the

Calorimeter Constant

we are going to talk about the

calorimeter constant okay

what's the calorimeter constant this is

going to be the energy that is absorbed

or released by the container when you

were doing some sort of thermodynamic

reaction I want you to pretend that

these are cups okay let's say that the

blue cup has cold water

the purple cup has hot water I pour the

hot water into the cold water you know

what's going to happen the cold water

will warm up the hot water will cool

down and they will come into thermal

equilibrium they'll meet at some

temperature in between the cold and the

hot temperature now when I pour this hot

water in here 100 percent of the energy

from the hot water does not go into the

cold water some of it goes into the cup

itself some of it goes into the

environment we are going to in our

experiments assume that all the energy

from this hot water goes to places that

it goes from the hot water into the cold

water and into the cup itself we'll

assume that we don't lose any to the

environment so when you're doing

experiments you want to be as quick as

possible turn pouring your hot water

into your cold water and cover it with a

lid as well once you pour it so the

question is well how much energy does

that cup absorb and that's the

calorimeter constant the abbreviation

for the calorimeter constant is C

subscript Cal so it's really kind of

like the specific key of the calorimeter

of the container that holds the reaction

I've also seen it called heat capacity

the heat capacity for the bond

calorimeter sometimes you'll see that

now notice what I put over here the

energy this my NTP absorbed or released

by the calorimeter itself so the energy

let's say absorbed by this calendar

meter it equals the calorimeter constant

times the change in temperature now why

this looks different we're used to same

q equals c m delta T there's no M

there's no massiveness we don't care

about the mass of

calorimeter we're taking that whole

calorimeter that cup or that bomb

calorimeter we're taking it as a unit

one unit so there's no ends there's no

mess and look at the units the unit's

totally reflect this so remember Q is

energy joules the calorimeter constant

if unit is joules per degree C energy

absorbed for every one degree the

temperatures increased so energy

absorbed for every one degree that the

temperatures increase multiply that by

the change of temperature and Celsius

cancels out and end up with Jools so

that's how the unit's work on this okay

to find the calorimeter constant you

might be in a lab where you have to find

the calorimeter constant of a Styrofoam

cup how much energy does that Styrofoam

cup absorbed for every one degree that

it increases what I've done is taken an

experiment I have hot water and I have

cold water I have 60 grams of the hot

water it I warmed it up to 95 degrees

and it's final temperature when it

reached thermal equilibrium was 55

degrees so again remember I took the hot

water and I poured it into the cold

water like this the cold water it was a

mass of 60 grams its initial temperature

was 20 degrees and again that thermal

equilibrium was 55 what I want to do is

find the calorimeter constant so let's

start with our formula if you can write

this formula your you can do the rest of

everything you the rest of the math so

I'm thinking transfer of energy what's

giving away the energy was absorbing the

energy was exothermic what's endothermic

well if I pour the hot water into the

cold water the hot water gives away its

energy that's exothermic negative the

cold water absorbs the energy thus

endothermic positive so the energy

released by the hot water okay has to

equal the energy absorbed by the cold

water and

and the calorimeter the cup itself

remember how he said they cut itself we

know in reality the cup is going to

absorb some of the energy now we can't

create or destroy energy so all the

energy released by that hot water it's

going to go to places remember we're

assuming we don't lose any to the

environment that hot water that hot that

energy is going to go into the cold

water and into the calorimeter so let's

go ahead and put in our formulas so the

energy our formula for the hot water

this is going to be negative c + delta T

now students make a mistake this is

where they make it they forget to carry

the negative they'll even write it right

here

they know it's exothermic but then they

forget to carry it that's important make

sure put a star to yourself that you

always include that negative carry that

negative use it mathematically equals so

remember this is for the hot equals CM

delta T for the cold plus our

calorimeter constant to find the energy

absorbed by that calorimeter is the

calorimeter constant C kal times the

change in temperature times the change

in temperature now this change of

temperature is going to match the cold

water because we're pouring the hot

water into the cold water that cup the

calorimeter would have come into thermal

equilibrium with the cold water it would

also begin at 20 degrees C and once the

hot water is poured into it it will also

warm up to 55 degrees C so delta T for

the calorimeter is the same as the cold

water because the cold water was in that

calorimeter

all right let's plug in numbers so we

have negative so specific heat of water

4.184 joules divided by grand conspiracy

times the mass with 60 grams times okay

change of temperature

remember final minus initial let's write

that as a reminder T final minus T

initial my final temperature 55 degrees

C - 95 degrees C okay final minus

initial equals we will have 4.184 for

the specific heat of water times its

mass with 50 grams little note on this

mm the easiest way to get the grams of

the water issues a graduated cylinder

you know that this density of water is

one gram per mil so if you have 50 mils

of water you have 50 grams of water just

a little tip there times our change of

temperature take final temperature 55

degrees C minus the initial was 20

degrees C plus I wanted to put the

calorimeter constant underneath the

calorimeter constant is our unknown it's

the one thing we're looking for

sea cow times change of temperature and

remember it matches the water so final

was 55 degrees minus the 20 degrees C

okay now we have a pretty good algebra

problem here let's begin um something

that I would do is I would well what I

tell my students is you can cancel out

the 4.1 84 if you subtract this whole

function across um let me do this for

you we are going to take four point I'm

going to do two all the algebra for you

so that you can see it because you might

not always have you might have a

different problem different specific

heat so let's go ahead and do this

remember order of operations we're going

to do 55 minus the 95 that's 40 55 minus

20 is 35 55 minus 20 is 35 now notice

over here it's a negative 55 minus 95 is

and negative 40 let's go ahead and

multiply so everybody have that negative

4.184

currently 24 times 60 times that

negative 40 and we get positive so this

is why it's really important that you

carry that negative 1 0 and 0 for 126

now little sign up sig figs I had 3 sig

figs here if you line up the sig figs 55

notice you truncate at this 10th place

so that would be 40 point 0 3 sig figs 3

sig figs I would round this when I

multiply I know that I would have 3 sig

figs right here

remember sig figs you have to be careful

is that when you change operations you

truncate so I have 3 sig figs here when

I multiply I can't have any more than 3

sig figs but we have 360 there this just

happened to work out nice okay let's

multiply over here are we on the 4.184

specific heat times 50 grams times 235

change of temperature

let me get seven thousand three hundred

and twenty-two and then I have my

unknown calorimeter constant times that

at 35 degrees C so let's subtract seven

thousand three hundred twenty-two from

both sides okay so doing sig figs on

this again I'm changing operations I

would have three sig figs here so I

would be truncating at the let's see

what would this be you guys the

hundredths place so when I truncate is

going to have to be at the hundreds

place when I when I subtract so let's go

ahead and subtract we will have let's

see here okay so if I have actually made

yet one more time

I have ten thousand sorry - the seven

three two - we get it's two six seven

eight but I know that we've got to

truncate

the hundreds place of the around that

would be 2,700 2,700 so whenever you

change operations between multiplication

division addition and subtractions

you've got to take time to truncate this

equals the see Cal times and remember

this was added three sig figs and this

is just a two sig figs so let's divide

both by the 35 so divided by 35 and we

get a calorimeter constant of 76 Oh

seventy six point five and I'm going to

do two sig figs so this is going to be

77 77 and my unit on that joules per

degree C okay what does that mean that

means when I pour the hot water into the

cold water all the energy is absorbed by

the water and for every one degree that

we increase 77 joules are absorbed by

the cold water by that excuse me by the

cups the calorimeter holding the cold

water so the energy goes to places into

the water and into the calorimeter for

every one degree increase 77 joules will

go into that old cup all right I want to

step back

overview really quick label everything

right your equation break that into your

equations plug in your numbers be

careful with three things be careful

with the sign be careful with your

algebra order of operations and be

careful with your sig figs and then

you'll get your final answer all right

calorimeter constant look at that

be proud of yourself as impressive good

work