Calorimetry

Summary

This lecture on calorimetry by Florence Joie F. Lacsa, explores the different concepts and equations related to calorimetry as part of thermochemistry. The discussion includes heat capacity, types of calorimeters, and detailed problem solving for different calorimetry scenarios. Students are taught how to derive and apply the equations related to calorimetry from the first law of thermodynamics, making it clear and relatable with practical examples. The lecturer emphasizes understanding the origin and application of equations rather than rote memorization.

Highlights

• Florence Joie F. Lacsa gives an engaging chemistry lecture on calorimetry and its significance. 🎓
• The lecture begins with defining heat capacity and its importance in calorimetry. 📊
• Students learn to distinguish between molar heat capacity and specific heat capacity. 📏
• There's a deep dive into the types of calorimeters: constant pressure and constant volume. ⚙️
• Florence elegantly explains crucial equations used in calorimetry, like q = mCΔT, emphasizing understanding over memorization. ✍️
• Problem-solving examples throughout solidify understanding of theoretical concepts. 🧠
• Fun fact: Stainless steel cookware heats faster than the water it contains. 🍳
• The lecture promises further exploration into thermodynamics, hinting at the exciting concepts of heat engines. 🚗

Key Takeaways

• Calorimetry is an essential part of thermochemistry, unlocking how we measure heat transfer in chemical reactions. 🔥
• Heat capacity is a key concept, helping to determine how much heat is needed to change a substance's temperature by a degree. 📈
• The coffee cup calorimeter is a tool for understanding processes at constant pressure, and the bomb calorimeter at constant volume. 💡
• Heat transfer equations, such as q = mCΔT, are derived from the first law of thermodynamics, ensuring they're applied correctly. 📚
• Understanding the principles behind calorimetry helps solve real-world problems efficiently, from cookware to combustion. 🚀

Overview

Florence Joie F. Lacsa's engaging lecture introduces students to the fascinating world of calorimetry, a fundamental topic in thermochemistry. This presentation covers the intricacies of heat capacity, elaborating on both its molar and specific types, setting the foundation for understanding how heat transfer influences chemical reactions.

Highlights from the lecture include comprehensive explanations of the coffee cup and bomb calorimeters. These devices are pivotal in experimental chemistry, allowing students to practice measuring heat changes in various chemical processes. By focusing on how these instruments work at either constant pressure or volume, the lecture deepens students' practical knowledge.

Florence marries theory with practice through problem-solving sessions, guiding students to derive and apply equations like q = mCΔT. This method ensures students appreciate the origins and significance of these equations within the broader context of the first law of thermodynamics, preparing them for advanced studies in chemical thermodynamics.

Chapters

• 00:00 - 00:30: Introduction to Calorimetry and Objectives The chapter titled 'Introduction to Calorimetry and Objectives' is presented by Jovi Laksa, a lecturer in Chemistry for Engineers. It serves as the concluding part of the discussion on thermal chemistry. Calorimetry, the subject of this chapter, is introduced as the last segment in thermochemistry. The chapter begins with a brief welcome and introduction by the lecturer.
• 00:30 - 01:00: Heat Capacity and its Importance The chapter titled 'Heat Capacity and its Importance' aims to equip readers with the ability to define heat capacity and discuss its significance. It focuses on solving problems related to the heat capacity of systems. The chapter also demonstrates how the first law of thermodynamics applies to calorimetry, enabling readers to tackle problems involving calorimetry and understand the heat produced from chemical reactions.
• 01:00 - 01:30: Experimental Technique and Definitions The chapter titled 'Experimental Technique and Definitions' introduces the concept of calorimetry, an essential experimental technique used to determine thermal properties. Calorimetry is grounded in the principles of the first law of thermodynamics. A key apparatus in calorimetry is called a calorimeter. The chapter also highlights the importance of understanding crucial definitions, such as heat capacity, which is described as the amount of heat needed to increase the temperature of a substance.
• 01:30 - 02:00: Heat Capacity and Specific Heat Heat capacity is defined as the amount of heat required to change the temperature of a system by one degree. Molar heat capacity refers to the heat capacity of one mole of a substance, while specific heat capacity (or specific heat) refers to the heat capacity on a per gram basis.
• 02:00 - 03:30: Temperature Change in Celsius and Kelvin The chapter discusses temperature change measurements in calorimetry, particularly focusing on the equivalence of temperature changes measured in Celsius and Kelvin. It explains why a one-degree change in Celsius corresponds to a one-degree change in Kelvin and highlights the conversion difference in the Fahrenheit scale, where a one-degree change in Celsius equates to a 1.8-degree change in Fahrenheit.
• 03:30 - 07:00: Calorimetry Equations and First Law of Thermodynamics The chapter covers the basics of calorimetry equations and the first law of thermodynamics. It begins with a discussion on the steam point and ice point of water on the Celsius scale.
• 07:00 - 09:00: Constant Volume Process and Derivation of Equations The chapter discusses the constant volume process and involves a derivation of equations related to this process. The transcript portion provided compares temperature scales, specifically noting the steam and ice points in Kelvin and Fahrenheit scales. The Kelvin scale is discussed with steam at 373.15 K and ice at 273.15 K, while the Fahrenheit scale lists steam at 212°F and ice at 32°F.
• 09:00 - 11:00: Constant Pressure Process and Enthalpy The chapter 'Constant Pressure Process and Enthalpy' discusses the concept of temperature differences in various units. It explains that the difference between steam point and ice point is 100 degrees for both Celsius and Kelvin, while for Fahrenheit, the difference is calculated as 212-32.
• 11:00 - 14:30: Specific Heat of Substances and Calorimeters The chapter discusses the concept of specific heat of substances and the use of calorimeters. It begins with an explanation of dividing a given temperature interval, in this case 180 degrees, into 100 equal parts. This process and its relation to degrees Celsius is mentioned, though the summary is brief and lacks complete detail due to the transcript provided.
• 14:30 - 15:00: Problem Solving in Calorimetry The chapter discusses the relationship between temperature scales, specifically focusing on the increment differences between Celsius, Kelvin, and Fahrenheit. It explains that a one-degree change in Celsius is equivalent to a one-degree change in Kelvin, but a 1.8-degree change in Fahrenheit.
• 15:00 - 18:00: Molar Heat of Combustion Problem This chapter focuses on the concept of molar heat of combustion, exploring the related problem-solving aspects. It introduces the equation q = mcpΔt, explaining its significance in calculating the change in heat. The chapter emphasizes the importance of understanding temperature change specifically, rather than just converting temperatures between different scales. This forms a fundamental part of the discussions around thermodynamics and heat energy transfers as applied in chemistry.
• 18:00 - 22:00: Insulated Container Problem with Iron Bolts The chapter begins by introducing a common equation used in calorimetry, emphasizing the importance of understanding its origin rather than just using it. It follows with a derivation of the equation q = mCpΔT, starting from the basic definition of specific heat capacity (c).
• 22:00 - 27:00: Glucose Combustion in Bomb Calorimeter The chapter discusses the concept of specific heat, defined as the amount of heat required to raise the temperature of a certain substance. The discussion is linked to the first law of thermodynamics, suggesting that it serves as a foundational principle in understanding the topic. The chapter appears to use visual aids, such as pen colors, to illustrate key points. The focus is on the practical application of these concepts, possibly using a bomb calorimeter to measure the combustion of glucose.
• 27:00 - 36:00: Coffee Cup Calorimeter and Heat of Solution The chapter begins with a discussion about the relationship between internal energy change (ΔU), heat exchange (ΔQ), and work done (ΔW) in thermodynamic systems. The equation ΔU = ΔQ + ΔW is introduced with a brief mention of notation change when using differentials: dU = dQ - PdV. This highlights the first law of thermodynamics, emphasizing the conservation of energy within the system. The instructor attempts to clarify these concepts further, possibly indicating that this segment sets the foundation for understanding calorimetry and related heat calculations in solutions.
• 36:00 - 43:00: Preparation for Exam and Study Tips The chapter discusses study tips and preparation strategies for exams. It delves into specific methods such as understanding constant volume processes in physics or engineering, referencing the equation related to changes in internal energy (du) and pressure-volume work (pdv). The text starts with explaining these concepts through equations and their implications during constant volume processes, indicating a focus on thermodynamics principles with practical examples.

Calorimetry Transcription

• 00:00 - 00:30 hello everyone this is jovi laksa your lecturer in chemistry for engineers we are now on our last part of our discussion under thermal chemistry so let me minimize myself okay so as mentioned but know what i'm having cursor okay this is a calorimetry which is the last part of thermochemistry so after successful completion of this
• 00:30 - 01:00 lesson you are expected to be able to define heat capacity and discuss its importance solve problems related to heat capacity of a system demonstrate the understanding of the first law of thermodynamics as applied in calorimetry and for you to be able to solve problems involving calorimetry okay the amount of heat resulting from a chemical reaction
• 01:00 - 01:30 can be determined experimentally through one important experimental technique known as calorimetry it uses the principle of the first law of thermodynamics the dubai the device that is used in calorimetry is called your calorimeter in the study of calorimetry there are important definitions that we need to understand first the heat capacity which is defined as the amount of heat required to raise the temperature of the
• 01:30 - 02:00 system by one degree heat capacity answers the question how much heat is needed to change the temperature of a substance by one degree when we talk about heat capacity of one mole of a substance that is what we call as the molar heat capacity and if it is in per gram basis we call it the specific heat capacity which is also known as the specific heat
• 02:00 - 02:30 since calorimetry since in calorimetry rather the change in temperature is usually encountered let us take a look on the reasons why a degree change in cell shoes is the same as a degree change in kelvin and a degree change in cell shoes is the same as or is equivalent to 1.8 degree change in the fahrenheit scale let us
• 02:30 - 03:00 recall on the ice point and steam point of water okay so you have water you have here the eye steam point steam point and then the ice point of water okay at the degree celsius scale the ice point
• 03:00 - 03:30 is 100 my steam point rather and the ice point is zero in the kelvin scale this is 373 point 15 and this is 273 point 15 okay while in the fahrenheit scale the stem point of water is 212 and the ice point is 32.
• 03:30 - 04:00 okay if you get the ha difference between the steam point and ice point okay for degree celsius this is 100 for um kelvin this is also 100 and for fahrenheit this is to 1 to -32 is
• 04:00 - 04:30 180 and then if we try to divide the interval into 100 okay divide the whole interval by 100 say you have this interval one [Music] two for example this is equally divided into a hundred we know for a fact that this one in the degree celsius is
• 04:30 - 05:00 1 degree and 4 kelvin is also 1 degree and for fahrenheit it is 1.8 degree this is the explanation of this one degree change in celsius is the same as the increment in kelvin and when it comes to fahrenheit this one degree change in celsius is equivalent to 1.8 degree
• 05:00 - 05:30 change in fahrenheit okay again we are talking about the change not on the conversion of one temperature reading into another scale okay another this q is equal to mcp delta t i don't know if you encounter this one during high school but this is
• 05:30 - 06:00 one of the most commonly used equations in the study of calorimetry okay but you know in the fields of science and engineering it is not enough for us to simply use equations without knowing where they came from so with this let us derive the equation this q is equal to mcp delta t from the definition of specific heat c okay
• 06:00 - 06:30 what color am i using it's white so specific heat is defined let's note it as c a specific hit is defined as the heat that is required to increase the temperature of a certain substance right so we we also know from the first law of thermodynamics so let me use another pen
• 06:30 - 07:00 let me try blue that your delta u is equivalent to your delta q plus delta w okay or this can be written as d u is equal to d q minus um p [Music] this of v d q
• 07:00 - 07:30 therefore is d u plus p dv now if we have a constant volume process okay that constant volume process let's make it cv is equal to
• 07:30 - 08:00 du plus p dv all over dt but since we have a constant volume process this pdv here is zero okay therefore our cv is reduced into du over
• 08:00 - 08:30 dt this is also written as cv is the partial derivative of u over the partial derivative of of t at constant volume okay this is d u equals sorry this is v d u which is this one is equal to cv dt okay
• 08:30 - 09:00 and you can integrate this one from temperature one to temperature two okay we know that at constant volume your q is equal to your change of internal energy d u which also equivalent to your cv dt okay integrated from t1 to t2 so this is our working equation if we have a constant
• 09:00 - 09:30 volume process so this is what you're going to use if you have a constant volume process how about if we have a constant pressure process okay constant b
• 09:30 - 10:00 process from here do you from from here you have a cp sequel to d u plus p dv over dt but okay
• 10:00 - 10:30 we know that we know that your dh or your exchange in enthalpy dh is equivalent to d u plus d of your pv dh is equivalent to your du
• 10:30 - 11:00 plus d p dv plus v this of p but since it is oh this is a constant volume constant pressure sorry we are in a constant pressure process so constant b but since this is a constant pressure process this term is zero okay with that
• 11:00 - 11:30 your du will give us dh minus p dv right so let's substitute this here cp therefore is equal to dh minus p
• 11:30 - 12:00 dv plus p dv all over dt okay therefore let's cancel out this one therefore you have cp is equal to dh over dt so this is your dh over dt okay
• 12:00 - 12:30 therefore your cp is the partial derivative of h over the partial derivative of t at constant pressure this can also be written as dh is equal to cp dt evaluated from t1 to t2 so we know that at constant pressure your heat is equal to your change of
• 12:30 - 13:00 enthalpy which is your cp dt evaluated from t1 to t2 so this is this one and this are the same so we can rewrite this one as q p is equal to cp delta t where in delta t is your t2 minus t1 the m there is written
• 13:00 - 13:30 because your cp most of the time is a specific heat okay which is in joule per gram so you need the mass to remove the grams so there that is how we arrive to that q is equal to mcp delta t equation which we are going to use every now and then remember this cp is the specific heat at constant
• 13:30 - 14:00 pressure okay um here are uh here is the list of specific hits of some substances at 298 kelvin so let us take a closer look at the specific hit of iron and water okay in iron we only need point 45 joules of heat to raise a temperature of one gram of iron by one kelvin while in water we need 4.18
• 14:00 - 14:30 joules of heat to raise one gram of water by one kelvin this shows that we need more heat to change the temperature of one gram of water compared to the amount of heat that is needed to change one gram of iron this is the reason why your stainless steel cookware hits faster than the wear it and then the water it contains by the
• 14:30 - 15:00 way steel is made up of iron and traces of carbon we have two types of calorimeters being used one is the constant pressure calorimeter which is used to measure the heat transferred in processes open to the atmosphere and it is also used to find the specific heat capacity of a solid that does not react with or dissolve in water should we have a
• 15:00 - 15:30 face to face class you're going to use this coffee cup calorimeter in the laboratory for your calorimetry experiment but since it is an online test you'll do this in a virtual laboratory the working principle behind the pressure the constant pressure calorimeter is that the heat of reaction in the solution is absorbed by the um the solution okay so it follows
• 15:30 - 16:00 the first law of thermal dynamics so let us derive the working equation that we are to use when we are dealing with constant pressure calorimetry okay as mentioned let me change my color pen color white oops
• 16:00 - 16:30 okay as mentioned you have the heat lost is equal to the hit gained he gained what's wrong with me um hit gained by the system okay again
• 16:30 - 17:00 because energy is conserved whatever is loss is gained by something so in a constant pressure calorimetry the one that is losing heat is the reaction so say q lost by the reaction is the heat gained
• 17:00 - 17:30 by the ah the heat gained by the solution okay we have the reaction and solution here gained by the
• 17:30 - 18:00 solution right so in short this is your q reaction or your heat of reaction is gained by your solution which is characterized by the mass of the solution times the heat capacity of the solution and the change of temperature of the solution in short
• 18:00 - 18:30 the heat of reaction is equal to your mcp delta t of your um solution so this will be your working equation when you are dealing with problems on constant pressure calorimetry okay let's now move on to the next type of calorimeter which is the constant volume
• 18:30 - 19:00 calorimeter yen which is used to measure the heat that is released in a combustion reaction this calorimeter has a bump chamber where the combustion process occurs the heat that is released from the combustion process is gained by the calorimeter the presence of the reactor chamber is called the bump is what makes it constant volume calorimeter the bomb calorimeter the working principle
• 19:00 - 19:30 principle behind the constant volume calorimeter is that the heat of combustion coming from the reaction of fuel and oxygen inside the device is absorbed by the calorimeter again you have a heat loss is gained by something the law of conservation of energy of course the heat loss comes from the heat loss from the combustion
• 19:30 - 20:00 reaction is the heat gained by the calorimeter okay so your heat of reaction okay lost is equivalent to the specific heat of your calorimeter times the delta t of the calorimeter
• 20:00 - 20:30 okay so this or you can rewrite this as the the heat of reaction is equal to secant delta t so when you are working with bomb calorimeters this is your working equation again as uh engineering students it's not good that you memorize these formulas or keep using some formulas without
• 20:30 - 21:00 understanding where they are coming from okay [Applause] animation let's solve problems okay problem number one the molar heat of combustion of naphthalene with a molecular weight of 128.17 gram per mole is negative 1 to 2 8.2 kilocalorie per mole if 0.300 grams of naphthalene is burned
• 21:00 - 21:30 in the calorimeter causes a rise in temperature of 2.050 degrees celsius what is the total heat capacity of the calorimeter so we have a given what kind of calorimeter are we dealing okay since um naphthalene okay underwent combustion we are dealing with a bomb
• 21:30 - 22:00 okay and with the bomb calorimeter we know that our working equation is negative q of the combustion reaction is equivalent to your sikhal delta t okay we are given the molar heat of combustion which is uh the heat of combustion this is the q of reaction itself is equivalent to negative one two to
• 22:00 - 22:30 eight point two kilo calorie per mole of your naphthalene and then the molecular weight of your naphthalene is one 28.17 gram per mole mass of naphthalene that was burned is 0.300 gram and the delta t is 2.0 50 degrees celsius
• 22:30 - 23:00 so again we have an afterline burning it gave uh with the heat of combustion of negative one two to eight point two kilo calories per mole but there are only three grams of naphthalene that was burned we are uh to we are required to calculate for the heat capacity of the calorimeter when the recorded change of temperature is 2.050 degrees celsius for the solution
• 23:00 - 23:30 okay you have your working equation q of reaction is equal to secant delta t okay required is your sequel which is your negative q of reaction divided by delta t right so secal is equivalent to the negative of
• 23:30 - 24:00 negative 1 2 2 8 8.2 kilo calorie per mole of your naphthalene divided by the change of temperature which is 2.050 degrees celsius okay so however we burned only 0.300 grams of
• 24:00 - 24:30 naphthalene okay to make the units consistent we will use the molecular weight of naphthalene to remove the gram and mole from the equation so there are 128.17 grams of naphthalene per one mole of naphthalene right so let's check on the units the mole cancels out
• 24:30 - 25:00 gram cancels out so you have kilo calorie per degree celsius as our um final unit so you have there c call is equivalent to one two two eight point two times point three divided by one twenty eight point seventeen divided by two point zero two point zero five
• 25:00 - 25:30 so this is one point i'm gonna sign fig three one point four zero kilo calorie per degree celsius okay there number two we have two iron bolts of
• 25:30 - 26:00 equal mass um one at 100 degrees celsius and the other at 55 degrees celsius they are placed in an insulated container assuming that the heat capacity of the calorimeter is negligible what is the final temperature inside the container the specific heat of iron is point four five zero joule per gram kelvin so given are two metals so have metal a
• 26:00 - 26:30 okay and then metal b okay your mass of a and it uh the both metals have masses in a way that mass of a is equal to mass of b the temperature of metal a is 100 degrees celsius and the temperature of metal b is 55 degrees celsius okay this two were placed in an
• 26:30 - 27:00 insulated container so it's some sort of uh calorimeter okay has insulated sha all right assuming that the heat capacity of the calorimeter is negligible okay we are asked to solve for the final temperature inside the container when the specific heat of your iron is 0.450
• 27:00 - 27:30 joule per gram kelvin okay so how are we going to solve for this problem okay so of course we need to start from our first law of thermodynamics which is q lost is equal to q gained
• 27:30 - 28:00 right here's the catch heat transfer is always from a higher temperature to a lower temperature so the one that is losing okay losing um energy is metal a because it has a higher temperature so can we call this one excuse of a and then the other one as qs of b
• 28:00 - 28:30 we know that q is equivalent to your mcp delta t so can i move it here so you have your m cp delta t of your metal a is equal to your positive m cp delta t of your metal b right and then let's try to expand this further you have negative
• 28:30 - 29:00 metal a cpa times the final temperature minus the temperature of metal a is equivalent to your metal b heat capacity of your metal b and then the final temperature minus the initial temperature which is your uh temperature of metal b okay
• 29:00 - 29:30 we know that metal a and metal b are uh i'm at mass a okay the mass of metal a and mass of metal b are the same so maybe we can cancel them out from the equation okay so let us rewrite the equations you have now cpa
• 29:30 - 30:00 tf minus cpa ta is equivalent to c p b t f minus c p b p s of b okay and let's try to us substitute the uh let's just combine
• 30:00 - 30:30 the like terms first so t f okay negative c p a t f then minus c p b t f is equivalent to negative c p b t b plus
• 30:30 - 31:00 right so we know that cpa and cpb are the same so you have their c p b d f minus c
• 31:00 - 31:30 p b t f is equivalent to negative c p b t b plus c p b t v so we have here negative 2 c p b t is equal to
• 31:30 - 32:00 cancel oh this is a sorry impossible so when you're solving problems please be mindful okay so this is a one month transpose
• 32:00 - 32:30 negative all right so we now have um negative c p b or that's a positive c p b is negative t b plus t s of a okay so let's cancel out the
• 32:30 - 33:00 cpb cpb is cancelled out so you have their t f is equivalent to negative t b plus t a over negative two t f is negative fifty five plus one hundred all over
• 33:00 - 33:30 two parameter it should be negative here a let me check so m a c p a d f a t f minus t a m b c p b t f minus t b so c p a d f minus
• 33:30 - 34:00 so this is plus so the numerat plus cpa okay the cpata so this becomes minus so then be very careful eraser [Music]
• 34:00 - 34:30 so this is my nose c plus don so c p b so minus minus it should be minus minus minus minus minus minus that is should be minus as well
• 34:30 - 35:00 minus minus so your tf is equal to negative 155 all over two so the final temperature will give you 77.5 degrees
• 35:00 - 35:30 shoes okay so this is degree celsius degrees degrees celsius sorry i forgot to put it all right so there so let me check kuntama negatives m a cpa delta t mcpd okay m a df minus so distributed minus minus minus
• 35:30 - 36:00 is a plus okay so there 77.5 degrees celsius is the final answer problem number three under constant volume conditions the heat of combustion of glucose is 15.57 kilojoule per gram a 3.5 gram sample of glucose
• 36:00 - 36:30 is burned in a bomb calorimeter the temperature of the calorimeter is increased from 20.9 degrees celsius to 24.7 degrees celsius what is the total heat capacity of the calorimeter if the size of the glucose sample has been exactly twice as large what would the temperature change of the color calorimeter have been so given again is a constant volume
• 36:30 - 37:00 calorimetry so we have a bum calorimeter right and then uh this bomb calorimeter here is where the glucose combustion is to take place
• 37:00 - 37:30 okay we're in the queue or the heat of combustion is negative 15.57 kilojoule per gram there is no negative sign here but we know that combustion reactions are exothermic all right and then the mass of glucose that was burned is 3.500 grams
• 37:30 - 38:00 the temperature of the calorimeter is increased from 20.94 degree celsius to a t2 that is 24.72 degrees celsius okay we are asked to solve for sikhal for letter a solution is c okay for letter a since we have a
• 38:00 - 38:30 bomb calorimeter our working equation is negative q of our reaction is equal to secant delta t therefore secal is equal to negative q of the reaction divided by delta t so we have negative negative 15.57 kilo joule per gram
• 38:30 - 39:00 divide this by your delta t which is 24.72 degrees celsius minus 20.94 degrees celsius okay your delta t is t2 minus t1 okay again this is your t2 minus t1 and we burn 3.500 grams so let's solve for our sequel
• 39:00 - 39:30 let's check on our units okay gram cancels out kilojoule per degree celsius is correct way of reporting the c call 15.57 times 3.5 divided by 24.72 minus 20.94 so the denominator is enclosed by parentheses so you have
• 39:30 - 40:00 14. 42 kilojoules per degree celsius now for letter b we are asked to solve for a temperature change if the glucose sample that was combusted is twice as large so again we have the working equation
• 40:00 - 40:30 negative q or x n is equal to secant delta t so delta t this time is equal to c i know q reaction over sequel that is
• 40:30 - 41:00 delta t of a is negative q reaction over c call so our q reaction is negative 15. 57 kilo joule per gram divided by 14.42 kilojoule per degree celsius this is the
• 41:00 - 41:30 one that we calculated from above and it is mentioned that the size of the glucose sample which is 3.500 grams is twice as large okay check on the unit so you should have a degree celsius there as your final unit let's say we're looking for a change in temperature so kilojoules cancels out gram
• 41:30 - 42:00 cancels out c will go up so your delta t is equivalent to 15.57 times 3.5 times two divided by fourteen point forty two you have seven point fifty six let's see seven seven point how many
• 42:00 - 42:30 how many i know significant figures for seven point five five eight degrees celsius eight 15.57 times three point five times two divided by pointing for forty two seven point five five eight degree
• 42:30 - 43:00 celsius okay so but if you're going to use the uh unrounded value for letter a is 15.57 times 3.5 divided by 24.72 minus 20.94 you have 114.41666 if you're going to use that you have 15.57 times 3.5 times 2 divided by answer you
• 43:00 - 43:30 will get exact 7.56 so whichever you will use i will consider it um i said during the quizzes i'm after the process not only final answer alone okay so again if you use the 14.4116 what you will be getting is 7.56 right problem number four you place 50 milliliters of 0.5 molar
• 43:30 - 44:00 naoh in a coffee cup calorimeter which means we are dealing with a constant pressure this time at 25 degrees celsius and carefully add 25 milliliters of 0.5 molar hcl or your hydrochloric acid also at 25 degrees celsius after steering the final temperature is 27.21 degree celsius calculate the heat of the solution in joules and the change in
• 44:00 - 44:30 enthalpy in kilojoules per mole of water produced assume that the total volume is the sum of the individual volumes and that the final solution has the same density and specific hit as water which is 1.00 gram per ml and 4.184 joule per gram kelvin respectively so let's illustrate the given
• 44:30 - 45:00 so you have sodium hydroxide solution that is 50.0 milliliters and has a concentration of 0.500 molar and then this one is placed in a coffee cup calorimeter okay
• 45:00 - 45:30 temperature is 25 degrees celsius and carefully we add hydrochloric acid of 25 milliliters and a concentration of 0.500 molar also temperature is 25 degrees celsius so
• 45:30 - 46:00 and then the solution was steered until the final temperature is 27.21 degrees celsius okay we are to calculate for the heat of the solution so you have your q solution here as our unknown
• 46:00 - 46:30 so we are required to solve for the heat of solutions can you still recall our working formula if we have a constant pressure calorimetry the heat loss by the solute by the reaction is gained by the solution okay
• 46:30 - 47:00 so there let me rewrite it you're asked to solve for the q of the solution is the one in question and then for letter b the delta h of the reaction in kilojoules per mole of water produced but now before we proceed to letter b
• 47:00 - 47:30 let's first solve letter one at a time so that it is not overwhelming okay your q of solution is equivalent to your m c p delta t of your solution okay remember you have your q loss of the reaction is equivalent to your q gained of the
• 47:30 - 48:00 solution all right so we need to solve for the mass of the solution or look for the mass of the solution the heat the cpu of the solution and the delta t of the solution okay do we have a given mass of the solution we don't have right but we have the density given here
• 48:00 - 48:30 okay and we know for a fact that density is muscle over volume therefore your mass is density times volume do you have the density of the solution yes it has because it was uh mentioned in the problem that that we can assume that the final solution has the same density as water which is one gram per ml okay how about the volume of the solution it also also mentioned in the problem
• 48:30 - 49:00 that assume that the total volume is the sum of the individual volumes of the finals uh volumes of of course the two solutions that makes up the final solution which is 50 ml plus 25 ml how about the cp in the problem it was also mentioned that we can assume that the specific uh the final solution has the same specific hit as water which is 4.184 joule per
• 49:00 - 49:30 gram kelvin how about delta t the initial temperature uh before uh before this neutralization reaction took place so you have an initial temperature of 25 degree celsius and the final temperature of 27.21 degrees celsius so we can simply um substitute the given or to the to the to the working equation so you have your heat
• 49:30 - 50:00 of solution is equivalent to your density volume cp delta t of your solution okay so you should know how to uh manipulate your equations coming from the different concepts that you've learned since you were in the event high school not high school even elementary because adding total volume so that is 50 plus 25 is basically a grade
• 50:00 - 50:30 i think nowadays it's a grade one mathematics okay so your density is 1.00 gram per ml your volume is 25 ml plus let's start about 50 ml okay heat capacity is 4.184
• 50:30 - 51:00 joule per gram kelvin your delta t is 27.21 minus 25 that is degree celsius okay let's check on our units um blue ml cancels out with this two the gram cancels out you have a joule a unit of energy how
• 51:00 - 51:30 about the c what are you going to the c well with uh what are you going to do with the temperature a while ago we discussed that the degree change okay in cell shoes is the same as the change in kelvin so you have kelvin kelvin cell shoes
• 51:30 - 52:00 cancelling out because you can do long cut naman you translate 27.21 or not translate you are going to convert 27.21 degree celsius to kelvin so add plus 273 minus degree celsius plus 273 point 15 to make it kelvin so kelvin and kelvin so it's basically the same so that was a shortcut
• 52:00 - 52:30 so let's now i think we are ready to compute for the q of our solution we have 1 times 25 plus 50 is 75 times 4.184 times 27.21 minus 25 which is 600 93.98
• 52:30 - 53:00 you have been three sig fig so you have 6 90 3 joules okay now let's go to letter b okay calculate the heat of the solution in joule and the change in enthalpy in kilojoule
• 53:00 - 53:30 for moles so for letter b we are asked to solve for the delta h per mole of h2o produced so basically what really what we really have is a neutralization reaction between your naoh or sodium hydroxide and hydrochloric acid so your n and sodium and chlorine will form
• 53:30 - 54:00 [Music] a soap then you have some water all that is produced okay so let's check if this uh equation confirms to our um law of conservation of mass so sodium is one oxygen one hydrogen there are two and chlorine one so okay so we are asked to solve for this mole
• 54:00 - 54:30 of h2o because it is a denominator we know that our delta h is our q of the solution or your q of your reaction right or your q of your or the negative q of your solution if you don't have mole h2o so it seems that we're going to do some stoichiometry uh stoichiometry
• 54:30 - 55:00 solving here okay but where can we get the number of moles of the reactants you can get it from the given 50 ml of 0.5 molar 25 ml of 0.5 molar okay i don't know if you studied
• 55:00 - 55:30 molarity when you were in high school okay the molarity of a solution is equivalent to the number of moles okay a mole solute over liter of solution so your solute this point is your naoh for solution 1 and the hcl for solution 2 so we can rewrite this as m is equal to
• 55:30 - 56:00 n over l liter of solution therefore your n is m times liter of solution n is liter of solutions and is your molarity times your liter of
• 56:00 - 56:30 solution so let's solve for us for the number of moles of naoh and hcl so for your n naoh you have there the molarity of point five zero zero mole naoh per liter of solution al almerusha is only 50 ml to cancel the liter there and the ml you need the conversion
• 56:30 - 57:00 factor 1000 ml is 2 1 liter okay then you have n for your hcl you have their 0.5 mole per liter times 25 ml times 1 liter over 1000 ml
• 57:00 - 57:30 okay so let's check on the units liter cancels out ml cancels out so you have mole liter and ml cancels out you have bowl remaining so let's solve 0.5 times 50 divided by 1000 is point zero
• 57:30 - 58:00 point zero two five mole of naoh and then for hcl you have point 25 times 0.5 times 25 divided by 1000 point zero this is point zero one two five moles of hcl
• 58:00 - 58:30 okay so let's see how many moles of water were produced okay i hope you still recall the limiting reactance the concept on limiting reactants we did this when uh yeah aside from your general chemistry we have this during our review on the stoichiometric application on combustion reactions so first let's assume an aoe naoh as
• 58:30 - 59:00 our limiting reactant so the mass of the water of an entire must of your h2o produce using your naoh yes point and aoh number of moles of naoh
• 59:00 - 59:30 it's point zero [Music] mol of naoh times one mole of naoh produces and one mole of h2o basically it produces 0.025 mole of water
• 59:30 - 60:00 if we make use of of hcl as the limiting reactants we have point zero one two five mole of your hcl times one mole of hcl in the working equation okay one mole of your hcl will produce one mole of water this is
• 60:00 - 60:30 0.0125 mol therefore the number of moles that was produced is your h2o because your hcl is the limiting reactant we can now solve for this one so your delta h is equivalent to your q of reaction okay you see your q of reaction
• 60:30 - 61:00 your q delta h is your q okay all over q of your solution all over mole of h2o so where this our um q of solution and then
• 61:00 - 61:30 oh yeah 693 there you have your 6 93 joe so this is 6 93 joules per 0.0125
• 61:30 - 62:00 mole of water so what is the answer you have 693 divided by .0125 which is fifty five thousand four hundred forty dollars oh my god uh kilo joule
• 62:00 - 62:30 problem oh my god sorry for that we are required to we are required to express our answer in kilo joules in [Music]
• 62:30 - 63:00 okay oh calculator yes 6 6 93 divided by point zero one two five divided by one thousand
• 63:00 - 63:30 divided by 1000 so it says 55.44 so delta h is equal to 55.40 by the way it's negative here okay because delta h is the q of the delta h is the q of our reaction therefore the q of solution
• 63:30 - 64:00 is a negative negative q so this is negative 55.44 kilojoules per mole there and i have three sig figs kilojoule kilojoule
• 64:00 - 64:30 per mole per mole h2o so this is the final answer again um man um your delta h the bar is equal to your
• 64:30 - 65:00 q of reaction your negative q of reaction is equal to your q of your solution okay therefore your delta h says q reaction is your negative q of your solution that is why negative okay and uh
• 65:00 - 65:30 and if we try to check on see and if you try to check on let's see let's check on the [Music] reaction of your naoh and hcl from the internet okay so we'll go heap of reaction
• 65:30 - 66:00 naoh and hcl is the reaction x or endo it is exothermic because uh hcl at naoh is um strong acid and strong baso okay so what's this
• 66:00 - 66:30 so whatever they say let me just remove that one okay so again this are the list of references that you can use or again you can look for other as references the mind in addition to this so thank you for listening so the coverage of your let's save it the coverage of our exam is our example
• 66:30 - 67:00 is from the start of thermodynamics so everything is integrated you don't know if i'm going to ask to solve for you for the combustion of reaction using thermochemical equation or the standard rate of formation muna or from a haslow before you put it in a calorimeter something like that so it's an
• 67:00 - 67:30 integrated uh sort of problem that is solvable for two hours it's not become solvable for two hours what i give kasim a student is times three to five of my time so if it is times three let's see us two hours is sixty times two that is 120 minutes divided by three if it's my multiplier is three i need to
• 67:30 - 68:00 solve the problem myself in 40 minutes but if my multiplier is five i need to solve it in 24 minutes so from 24 minutes to 40 minutes or so problem two hours so there is a multiplier between a teacher and a student so practice a teaching i don't know if everyone knows that but i am
• 68:00 - 68:30 practicing it so um but times three you multiply or it's easy but times five it's difficult but times four it's moderate so maximum now 40 minutes the muscle core minimum of 24 depending but it will should be answerable by two hours okay for the students of course that is the standard but uh it depends if you study because it can be that although it is answerable for
• 68:30 - 69:00 two hours three hours two hours i will give you one hour to submit because you need to submit the solution of course so there um but before the mind the quiz i'll give you a week to review pa um
• 69:00 - 69:30 and um i may call a synchronous meeting on monday to give you the chance to ask some questions that you need to ask prior to the quiz so so there um we're done with safety we're done with thermal chemistry so after this we are to have your laboratory now on
• 69:30 - 70:00 calorimetry and of course your chris before we proceed to your chemical thermodynamics it talks norman about the um second law of thermodynamics and its application to the um hit engines just an introduction to the hit engine so so yeah um i think syllabus so
• 70:00 - 70:30 see you see you monday and tuesday for my tuesday classes bye