humidity. Develop a preliminary design of an air-conditioning and distribution system that satisfies National Institute of Health (NIH) standards for animal facilities. Assume a biological safety level of one (BSL-1), and that two thirds of the floor space is devoted to animal care. Since an interruption in ventilation or air conditioning could place the laboratory animals under stress and compromise the research under way in the facility, account for redundancy in your design.

See all solutions

Recommended explanations on Physics Textbooks

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.