Slides for the eLearning course Separation and purification processes in biorefineries (https://open-learn.xamk.fi) in IMPRESS project.
Section: Distillation
Subject: 0.3 Basic concepts of distillation
its the ppt about giving basic information about thermodynamic which is part of the chemical engineering. it includes all the laws and other features related thermodynamics like entropy-temperature and pressure and other things
Slides for the eLearning course Separation and purification processes in biorefineries (https://open-learn.xamk.fi) in IMPRESS project.
Section: Distillation
Subject: 0.3 Basic concepts of distillation
its the ppt about giving basic information about thermodynamic which is part of the chemical engineering. it includes all the laws and other features related thermodynamics like entropy-temperature and pressure and other things
Second law of thermodynamics (and third law of thermodynamics) as taught in introductory physical chemistry (including general chemistry). Covers concepts such as entropy, Gibbs free energy, and phase equilibrium.
Excess gibbs free energy models,MARGULES EQUATION
,REDLICH-KISTER EQUATION,VAN LAAR EQUATION
,WILSON AND “NRTL” EQUATION
,UNIversal QUAsi Chemical equation
Armfield Gas Absorption Column ExperimentHadeer Khalid
The absorption of CO2 from air to water was studied in Gas absorption column built by Armfield company. Lab report and experiment was part of Separation Lab.
Slides for the eLearning course Separation and purification processes in biorefineries (https://open-learn.xamk.fi) in IMPRESS project.
Section: Distillation
Subject: 1.1 Vapor Liquid Equilibrium
Types of Distillation & column internalsBharat Kumar
More:- https://chemicalengineeringworld.com
Distillation is a method of separating the components of a solution which depends upon distribution of the substances between a gas and liquid phase, applied to cases where all components are present in both phases.
* What is distillation ?
* Types of Distillation
* Batch Distillation
* Azeotropic Distillation
* Flooding
* Priming
* Coning
* Weeping
* Dumping
* Packed Column
* Tray column
* Reflux Ratio
* Relative volatility
* Distillation column
Distillation is a method of separating mixtures based on differences in volatility (volatility is the tendency of a substance to vaporize. Volatility is directly related to a substance's vapor pressure.) of components in a boiling liquid mixture. Distillation is a unit operation, or a physical separation process, and not a chemical reaction
Slides for the eLearning course Separation and purification processes in biorefineries (https://open-learn.xamk.fi) in IMPRESS project (https://www.spire2030.eu/impress).
Section: Distillation
Subject: 1.2 Flash distillation.
Therminol 66 Synthetic Heat Transfer Fluid. Cotizado en Bolivares en Venezuela.
Therminol Venezuela.
Therminol 66 is the world’s most popular high temperature, liquid phase heat transfer fluid. Suitable for operation up to 650° F (345° C), Therminol 66 is pumpable to 27° F (-3° C) and delivers exceptional performance.
Second law of thermodynamics (and third law of thermodynamics) as taught in introductory physical chemistry (including general chemistry). Covers concepts such as entropy, Gibbs free energy, and phase equilibrium.
Excess gibbs free energy models,MARGULES EQUATION
,REDLICH-KISTER EQUATION,VAN LAAR EQUATION
,WILSON AND “NRTL” EQUATION
,UNIversal QUAsi Chemical equation
Armfield Gas Absorption Column ExperimentHadeer Khalid
The absorption of CO2 from air to water was studied in Gas absorption column built by Armfield company. Lab report and experiment was part of Separation Lab.
Slides for the eLearning course Separation and purification processes in biorefineries (https://open-learn.xamk.fi) in IMPRESS project.
Section: Distillation
Subject: 1.1 Vapor Liquid Equilibrium
Types of Distillation & column internalsBharat Kumar
More:- https://chemicalengineeringworld.com
Distillation is a method of separating the components of a solution which depends upon distribution of the substances between a gas and liquid phase, applied to cases where all components are present in both phases.
* What is distillation ?
* Types of Distillation
* Batch Distillation
* Azeotropic Distillation
* Flooding
* Priming
* Coning
* Weeping
* Dumping
* Packed Column
* Tray column
* Reflux Ratio
* Relative volatility
* Distillation column
Distillation is a method of separating mixtures based on differences in volatility (volatility is the tendency of a substance to vaporize. Volatility is directly related to a substance's vapor pressure.) of components in a boiling liquid mixture. Distillation is a unit operation, or a physical separation process, and not a chemical reaction
Slides for the eLearning course Separation and purification processes in biorefineries (https://open-learn.xamk.fi) in IMPRESS project (https://www.spire2030.eu/impress).
Section: Distillation
Subject: 1.2 Flash distillation.
Therminol 66 Synthetic Heat Transfer Fluid. Cotizado en Bolivares en Venezuela.
Therminol Venezuela.
Therminol 66 is the world’s most popular high temperature, liquid phase heat transfer fluid. Suitable for operation up to 650° F (345° C), Therminol 66 is pumpable to 27° F (-3° C) and delivers exceptional performance.
The chemical energy of a system is changed as a result of a reaction. Calorimetry. Heat of combustion . Calculation of caloric content of sucrose or food. Combustion reaction.
Discusses macroscopic and microscopic properties of solids and liquids.
**More good stuff available at:
www.wsautter.com
and
http://www.youtube.com/results?search_query=wnsautter&aq=f
1. Understand that Energy is exchanged or transformed in all chemical reactions and physical changes of matter. As a basis for understanding this concept: (a) Students know how to describe temperature and heat flow in terms of the motion of molecules (or atoms) and (b) Students know chemical processes can either release (exothermic) or absorb (endothermic) thermal energy.
LF Energy Webinar: Electrical Grid Modelling and Simulation Through PowSyBl -...DanBrown980551
Do you want to learn how to model and simulate an electrical network from scratch in under an hour?
Then welcome to this PowSyBl workshop, hosted by Rte, the French Transmission System Operator (TSO)!
During the webinar, you will discover the PowSyBl ecosystem as well as handle and study an electrical network through an interactive Python notebook.
PowSyBl is an open source project hosted by LF Energy, which offers a comprehensive set of features for electrical grid modelling and simulation. Among other advanced features, PowSyBl provides:
- A fully editable and extendable library for grid component modelling;
- Visualization tools to display your network;
- Grid simulation tools, such as power flows, security analyses (with or without remedial actions) and sensitivity analyses;
The framework is mostly written in Java, with a Python binding so that Python developers can access PowSyBl functionalities as well.
What you will learn during the webinar:
- For beginners: discover PowSyBl's functionalities through a quick general presentation and the notebook, without needing any expert coding skills;
- For advanced developers: master the skills to efficiently apply PowSyBl functionalities to your real-world scenarios.
Essentials of Automations: Optimizing FME Workflows with ParametersSafe Software
Are you looking to streamline your workflows and boost your projects’ efficiency? Do you find yourself searching for ways to add flexibility and control over your FME workflows? If so, you’re in the right place.
Join us for an insightful dive into the world of FME parameters, a critical element in optimizing workflow efficiency. This webinar marks the beginning of our three-part “Essentials of Automation” series. This first webinar is designed to equip you with the knowledge and skills to utilize parameters effectively: enhancing the flexibility, maintainability, and user control of your FME projects.
Here’s what you’ll gain:
- Essentials of FME Parameters: Understand the pivotal role of parameters, including Reader/Writer, Transformer, User, and FME Flow categories. Discover how they are the key to unlocking automation and optimization within your workflows.
- Practical Applications in FME Form: Delve into key user parameter types including choice, connections, and file URLs. Allow users to control how a workflow runs, making your workflows more reusable. Learn to import values and deliver the best user experience for your workflows while enhancing accuracy.
- Optimization Strategies in FME Flow: Explore the creation and strategic deployment of parameters in FME Flow, including the use of deployment and geometry parameters, to maximize workflow efficiency.
- Pro Tips for Success: Gain insights on parameterizing connections and leveraging new features like Conditional Visibility for clarity and simplicity.
We’ll wrap up with a glimpse into future webinars, followed by a Q&A session to address your specific questions surrounding this topic.
Don’t miss this opportunity to elevate your FME expertise and drive your projects to new heights of efficiency.
Neuro-symbolic is not enough, we need neuro-*semantic*Frank van Harmelen
Neuro-symbolic (NeSy) AI is on the rise. However, simply machine learning on just any symbolic structure is not sufficient to really harvest the gains of NeSy. These will only be gained when the symbolic structures have an actual semantics. I give an operational definition of semantics as “predictable inference”.
All of this illustrated with link prediction over knowledge graphs, but the argument is general.
UiPath Test Automation using UiPath Test Suite series, part 4DianaGray10
Welcome to UiPath Test Automation using UiPath Test Suite series part 4. In this session, we will cover Test Manager overview along with SAP heatmap.
The UiPath Test Manager overview with SAP heatmap webinar offers a concise yet comprehensive exploration of the role of a Test Manager within SAP environments, coupled with the utilization of heatmaps for effective testing strategies.
Participants will gain insights into the responsibilities, challenges, and best practices associated with test management in SAP projects. Additionally, the webinar delves into the significance of heatmaps as a visual aid for identifying testing priorities, areas of risk, and resource allocation within SAP landscapes. Through this session, attendees can expect to enhance their understanding of test management principles while learning practical approaches to optimize testing processes in SAP environments using heatmap visualization techniques
What will you get from this session?
1. Insights into SAP testing best practices
2. Heatmap utilization for testing
3. Optimization of testing processes
4. Demo
Topics covered:
Execution from the test manager
Orchestrator execution result
Defect reporting
SAP heatmap example with demo
Speaker:
Deepak Rai, Automation Practice Lead, Boundaryless Group and UiPath MVP
DevOps and Testing slides at DASA ConnectKari Kakkonen
My and Rik Marselis slides at 30.5.2024 DASA Connect conference. We discuss about what is testing, then what is agile testing and finally what is Testing in DevOps. Finally we had lovely workshop with the participants trying to find out different ways to think about quality and testing in different parts of the DevOps infinity loop.
"Impact of front-end architecture on development cost", Viktor TurskyiFwdays
I have heard many times that architecture is not important for the front-end. Also, many times I have seen how developers implement features on the front-end just following the standard rules for a framework and think that this is enough to successfully launch the project, and then the project fails. How to prevent this and what approach to choose? I have launched dozens of complex projects and during the talk we will analyze which approaches have worked for me and which have not.
Slack (or Teams) Automation for Bonterra Impact Management (fka Social Soluti...Jeffrey Haguewood
Sidekick Solutions uses Bonterra Impact Management (fka Social Solutions Apricot) and automation solutions to integrate data for business workflows.
We believe integration and automation are essential to user experience and the promise of efficient work through technology. Automation is the critical ingredient to realizing that full vision. We develop integration products and services for Bonterra Case Management software to support the deployment of automations for a variety of use cases.
This video focuses on the notifications, alerts, and approval requests using Slack for Bonterra Impact Management. The solutions covered in this webinar can also be deployed for Microsoft Teams.
Interested in deploying notification automations for Bonterra Impact Management? Contact us at sales@sidekicksolutionsllc.com to discuss next steps.
Smart TV Buyer Insights Survey 2024 by 91mobiles.pdf91mobiles
91mobiles recently conducted a Smart TV Buyer Insights Survey in which we asked over 3,000 respondents about the TV they own, aspects they look at on a new TV, and their TV buying preferences.
State of ICS and IoT Cyber Threat Landscape Report 2024 previewPrayukth K V
The IoT and OT threat landscape report has been prepared by the Threat Research Team at Sectrio using data from Sectrio, cyber threat intelligence farming facilities spread across over 85 cities around the world. In addition, Sectrio also runs AI-based advanced threat and payload engagement facilities that serve as sinks to attract and engage sophisticated threat actors, and newer malware including new variants and latent threats that are at an earlier stage of development.
The latest edition of the OT/ICS and IoT security Threat Landscape Report 2024 also covers:
State of global ICS asset and network exposure
Sectoral targets and attacks as well as the cost of ransom
Global APT activity, AI usage, actor and tactic profiles, and implications
Rise in volumes of AI-powered cyberattacks
Major cyber events in 2024
Malware and malicious payload trends
Cyberattack types and targets
Vulnerability exploit attempts on CVEs
Attacks on counties – USA
Expansion of bot farms – how, where, and why
In-depth analysis of the cyber threat landscape across North America, South America, Europe, APAC, and the Middle East
Why are attacks on smart factories rising?
Cyber risk predictions
Axis of attacks – Europe
Systemic attacks in the Middle East
Download the full report from here:
https://sectrio.com/resources/ot-threat-landscape-reports/sectrio-releases-ot-ics-and-iot-security-threat-landscape-report-2024/
Epistemic Interaction - tuning interfaces to provide information for AI supportAlan Dix
Paper presented at SYNERGY workshop at AVI 2024, Genoa, Italy. 3rd June 2024
https://alandix.com/academic/papers/synergy2024-epistemic/
As machine learning integrates deeper into human-computer interactions, the concept of epistemic interaction emerges, aiming to refine these interactions to enhance system adaptability. This approach encourages minor, intentional adjustments in user behaviour to enrich the data available for system learning. This paper introduces epistemic interaction within the context of human-system communication, illustrating how deliberate interaction design can improve system understanding and adaptation. Through concrete examples, we demonstrate the potential of epistemic interaction to significantly advance human-computer interaction by leveraging intuitive human communication strategies to inform system design and functionality, offering a novel pathway for enriching user-system engagements.
4. THERMODYNAMICS Why study thermodynamics? 1. All chemical reactions result in heat transfer 2. Understanding heat transfer properties is important for building materials 3. Food is evaluated by the amount of energy it releases
5. THERMODYNAMICS Heat – the transfer of energy due to contact Thermal energy – the energy of an object directly related to temperature Temperature – measure of internal energy of an object due to particle motion (kinetic energy)
6. THERMODYNAMICS FIRST LAW OF THERMODYNAMICS The total amount of energy in the universe is constant (conservation of energy). Energy is not created or destroyed.
7. THERMODYNAMICS Different types of matter require different amounts of heat transfer to change the same amount of temperature. If provided the same amount of heat, which substance will feel the hottest? Water is unusual in that it can absorb and release a lot of heat without the temperature changing drastically.
9. THERMODYNAMICS Specific Heat Capacity – the amount of heat transfer required to change the temperature of one gram of a substance one degree Celsius or Kelvin c = q / m Δ T Specific heat capacity (J/g·ºC) Quantity of heat transferred (J) Mass (g) Temperature (ºC) Normally, you see the formula written as such: q = mc Δ T
10. THERMODYNAMICS Heat Capacity – The amount of energy required to raise the temperature of a sample by one degree Celsius or Kelvin. The units are J/°C. Heat capacity can also be given for a specific quantity of substance (i.e. 100g). The units are J
11. THERMODYNAMICS Heat Capacity – the amount of heat transfer required to change the temperature of a substance one degree Celsius or Kelvin c = q / Δ T heat capacity (J/ºC) Quantity of heat transferred (J) Temperature (ºC) Normally, you see the formula written as such: q = c Δ T Notice how the formula is almost identical to the previous formula, except it does not have mass (m)
12. THERMODYNAMICS Example 1: If a gold ring with a mass of 5.5 g changes in temperature from 25.0°C to 28.0°C, how much heat energy, in Joules, has it absorbed? The value of the specific heat capacity of gold is 0.129 J/(g·ºC) . Given: m = 5.5g Δ T = 28.0ºC – 25.0ºC = 3.0ºC c = 0.129 J/(g·ºC) c = q / m Δ T q = mc Δ T = (5.5g) x (0.129 J/g·ºC) x (3.0ºC) = 2.1285 J = 2.1 J Therefore the amount of heat absorbed is 2.1 J
13. THERMODYNAMICS Example 2: What would be the final temperature if 250.0 J of heat were transferred into 10.0g of methanol (c = 2.9 J / g • °C) initially at 20. °C? Given: m = 10.0g Δ T = x - 20ºC c = 2.9 J/(g·ºC) c = q / m Δ T Δ T = q / mc = 250.0 J / 10.0g x 2.9 J/g·ºC = 8.62 ºC Therefore the final temperature is 29ºC Δ T = x - 20 ºC 8.62ºC = x - 20ºC = 28.62ºC = 29ºc
14. THERMODYNAMICS Example 3: The temperature of a 250g sample of water is changed from 25.0°C to 30.0°C. How much energy was transferred into the water to cause this change? Calculate your answer in J, kJ, calories, and kilocalories. The heat capacity of water is 4.2J/g∙°C Let’s analyze the question… You’re solving for q, but you don’t know the following: What is a kJ? What is a calorie? What is a kilocalorie? 1kJ =1000J 1cal =4.2J 1kcal =4200J or 4.2kJ A calorie is the specific amount of heat required to raise the temperature of 1g of water by 1°C.
15. THERMODYNAMICS Example 3: The temperature of a 250g sample of water is changed from 25.0°C to 30.0°C. How much energy was transferred into the water to cause this change? Calculate your answer in J, kJ, calories, and kilocalories. The heat capacity of water is 4.2J/g∙°C Given: m = 250g Δ T = 30.0 ºC – 25.0ºC = 5.0°C c = 4.2 J/g·ºC q = mc Δ T = (250g) x (4.2 J/g·ºC) x (5°C) = 5250J = 5250 J = 5.25 kJ = 1.25 kcal = 1250 cal Therefore the energy transferred into the water is 5.2x10 3 J, 5.2kJ, 1.2kcal, and 1.2x10 3 cal
16. THERMODYNAMICS Calorimetry - the measure of heat change due to a chemical reaction Chemical reaction occurs inside
17. THERMODYNAMICS Calorimetry System – all objects that are being studied (usually the chemical reaction) systems typically are defined by boundaries Surroundings – all objects that are not the system
18. THERMODYNAMICS Calorimetry System and Surroundings: When solid NaOH is dissolved in water, heat energy is released. System : NaOH (s) Na + (aq) + OH - (aq) Surroundings : Water Na + OH - Heat energy
19. THERMODYNAMICS Calorimetry Open system - heat and matter may be transferred between the system and the surroundings Where is the system in the picture?
20. THERMODYNAMICS Calorimetry Closed system - no matter transfer occurs between the system and the surroundings Heat may transfer between system and surroundings
22. THERMODYNAMICS Calorimetry Bomb calorimeter - reaction chamber allows heat transfer to the surrounding water, all contained within an insulated container
24. THERMODYNAMICS Calorimetry Exothermic reaction – Reaction where the system releases heat to the surroundings - ALL calculations involving these reactions will have a negative answer Endothermic reaction – Reaction where the system absorbs heat from the surroundings - ALL calculations involving these reactions will have a positive answer
25. THERMODYNAMICS Calorimetry Let’s analyze the question… Normally, we solve for the total change in heat. This time, we solve for the same thing, but PER MOLE. So we simply use the specific heat formula to solve for q, and divide the answer by the number of moles (n) q = c Δ T Then, divide q by n q n Example 4: When 1.02g of steric acid, C 18 H 36 O 2 , was burned completely in a bomb calorimeter, the temperature of the calorimeter rose by 4.26°C. The heat capacity of the calorimeter was 9.43 kJ/°C. Calculate the total heat of combustion of steric acid in kJ / mol. Is this heat capacity or specific heat cpacity? # of moles
26. THERMODYNAMICS Calorimetry Example 4: When 1.02g of steric acid, C 18 H 36 O 2 , was burned completely in a bomb calorimeter, the temperature of the calorimeter rose by 4.26°C. The heat capacity of the calorimeter was 9.43 kJ/°C. Calculate the total heat of combustion of steric acid in kJ / mol. Given: m = 1.02g n = m/M = 1.02g / (12.011 x 18 + 1.008 x 36 + 16.00 x 2)g/mol = 1.02g / 284.486g/mol = 0.003585414 mol Δ T = 4.26ºC c = 9.43 kJ/ºC Because this question is asking for q per mole Why do we need to solve for moles?
27. THERMODYNAMICS Calorimetry Given: m = 1.02g n = 0.003585414 mol Δ T = 4.26ºC c = 9.43 kJ/ºC q = c Δ T = (9.43 kJ/ºC) x (4.26ºC) = 40.1718kJ q = 40.1718kJ n 0.003585414mol = 11204.23 kJ/mol Example 4: When 1.02g of steric acid, C 18 H 36 O 2 , was burned completely in a bomb calorimeter, the temperature of the calorimeter rose by 4.26°C. The heat capacity of the calorimeter was 9.43 kJ/°C. Calculate the total heat of combustion of steric acid in kJ / mol.
28. THERMODYNAMICS Calorimetry Given: m = 1.02g n = 0.003585414 mol Δ T = 4.26ºC c = 9.43 kJ/ºC q = 11204.23 kJ/mol Example 4: When 1.02g of steric acid, C 18 H 36 O 2 , was burned completely in a bomb calorimeter, the temperature of the calorimeter rose by 4.26°C. The heat capacity of the calorimeter was 9.43 kJ/°C. Calculate the total heat of combustion of steric acid in kJ / mol. Don’t forget scientific notation and significant digits! = 1.12 x 10 4 kJ/mol Therefore the total heat of combustion is -1.12 x 10 4 kJ/mol Don’t forget that this is an exothermic reaction! = - 1.12 x 10 4 kJ/mol
29. THERMODYNAMICS Calorimetry Example 5: 175g of water was placed in a coffee cup calorimeter and chilled to 10.°C. Then 4.90 g of sulfuric acid was added at 10.°C and the mixture was stirred. The temperature rose to 14.9°C. Assume the specific heat capacity of the solution is 4.2 J/g •°C. Calculate the heat produced in kJ and the heat produced per mole of sulfuric acid. Let’s analyze the question… You’re given the mass of water and sulfuric acid. Since both are placed in the calorimeter to form the “solution”, then their masses must be added together. m = 175g + 4.90g = 179.9g H 2 O + H 2 SO 4
30. THERMODYNAMICS Calorimetry Example 5: 175g of water was placed in a coffee cup calorimeter and chilled to 10.°C. Then 4.90 g of sulfuric acid was added at 10.°C and the mixture was stirred. The temperature rose to 14.9°C. Assume the specific heat capacity of the solution is 4.2 J/g •°C. Calculate the heat produced in kJ and the heat produced per mole of sulfuric acid. Let’s analyze the question a bit more… The question is asking for q, but also q per mole. What “mole” is it referring to? It’s asking for q per mole of H 2 SO 4 . So you must solve for the moles of H 2 SO 4 .
31. THERMODYNAMICS Calorimetry Example 5: 175g of water was placed in a coffee cup calorimeter and chilled to 10.°C. Then 4.90 g of sulfuric acid was added at 10.°C and the mixture was stirred. The temperature rose to 14.9°C. Assume the specific heat capacity of the solution is 4.2 J/g •°C. Calculate the heat produced in kJ and the heat produced per mole of sulfuric acid. Given: m = 179.9g n = m/M = 4.90g / (1.008 x 2 + 32.065 x 1 + 16.00 x 4)g/mol = 4.90g / 98.081g/mol = 0.049958708 mol Δ T = 14.9ºC - 10ºC = 4.9ºC c = 4.2 J/g·ºC
32. THERMODYNAMICS Calorimetry Example 5: 175g of water was placed in a coffee cup calorimeter and chilled to 10.°C. Then 4.90 g of sulfuric acid was added at 10.°C and the mixture was stirred. The temperature rose to 14.9°C. Assume the specific heat capacity of the solution is 4.2 J/g •°C. Calculate the heat produced in kJ and the heat produced per mole of sulfuric acid. Given: m = 179.9g n = 0.049958708 mol Δ T = 4.9ºC c = 4.2 J/g·ºC q = mc Δ T = (179.9g) x (4.2 J/g·ºC) x (4.9ºC) = 3702.342 J q = 3702.342 J n 0.049958708mol = 74108.04138 J/mol = 74 kJ/mol This is your first answer! This is your second answer!
33. THERMODYNAMICS Calorimetry Example 5: 175g of water was placed in a coffee cup calorimeter and chilled to 10.°C. Then 4.90 g of sulfuric acid was added at 10.°C and the mixture was stirred. The temperature rose to 14.9°C. Assume the specific heat capacity of the solution is 4.2 J/g •°C. Calculate the heat produced in kJ and the heat produced per mole of sulfuric acid. Given: m = 179.9g n = 0.049958708 mol Δ T = 4.9ºC c = 4.2 J/g·ºC q = 3702.342 J = 3.7kJ q = 74kJ/mol n Therefore the heat of reaction is -3.7kJ and the heat per mole of sulfuric acid is -74kJ/mol Remember that this is an exothermic reaction!!
34. THERMODYNAMICS Calorimetry Example 6: The reaction of HCl and NaOH is exothermic. A student placed 50.0 mL of 1.00 M HCl at 25.5°C in a coffee cup calorimeter and then added 50.0 mL of 1.00 M NaOH also at 25.5°C. The mixture was stirred and the temperature quickly increased to 32.4°C. What is the heat of the reaction in J/mol of HCl? The heat capacity of H 2 O is 4.2 J/g•°C. , and its density is 1.00g/mL. Let’s analyze the question… The reaction that occurs is HCl (aq) + NaOH (aq) H 2 O (l) + NaCl (aq) + energy Since we have equal volumes of both, then the total volume is 100.0mL. With a density of 1.00g/mL, the mass of the water is 100.0g. Once again, we are solving for q, but per mole.
35. THERMODYNAMICS Calorimetry Example 6: The reaction of HCl and NaOH is exothermic. A student placed 50.0 mL of 1.00 M HCl at 25.5°C in a coffee cup calorimeter and then added 50.0 mL of 1.00 M NaOH also at 25.5°C. The mixture was stirred and the temperature quickly increased to 32.4°C. What is the heat of the reaction in J/mol of HCl? The heat capacity of H 2 O is 4.2 J/g•°C, and its density is 1.00g/mL. Given: m = 100.0g c = 4.2 J/g•°C Δ T = 32.4°C-25.5°C = 6.9°C n HCl = CV = 1.00M x 0.0500L = 0.0500mol Remember to convert mL to litres
36. THERMODYNAMICS Calorimetry Example 6: The reaction of HCl and NaOH is exothermic. A student placed 50.0 mL of 1.00 M HCl at 25.5°C in a coffee cup calorimeter and then added 50.0 mL of 1.00 M NaOH also at 25.5°C. The mixture was stirred and the temperature quickly increased to 32.4°C. What is the heat of the reaction in J/mol of HCl? The heat capacity of H 2 O is 4.2 J/g•°C, and its density is 1.00g/mL. Given: m = 100.0g c = 4.2 J/g•°C Δ T = 6.9°C n HCl = 0.0500mol q = mc Δ T = (100.0g) x (4.2 J/g·ºC) x (6.9ºC) = 2898J q = 2898J n 0.0500mol = 57960J/mol = -5.8x10 4 J/mol Therefore the heat of the reaction is - 5.8x10 4 J/mol of HCl
37. THERMODYNAMICS Calorimetry Example 7: A sample of 2.65 g of carbon in the form of graphite was burned in oxygen in a bomb calorimeter. The temperature of the calorimeter increased from 22.25 o C to 30.55 o C. The calorimeter itself is made of 3.000 kg of lead (c lead= 300 J/g o C) and contains 2000.0 mL of water (4.18 J/ g o C). Determine the molar enthalpy of this sample of carbon’s combustion. Let’s analyze the question… What are the two components of the calorimeter? How does this affect the heat capacity of the calorimeter?
38. THERMODYNAMICS Calorimetry Example 7: A sample of 2.65 g of carbon in the form of graphite was burned in oxygen in a bomb calorimeter. The temperature of the calorimeter increased from 22.25 º C to 30.55 º C. The calorimeter itself is made of 3.000 kg of lead (c lead= 300 J/g o C) and contains 2000.0 mL of water (4.18 J/ g o C). Determine the molar enthalpy of this sample of carbon’s combustion. Given: Δ T = 30.55ºC – 22.25ºC = 8.3ºC m calorimeter = 3000.g + 2000.0g = 5000.g m% of the calorimeter: 60% lead, 40% water c calorimeter = 60% x 300J/gºC + 40% x 4.18J/gºC = 181.672J/gºC OR c calorimeter = 3000.g x 300J/gºC + 2000.0g x 4.18J/gºC = 908360J/ºC specific heat capacity heat capacity
39. THERMODYNAMICS Calorimetry Example 7: A sample of 2.65 g of carbon in the form of graphite was burned in oxygen in a bomb calorimeter. The temperature of the calorimeter increased from 22.25 º C to 30.55 º C. The calorimeter itself is made of 3.000 kg of lead (c lead= 300 J/g o C) and contains 2000.0 mL of water (4.18 J/ g o C). Determine the molar enthalpy of this sample of carbon’s combustion. Given: Δ T = 8.3ºC m calorimeter = 5000.g C calorimeter = 908360J/ºC n = m/M = 2.65g / 12.011g/mol = 0.220631088mol
40. THERMODYNAMICS Calorimetry Example 7: A sample of 2.65 g of carbon in the form of graphite was burned in oxygen in a bomb calorimeter. The temperature of the calorimeter increased from 22.25 º C to 30.55 º C. The calorimeter itself is made of 3.000 kg of lead (c lead= 300 J/g o C) and contains 2000.0 mL of water (4.18 J/ g o C). Determine the molar enthalpy of this sample of carbon’s combustion. Given: Δ T = 8.3ºC c calorimeter = 908360J/ºC n=0.220631088mol q = c Δ T = (908360J/ºC) x (8.3ºC) = 7539388 J q = 7539388 J n 0.220631088mol = 34171920.5J/mol = -3.42x10 4 KJ/mol Therefore the molar enthalpy is -3.42x10 4 kJ/mol
41. THERMODYNAMICS Calorimetry Example 8: How much water at 100.0 ºC must be added to 80.0 g of water at 25.0 ºC to give a final temperature of 85.0 ºC? (3 marks; T/I) Given: T f = 85ºC T 1 = 25ºC, m 1 =80.0g T 2 = 100ºC, m 2 = x 80.0 x 25ºC + x x 100ºC = 85ºC 80.0+ x 80.0+ x 2000 + 100 x = 6800 + 85 x 15 x = 4800 x = 320g Therefore 320g of water must be added