- [Instructor] Calorimetry refers to the measurement
of heat flow.
And there are many different types of calorimeters.
In this case, we're looking at a constant volume calorimeter
which is also called a bomb calorimeter.
Let's look at how a bomb calorimeter works.
First, the sample to be combusted is placed
in a container that has some oxygen.
And then there's some ignition wires that go into
this little container here, and the sample is ignited
and heat is given off by the combustion reaction.
So heat is being transferred from our sample
to the water in our containers.
Let me go ahead and draw some water in here.
So imagine we have some water,
and it's important to know that the container is rigid.
So the walls are very, very solid and cannot move.
There's also something to stir the water
and since heat is being transferred
from the combustion reaction to the water,
the temperature of the water will increase
which we can see on the thermometer.
Now that we understand how a bomb calorimeter works,
let's think about that heat
that's being transferred from the combustion reaction
to the water, so that heat is q.
Let's go back to the first law of thermodynamics,
which says that the change in the internal energy
of the system is equal to q plus w,
where q is the heat that's transferred
and w is the work done.
Let's say we do this combustion reaction in a container
with a movable piston.
And the combustion reaction is performed
under the constant pressure of the atmosphere.
So this time, when we do the combustion reaction,
we will transfer some heat.
So heat is being transferred from the combustion reaction
and we would also produce some gases, which would push up
on the piston and so the piston would move up,
and since the piston's moving,
work is being done by the combustion reaction.
In this case, the heat that's transferred q
is done under constant pressure,
and so we can write qp here and by definition,
the heat that's transferred at constant pressure,
that's the change in the enthalpy delta H.
So for this example with the container
with the movable piston,
when we did our combustion reaction,
the heat that's transferred at constant pressure is equal
to the enthalpy delta H, the change in enthalpy delta H,
and as the gases expand and pushing the piston work is done.
Let's compare the example with the movable piston
to our bomb calorimeter.
Our bomb calorimeter has rigid walls
and therefore no work can be done.
So the work done is equal to zero.
When we plugged that into the first law of thermodynamics,
we find that the change in the internal energy delta E
is equal to the heat transferred q.
And since this is a constant volume
calorimeter right, the walls are rigid.
We can write q sub v here.
So this heat that's transferred from our combustion reaction
in this case is not equal to the change in the enthalpy.
It's equal to the change in the internal energy delta E.
So the heat that's transferred
at constant pressure is equal to the change
in the enthalpy delta H,
while the heat that's transferred at constant volume
is equal to the change in the internal energy delta E.
To do a constant volume calorimetry problem,
we need to know the heat capacity of the calorimeter
which is symbolized by C with a subscript cal.
To find the heat capacity of the calorimeter,
we need to combust something that we know the exact amount
of heat for them.
For example, if you combust exactly one gram
of benzoic acid,
you'll get 26.38 kilojoules released of energy.
So let's say we have a 0.2350 gram sample of benzoic acid.
And we put that in our calorimeter
and we go ahead and combust the benzoic acid.
And we find that the temperature increases
by positive 1.642 degrees Celsius.
To find the heat capacity for the calorimeter,
first we take our known amount
which is 26.38 kilojoules per gram
and we multiply that by how much benzoic acid
we used in our calorimeter which was 0.2350 grams.
And so the grams will cancel out
and this is equal to 6.199 kilojoules.
Next we divide this by our temperature change
which was positive 1.642 degrees Celsius.
And this gives us the heat capacity
of our calorimeter, which turns out to be 3.3775 kilojoules
per degree Celsius.
Now that we know the heat capacity
for our specific calorimeter,
we can use this value to calculate the heat
of combustion for another substance.
So the heat of combustion for another substance
or just q would be equal to the negative
of the heat capacity of the calorimeter
times the change in the temperature of the water
in that calorimeter.
Let's say our goal is to calculate the heat
of combustion of caffeine in kilojoules per mole.
So we take 0.265 grams of caffeine.
We put that in our calorimeter, we combust it
and we find the temperature of the water increases
by positive 1.525 degrees Celsius.
So to calculate q, q is equal to the negative
of the heat capacity of the calorimeter,
which is 3.775 kilojoules per degree Celsius.
And we multiply that by the temperature change
which has 1.525 degrees Celsius.
So degrees Celsius cancels out
and this gives us negative 5.757 kilojoules.
And technically this is the heat transfer
to a constant volume so we could even write q sub v in here,
and remember this is equal to the change
in the internal energy of our system.
So this is qv is equal to delta E.
Since our goal is to find the heat of combustion
of caffeine in kilojoules per mole,
next we need to take our grams of caffeine,
which is 0.265 grams, and divide that
by the molar mass of caffeine.
And so grams will cancel out and give us moles of caffeine.
So this calculation is equal to 1.36
times 10 to the negative third moles of caffeine.
Now, all we have to do is divide our heat,
which is negative 5.757 kilojoules by our moles
of 1.36 times 10 to the negative third moles,
to give us a final value of negative 4.23
times 10 to the third kilojoules per mole
with a negative sign, meaning heat is given off.
So we can say that this is the value,
this is the change in the internal energy
for our reaction in kilojoules per mole.
And often the change in enthalpy is about the same
as the change in the internal energy.
So we can say that this is approximately equal
to the change in the enthalpy for the reaction as well.