For more information, please see full course syllabus of AP Chemistry

For more information, please see full course syllabus of AP Chemistry

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### Calorimetry

- Specific Heat: Amount of heat(Energy) required to raise the temp of 1 g of a substance by 1°C. Unit is J/g°C.
- Heat lost/gained by a substance in a process is

q=mCΔT

m is mass in grams

C is Heat Capacity of the substance

ΔT is the Temperature Change experienced by the substance. - Constant Pressure Calorimetry: Heat lost/gained by one substance equals heat gained/lost by other substance. One of the substances is usually water:

q1 = −q2 - Constant Volume Calorimetry: Mass and Heat Capacity are combined into a single Heat Capacity for the apparatus, Cm, so the equation becomes:

q = CmΔT

### Calorimetry

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Thermochemistry 0:21
- Heat Capacity
- Molar Heat Capacity
- Constant Pressure Calorimetry
- Example 1
- Constant Volume Calorimetry
- Example 2
- Example 3

### AP Chemistry Online Prep Course

### Transcription: Calorimetry

*Hello, and welcome back to Educator.com; welcome back to AP Chemistry.*0000

*Today, we're going to be talking about calorimetry; we are going to be finishing off our discussion of thermochemistry.*0004

*We have talked about enthalpy, and a little bit about Hess's law, and things like that; we are going to finish it off with a discussion of calorimetry today.*0011

*So, let's just jump on in and see what we can do.*0018

*OK, so let's recall that, at constant pressure, the ΔH, the enthalpy, is actually equal to the heat of the reaction--which is very, very convenient for chemistry.*0022

*We do things at constant pressure so, when we talk about the heat and the enthalpy, they are the same.*0032

*Let's just write that down to begin with: recall that, at constant pressure, ΔH is equal to the q.*0038

*Remember energy: you can transfer energy to and from a system...or the change in energy of a system--the internal energy of a system, ΔE...there are two ways that energy can be changed.*0053

*It can either transfer as heat in or out of the system, or as work; in other words, you can do work on the system, or work out of the system.*0066

*So, q was heat; w was work, and it was usually pressure-volume work; and at the constant pressure, as it turns out, the enthalpy is equal to the heat of reaction.*0074

*We speak about the heat of a particular reaction: 2 H _{2} + O_{2} goes to 2 H_{2}O; there is a certain heat that is given off--it is the enthalpy; it's written as ΔH=...negative for exothermic, positive for endothermic, things like that.*0085

*Now, let us define something called the heat capacity.*0100

*No, I'm not going to write "definition"; I just said that it is a definition.*0106

*Heat capacity: the heat capacity is basically the heat absorbed by a material, divided by the increase in temperature.*0110

*For example, let's say you had absorbed 20 Joules of heat--some material, we don't know what it is--and the rise in temperature was, let's say, 4 degrees Celsius.*0134

*Well, of course, you know, when you do division, basically what you are doing with all division is taking the denominator to 1; so, this is going to be Joules per 1 degree Celsius.*0147

*That is the whole idea behind units; so 20 over 4--you get 5 Joules per degree Celsius.*0156

*That means, for every 5 Joules of energy that you put into this thing, the temperature of that thing is going to rise 1 degree Celsius.*0163

*That is a general definition of heat capacity: heat absorbed, divided by the increase in temperature.*0170

*OK, now, temperature is going to be expressed sometimes in degrees Celsius, degree Kelvin--the scales for degree Celsius and degree Kelvin are the same, so the actual gradation is the same; they just differ by 273; so it's not a problem.*0176

*Sometimes, you will see J/K; sometimes you will see J/ΔC.*0193

*All right, now different substances have different heat capacities, and this is what is kind of interesting.*0198

*Now, as a standard, in order for us to have something to discuss...as a standard, we choose 1 gram of a substance.*0204

*When we choose 1 gram of a particular substance to apply a certain amount of heat to, to see how much the temperature rises, it is called the specific heat--so not just the heat capacity, but the specific heat capacity.*0217

*So, as a standard, we choose 1 gram of a substance, and call it the specific heat capacity.*0228

*Now, heat capacity is Joules per degree Celsius; well, when we choose one gram of that, it's going to be Joules per degree Celsius per gram.*0252

*The unit for specific heat capacity: the unit is Joules per degree Celsius per gram of substance, which is often written as Joules over gram-degree Celsius; that is the unit of heat capacity.*0263

*OK, there is also molar heat capacity, and it is exactly what you think it is.*0283

*It is going to be the amount of energy that I need to...the amount of energy that one mole of something absorbs, and rises by 1 degree Celsius.*0295

*Or, if I want another way of saying it, if I want to raise the temperature of 1 mole of something by 1 degree Celsius, or 1 degree Kelvin, how much heat do I have to put into it?*0305

*1 gram; 1 mole; we can work with both; so the unit of that is, of course, Joules per degree Celsius per mole, which is equivalent to (the way you will see it is) mole-degree Celsius.*0314

*And again, this can be a K instead of a C; it just depends on what the problem is asking for.*0333

*Specific heat capacity/molar heat capacity: the amount of heat required to raise a particular amount of something 1 degree Celsius.*0337

*OK, now let's talk about calorimetry.*0347

*There are two types of calorimetry: there is constant-pressure calorimetry, and there is constant-volume calorimetry.*0354

*We're going to start with constant-pressure calorimetry.*0359

*Constant-pressure calorimetry measures heats of reaction (you remember that little rxn; it's a shorthand for "reaction"), or the enthalpies of reaction, for reactions occurring in solution--so in aqueous solution or some kind of solution; it doesn't necessarily need to be water as the solvent--it could be anything else--but something that happens in solution.*0366

*If I have hydrochloric acid, and I mix it with sodium hydroxide, and I stir it all up to...you know that it's an acid-base neutralization reaction...it is going to give off some heat.*0409

*Well, in calorimetry, what you are doing is: that reaction is taking place in solution, so the heat that that reaction either gives off or absorbs is going to be reflected in the temperature of the solution.*0422

*We actually use the solution to measure decrease in temperature, and because we know the particular heat capacity of water, it is an indirect way of measuring the actual amount of heat that is being given off, or released, from the particular reaction.*0436

*It will make more sense when we do a little bit of a problem.*0452

*Now, recall that energy is conserved; so, if a reaction gives off heat (exothermic), the solution gets warmer.*0456

*You know that from experience; if a particular reaction is taking place in solution, the solvent that is containing it is like some sort of a container, if you will.*0489

*The reaction itself gives off heat; well, the water absorbs that heat, so the heat given off is equal to the heat absorbed.*0497

*That is sort of the fundamental idea behind calorimetry.*0504

*Calorimetry is a device that measures heats given off or absorbed, using a particular medium that we know the properties of; and that gives us the heat of a particular reaction.*0507

*OK, so we measure the rise in the temperature or the decline in the temperature, and then we calculate heat absorbed or released.*0521

*OK, now, energy released equals energy absorbed.*0529

*If a system is releasing heat, the surroundings are absorbing that heat.*0544

*If a system is absorbing heat, it is the surroundings that are giving off that heat; it's a trade-off, and it happens at the boundary.*0548

*That is why we talk about systems and surroundings and universes.*0556

*OK, now, this is the particular equation that we are interested in: mCΔt.*0561

*q is the energy or the heat; so energy or heat...heat and energy are interchangeable.*0571

*This right here--this is the mass of the solution, or the mass of the particular thing that you are discussing--it could be a solid; it could be a liquid--its mass, generally in grams.*0582

*This C right here is the symbol for heat capacity--specific heat capacity.*0594

*Heat capacity: I'll say it's heat capacity, and I'll put "specific," because more often than not, it is specific heat capacity.*0601

*Remember, we are talking about a gram of something; this is why this mass is in grams.*0610

*This is going to be specific heat capacity, and this ΔT (I'm sorry, this shouldn't be a small t) is a change in temperature.*0614

*So, that is the change in temperature; now, mind you, it is not the beginning or the ending temperature--it is the actual change in the temperature.*0625

*It is the final temperature, minus the initial temperature--that is the definition of Δ.*0633

*It is the change in temperature; so, if a particular problem says the temperature went from 25 to 30, our ΔT in this case is 5.*0639

*Or, if they say (a problem could say) there was a temperature increase of 4.7 degrees, our ΔT is 4.7; so this is the change.*0648

*OK, so once again: mass, times heat capacity, times the change in the temperature, is equal to the energy; let's make sure that the units match here.*0658

*Mass is going to be in grams; the heat capacity is in Joules per gram per degree Celsius; and the change in temperature is going to be in degrees Celsius.*0667

*Degrees Celsius cancels degrees Celsius; gram cancels gram, leaving you Joules--sure enough, Joules: heat or energy is in Joules.*0679

*This equation, right here, is our fundamental equation for calorimetry.*0688

*Anything having to do with heat capacity, work, the heat absorbed or released by a substance (liquid, solid, whatever) is equal to the mass of that substance, times its heat capacity, times the change in temperature that it experiences.*0692

*It's fantastic: 1, 2, 3, 4: you have 4 variables--given any three of them, you can find the fourth.*0711

*Let's say you have the heat, the mass, and the change in temperature; you can calculate the heat capacity.*0718

*Let's say you have the heat capacity, the temperature, and the q; you can calculate the mass.*0723

*Whatever it is--like any equation, you can rearrange it to get what it is that you need.*0728

*You will run into different problems where you have to solve for each one of those, so it isn't just solving for one or the other; this is just how it is arranged.*0733

*You will rearrange it accordingly.*0742

*OK, well, let's just go ahead and do a problem, and I think it will start to come together.*0745

*Let's do an example: we have: 50 milliliters of a 1 Molar hydrochloric acid solution (and this m with a line on it--that is just an older symbol for molarity; it is just something that I'm used to; you will see the capital M--just as a reminder) is mixed with 50 milliliters of a 1 Molar sodium hydroxide solution, NaOH.*0755

*OK, temperature rises from 25 degrees to 31.9 degrees (this is Celsius).*0794

*Calculate the heat released...let's see...and the molar heat released.*0811

*OK, so 50 milliliters of a 1 Molar hydrochloric acid solution is mixed with 50 milliliters of a 1 Molar sodium hydroxide solution; the temperature rises from 25 to 31.9 (let's make this 31; sorry, it looks like a 37 here); we want you to calculate the heat released in this reaction when I mix these, and the molar heat released (in other words, the heat released per formation of 1 mole of whatever it is that is actually reacting here).*0836

*Let's just write a reaction; this is chemistry--you always want to start with a reaction, if you can, which...more often than not, you are going to be starting with some reaction.*0867

*We are putting together hydrochloric acid, and we are putting together sodium hydroxide.*0875

*We have H and we have OH; this is a neutralization reaction.*0880

*Any time you have an H and an OH, they are going to seek each other out, and they are going to form liquid water.*0885

*Of course, what you have left over is the sodium chloride.*0892

*Once you actually write it out, then you balance it; in this particular case, everything is balanced, so our reaction is HCl + NaOH → H _{2}O, which I am actually going to rewrite a little differently, simply because I like to.*0895

*It's up to you; you can write H _{2}O; I have always been a big fan of writing it HOH, because I know that this H (let me do this in red)--this H comes from here, and this OH comes from here.*0908

*I like to keep them separate, especially when I am balancing things that aren't 1 to 1; it makes it a little bit easier to balance.*0921

*So, HOH is perfectly acceptable--unless your teacher says otherwise, but hopefully, they will be OK with it.*0928

*50 milliliters of a 1 Molar--all right, so let's see what we have here.*0935

*Well, they want the heat released, so q=mCΔT.*0942

*Well, we need to find the mass of the solution; so the heat that is going to be released in this reaction (let me draw a little picture here)--this reaction takes place throughout this solution.*0952

*The solution is 100 milliliters of solution; well, the heat released into it is going to go into the water.*0963

*The water is going to rise in temperature; that is what is happening here.*0970

*The mass of the water--well, 100 milliliters; if you have 100 milliliters, times...water is 1 gram per milliliter, so you have a mass of 100 grams.*0976

*This mass has to be expressed in grams.*0988

*We have 100 grams of water; and again, heat released is heat absorbed.*0991

*Heat released by the reaction is heat absorbed by the water, by the solution.*0995

*Well, we have 100 grams of water; now, the heat capacity of water happens to be 4.18 Joules per gram per degree Celsius.*1000

*What that means is that, if I have 1 gram of water, and if I want to raise it by 1 degree Celsius, I have to put in (I have to heat it by) 4.18 Joules.*1010

*It is a very high heat capacity; that is what makes water the extraordinary substance that it does--one of the things is the fact that it has such a high heat capacity.*1019

*The change in temperature, now: 25 to 31.9--what is that--6.9 degrees Celsius?*1027

*OK, so now, Celsius cancels Celsius; gram cancels gram; we do our multiplication, and, if I am not mistaken, we end up with (if I have done my arithmetic right) 2884.2 Joules.*1034

*Now, you can express this as kilojoules if you want; that is going to be 28.9 kilojoules; but I'm just going to go ahead and leave it as Joules here--it's not a problem.*1054

*That is how much heat is released, based on everything that they gave us.*1062

*Rise in temperature: the heat released by the reaction is absorbed by the water; this is what calorimetry is about; usually, water is the particular...because we know the heat capacity of water, we can measure it, so this reaction released this much heat.*1069

*Now, this was part A; now we want to find the molar heat; OK.*1086

*This is: now we want to find the amount of Joules per mole of whatever it is we are discussing.*1093

*In this particular case, this particular reaction is a neutralization reaction.*1100

*The reaction that is taking place is: H ^{+} is getting together with OH^{-} to form H_{2}O.*1104

*What we want to find is the amount of heat that is going to be released per mole of water formed.*1112

*That is what this is--as is: with 50 milliliters of a 1 Molar HCl, 50 milliliters of a 1 Molar NaOH, what we have is 28.9 kilojoules.*1122

*Now, we want to know what that is per mole of water formed.*1131

*OK, here is where we have to calculate how much is actually formed.*1135

*Well, 50 milliliters is 0.050 liters; times 1 mole per liter--that gives me 0.050 moles of HCl, which gives me 0.050 moles of H ^{+}, right?*1141

*One molecule of HCl releases 1 hydrogen ion.*1159

*Now, we do the same thing for the hydroxide.*1164

*We have 0.050 liters of the sodium hydroxide, and it is also 1 Molar; liters cancel; that gives me 0.050 mol of OH ^{-}, because 1 mole of sodium hydroxide releases 1 mole of hydroxide.*1167

*Well, this is nice: the equation is 1 to 1, so there is nothing limiting here; so every H is going to join with an OH--nothing is going to be left over, and we have 1:1:1.*1184

*We are not going to have .1 here; we don't add these; it says 1 mole of H ^{+} and 1 mole of OH^{-} goes to 1 mole of water.*1197

*We're going to form 0.050 moles of H _{2}O.*1204

*OK, so now, we can go ahead and finish the problem.*1213

*We said that we ended up releasing 2884.2 Joules in this reaction.*1219

*Well, the amount of water formed was 0.050 moles; so again, when you divide, you are taking the denominator, and you are turning it to 1.*1229

*So, this becomes 57684 Joules per mole, which is equivalent to 57.7 kilojoules per mole(when the numbers start to get this large, generally they will express it in kilojoules)--per mole of water formed.*1244

*So, a particular set of conditions gives you the amount of heat, just in Joules.*1270

*Then, if you adjust that by putting the number of moles of the thing that you actually created, or that you are talking about, in the denominator, that gives you the molar heat released.*1276

*There you have it; OK.*1289

*Now, that was constant-pressure calorimetry; and again, constant-pressure calorimetry takes place in solution, because you are not really changing the pressure of the environment when you are pouring--when you are working in a solution.*1295

*The pressure doesn't change; the pressure is the atmospheric pressure on that solution, so everything works out really nicely.*1307

*Well, there is also something called constant-volume calorimetry, and so let's talk about that a little bit.*1314

*Constant-volume calorimetry: this takes place in something called a bomb calorimeter.*1324

*So basically, what a bomb calorimeter is (I'll give you a quick...): it is this container, and it is full of water, and inside the container, there is another container.*1336

*That is connected to the outside; and what we do is: we put some compound in there; we ignite it; and we burn it--we combust it completely--we blow it up, basically.*1353

*Now, because it is constant volume--the volume doesn't change--we're keeping a constant volume, the pressure rises; the heat rises; the heat escapes into the surrounding medium.*1365

*Well, what we have done with a standard: we have actually already calculated the heat capacity for the entire calorimeter, which involves the metal, any parts, the solution, the water--everything in there.*1382

*So, for constant-volume calorimetry, it isn't q=mCΔT; it is actually C _{m}ΔT; this C_{m} actually includes (so I'll put it over here--q=mCΔT)...this C_{m} is the heat capacity of the entire bomb--of the calorimeter itself.*1396

*It includes--it has already taken into account--the heat capacity of the water, the mass of the water, the heat capacity of the metal, the mass of the metal...everything is taken into account; so this one is actually going to be Joules per degree Celsius, or Joules per degree Kelvin.*1422

*This is slightly different; so, for constant-volume calorimetry, this is our equation.*1441

*The mass is accounted for, which is why they ended up putting it as a subscript on the heat capacity; so now, it's just a symbol reminding you that we are talking about constant-volume calorimetry.*1448

*The principle is the same: so you are just taking something--in this case, you are blowing it up; you are completely combusting it, seeing how much heat it releases or...well, yes, it's going to release heat.*1460

*The heat passes through the bomb into the solution; the temperature of the solution rises; we measure that temperature change, and then, we just use this.*1470

*Same thing; so, let's do an example.*1480

*OK, let's do this one in red.*1484

*Example (let's see): 0.1964 grams of quinone (which is C _{6}H_{4}O_{2}) is placed in a bomb calorimeter; the bomb calorimeter is known to have a heat capacity of 1.56 (let me actually write this over here, so I can include the units)...C_{m} = 1.56 (not 7) kilojoules per degree Celsius.*1488

*So here, they are actually expressing the specific heat in kilojoules per degree Celsius.*1553

*Now, the rise in temperature is (I don't know where these lines are coming from; OK) 3.2 degrees Celsius.*1559

*Now, the rise in temperature--this is the ΔT, right here.*1575

*Calculate (whoa, that was interesting; we don't want that; OK) the heat of combustion per gram and per mole of quinone.*1584

*We want to calculate the heat of combustion; we want to calculate the enthalpy of this combustion reaction.*1619

*Quinone--so, just to let you know, we have this thing quinone--I'm just going to write out the reaction with names.*1625

*Combustion: Quinone + O _{2} → CO_{2} + H_{2}O.*1632

*All combustion reactions are the same; O _{2} is the other reactant; they produce carbon dioxide, and they produce water; this is a standard combustion reaction.*1638

*We want to find out what the heat is per gram of quinone, and per mole of quinone.*1647

*OK, so we use our constant-volume equation, which is C _{m}ΔT.*1652

*Well, we know what C _{m} is; it is 1.56 kilojoules per degree Celsius.*1665

*We want to multiply that by 3.2 degrees Celsius.*1673

*What you end up with, when you multiply this out: you get 4.99 kilojoules.*1678

*OK, 4.99 kilojoules is released; now, A: per gram--so 4.99 kilojoules was released, and we said that there were .1964 grams of this quinone, so we take the 4.99 kilojoules, and we divide it by the mass, 0.1964 grams.*1686

*We want Joules per gram; when you divide by something, you are making the denominator 1.*1718

*So, you end up with 25.4 kilojoules per gram; that is how much heat is released.*1725

*If you have a gram of quinone--not a lot--and you combust it, it releases--the bonds in quinone release--25.4 kilojoules of heat; that is a lot of heat.*1734

*Now, per mole: well, we just have to now change this to moles.*1748

*Well, we know how to do that: we use the molar mass, grams per mole.*1752

*We go to 25.4 kilojoules per gram, times grams per mole, the molar mass; and now, we look up quinone--the molecular formula.*1755

*It ends up being 108 grams per mole; we end up doing the multiplication; we get 2743 kilojoules per mole.*1771

*108 grams of quinone--that is 1 mole--it releases 2,743 kilojoules (in other words, 2 million, 743 thousand Joules) of heat.*1783

*That is a very, very, very powerful molecule; there is a lot of energy stored in those bonds.*1796

*So again, basic calorimetry: constant-pressure calorimetry--let's just see--constant P: you have that the heat comes from mass, times the heat capacity, times the change in temperature.*1802

*And you have constant volume--Constant V--which is q=C _{m}ΔT.*1819

*The mass has been accounted for; it is just one calorimeter; you have the bomb; everything is accounted for.*1828

*Now, this equation could be used for anything; as far as calorimetry is concerned, the heat capacity--notice, we have been using this for water, 4.18 Joules per gram-degree Celsius.*1834

*But, water doesn't have to be the thing that I am heating up; this equation is perfectly valid for anything that you heat up, whether it is iron, concrete...anything at all; water, mercury...*1845

*Each different substance has its own heat capacity, so there you have it.*1857

*OK, let's close off with one more problem, and it will sort of tie everything together.*1864

*OK, a 28.2-gram sample of nickel at 99.8 degrees Celsius is placed in--is dropped in, if you will--150.0 grams of water at 23.5 degrees Celsius.*1872

*So, the nickel is at 99.8 degrees Celsius; we drop it in water that is at 23.5 degrees Celsius.*1909

*That is what this says.*1915

*After thermal equilibrium is reached (in other words, after we allow everything to settle down, where now no more heat is being transferred between the metal and the water)--after thermal eq... (oh, let me write it out, actually--after thermal equilibrium is reached), the temperature of the solution is 25 degrees Celsius.*1919

*What is C--what is the heat capacity for nickel?*1952

*So here, we are taking hot nickel and dropping it in water; the heat is released into the water, so the water rises in temperature.*1959

*Now, from that, we want to be able to measure the heat capacity of the nickel.*1969

*OK, let's draw a little schematic of what is actually going on.*1974

*OK, I have the nickel here at a temperature of 99.8; this is actually a good way to do this, because you can actually see it pictorially--what is going on.*1978

*We have the water, which is at 23.5 degrees Celsius; this is degrees Celsius; now, this one drops in temperature; this one rises in temperature; and at some point, when thermal equilibrium is reached, they are both going to be at 25 degrees Celsius.*1988

*That is where we ended up; that is the thermal equilibrium.*2005

*OK, now, energy that is lost by the nickel is energy that is absorbed by the water.*2010

*So, in other words, the heat that the nickel gives off is equal to the heat that the water absorbs.*2023

*They are equal to each other; OK.*2037

*I'm going to run this calculation one way, and then I'm going to do something slightly different--at the end, I'm going to talk about this little negative sign that is going to show up out of nowhere.*2041

*So, q of nickel is mCΔT, and q of H _{2}O is mCΔT; so now, let me see; let's go ahead and put our values in.*2049

*We have 28.2 grams of nickel; we are looking for C--that is what we want--and ΔT is 25-99.8 (I don't know if I should do it this way or not; OK, that is fine); 25 minus 99.8--that is equal to the mass of the water, which was 150, and the specific heat of water is 4.18, and the change in temperature is 25 minus 23.5.*2077

*25 minus 23.5: that is the ΔT for the water--it goes from 23.5 to 25, and this one goes from (the final minus initial) 25 from the 99.8; that is why we have set it up this way.*2124

*OK, so let's do 28.2C, times (25 minus 99.8), equals 940.5 Joules; when we run through this math, we multiply this by this, divide through; we should end up with .446.*2141

*OK, this is the heat capacity for the nickel; this is in Joules per gram per degree Celsius.*2177

*OK, we said that the (so let's stop and take a look at this) heat capacity is not a negative number; this is just a mathematical artifact here, based on how we did the problem.*2187

*When you are doing this particular kind of problem, what you end up having to do is--because the heat absorbed is equal to the heat released, well, as it turns out, because one is releasing heat, it is losing, and the other one is gaining that heat, so it's positive.*2200

*So, as it turns out, the sum of those is actually equal to 0; that is the whole idea.*2227

*From one perspective, the loss is equal to the gain; therefore, to actually make them equal, you really have to stick a negative sign in front of it.*2234

*Or, you can just run this particular calculation as-is, just leaving everything: mCΔT, mCΔT.*2245

*Because one is dropping in temperature and one is rising in temperature, the Δ (because it's defined as final minus initial)--one of these is going to be a negative number.*2254

*That is where this little negative sign shows up; this is an artifact of the mathematics, based on how you did it.*2264

*If you want to run this same problem and not have to worry about this sign at the end, you can just say the q of one equals negative (the q of the other), because that is what it is.*2273

*One of them is a positive quantity--it is being absorbed; the other one is a negative quantity--it is being released.*2287

*It just depends on your perspective.*2294

*As far as the mathematics is concerned, one is the negative of the other.*2296

*So, you can go ahead and put the negative sign in here, and then run this calculation; or you can just run this calculation, doing final minus initial for nickel, final minus initial for water--leaving everything as it is, based on the definition of Δ.*2300

*And then, if you end up with a negative sign, just go ahead and drop that negative sign; the negative sign is a physical feature of the problem--it doesn't necessarily reflect the actual mathematics of the problem.*2317

*Heat capacity is a positive capacity, so we will just take the .446.*2329

*As long as you keep that in mind with problems of this nature, where something is dropping in temperature and something is rising in temperature, the final is going to be--one of them is going to be--negative (the ΔTs).*2334

*We have to account for that negative; you can account for it here, in the beginning, or you can account for it at the end.*2345

*I hope that that makes sense; we may actually be returning to a particular problem like this, maybe, to talk about it a little bit more, because sometimes there is a little bit of confusion regarding this.*2353

*OK, thank you for joining us for AP Chemistry, and thank you for coming to see us here at Educator.com.*2362

*Take good care; goodbye.*2367

0 answers

Post by Julia Zhu on April 23, 2017

Hi Prof. Raffi,

In my AP Chem class, we say that qlost=-qgained, which means that c for example three would be positive. The link to the Chemistry textbook we use for this is http://www.nonsibihighschool.org/advancedch13.php. Why would this not be correct? Thank you so much!

Best,

Julia

1 answer

Last reply by: Professor Hovasapian

Thu Nov 3, 2016 9:44 PM

Post by Warwick Shaw on October 31, 2016

In my AP Chem class we use the term calorimeter constant, is this the same as C sub m?

Also, my class is asked to calculate the q of solution as the -(q of aq + q of cal) when using a calorimeter. So I am asked to use your constant volume equation to find q of cal and plug it into this equation. Is that the purpose of your constant volume equation, to find the heat lost or gained by the calorimeter?

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Last reply by: Professor Hovasapian

Sat Aug 8, 2015 9:58 PM

Post by Jim Tang on August 8, 2015

Are we gonna have to know how to calculate C_m values?

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Last reply by: Jim Tang

Sun Aug 9, 2015 2:31 AM

Post by Jim Tang on August 8, 2015

Did you copy wrong? Where does 2884.2 come from?

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Last reply by: Professor Hovasapian

Fri Dec 26, 2014 3:17 AM

Post by Bryan M on December 25, 2014

In Example 1, why would 100 ml of the solution be equivalent to 100 ml of water. The 100 ml of solution contains HCL and NaOH, does that not effect the conversion to 100 gram of water because the solution is not 100 percent water? Thanks for the help and clarification.

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Last reply by: Professor Hovasapian

Wed Nov 19, 2014 5:44 AM

Post by Burhan Akram on November 17, 2014

Hello Prof. Raffi,

Are you planning on teaching anything related to Physics anytime soon?

Regards,

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Last reply by: Professor Hovasapian

Thu Oct 30, 2014 1:15 AM

Post by Delaney Kranz on October 27, 2014

Hello...in the first example, when you were looking at molar heat, why did you only work with the HOH? What about the molar heat required for the chemicals?

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Last reply by: Professor Hovasapian

Sun Jul 15, 2012 9:25 PM

Post by Ciara Flynn on April 9, 2012

How do you know which substance in a reaction will be "absorbing" the heat? I.e., which substance's mass do you calculate? And will they always give you the heat capacity of the substance, or should we memorize the common ones?

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Last reply by: Professor Hovasapian

Sun Jul 15, 2012 11:14 PM

Post by NGAWANG TSERING on March 8, 2012

hi...for constant pressure calorimetry, u told that molar heat capacity = j/ degree celsius x mol but example 1... to find molar heat you just divide 2884.2/ 0.050 mol ...why u didn't put degree celsius