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### Course: AP®︎/College Chemistry > Unit 6

Lesson 3: Heat capacity and calorimetry# Constant-volume calorimetry

Constant-volume calorimetry is used to measure the change in internal energy, Δ

*E*, for a combustion reaction. In this technique, a sample is burned under constant volume in a device called a bomb calorimeter. The amount of heat released in the reaction can be calculated using the equation*q*= -*C*Δ*T*, where*C*is the heat capacity of the calorimeter and Δ*T*is the temperature change. Because the combustion occurs at constant volume,*q*is equal to Δ*E*for the reaction. Created by Jay.## Want to join the conversation?

- At5:28, why is there a negative sign in front of Ccal for the equation of q reaction?

Also, why is there no mass included in the equation for q reaction (just like there is in q = mc∆T). Is it because the general heat capacity for the calorimeter is used in place of the specific heat capacity (which is dependent on mass)?(8 votes)- There's a negative sign since the signs of both sides of the equation must be different. This is because it's describing the energy change of the reaction based on the temperature change of the calorimeter. The energy released by the reaction is the amount of energy gained by the calorimeter. So the energy of the reaction decreases as the energy of the calorimeter increase. Mathematically we can show this by having the changes in heat having different signs.

For your second question, yes you're correct. Here we're using heat capacity as opposed to specific heat capacity to calculate the heat. Heat capacity requires a temperature difference only, while specific heat capacity requires a temperature difference and a mass.

Hope that helps.(11 votes)

- Why does ∆H require a constant pressure?(5 votes)
- It's proved mathematically using a few substitutions of thermodynamic equations.

So the enthalpy of a system is defined as the sum of its internal energy and the product of pressure and volume.

H = E + PV

Where H is enthalpy, E is internal energy, P is pressure, and V is volume. And usually we're concerned with the change in enthalpy for a system so we use Greek deltas to indicate change. Except for pressure because we're assuming constant pressure here.

ΔH = ΔE + PΔV

Internal energy is also the sum of heat and work (first law of thermodynamics) so: ΔE = q + w

We also know that pressure-volume work is the work done by a system against the surroundings. It results in a change in volume at a certain pressure (such as atmospheric pressure). And since it is work done by the system, it is negative. So pressure-volume work is defined as: -w = PΔV (or w = -PΔV)

Substituting these definitions for internal energy (1) and pressure-volume work (2) into the previous equation for enthalpy yields:

ΔH = ΔE + PΔV

(1) ΔH = q + w + PΔV

(2) ΔH = q + -PΔV + PΔV

So, ΔH = q

And so enthalpy at constant pressure is simply the heat transferred into/out of the system. So enthalpy used in most chemistry applications is assumed to occur at constant pressure to simplify the math.

Hope that helps.(5 votes)

- for5:18, how is the heat capacity of the calorimeter the same for when it has caffeine vs benzoic acid? Wouldn't the different amounts of the different substances completely change how much heat is needed per C?(1 vote)
- The heat capacity of the calorimeter depends on the type of container used. Since they use the same calorimeter for both reactions, the heat capacity of the calorimeter would be unchanged.(2 votes)

- Why is heat transfer at constant volume not equal to ∆H?

What ∆H does actually represent?(1 vote) - So if ∆E = q and q = ∆H, why did he say (at7:40) that ∆H ~ ∆E and not ∆H = ∆E?

Is it because some of the energy was still lost from work?(1 vote) - In a closed volume, a change in energy is inevitably going to change the pressure (the gas law). Why is the change in pressure not considered in the equation q= -c* delta-T? Is it part of the calibration?(1 vote)
- At7:45, Jay said 'often the change in enthalpy is about the same as the change in internal energy'. I wonder why the statement is true. My guess is that there is only a small amount of sample participating in the rxn, so the change in pressure is negligible. Is that correct?(1 vote)
- Mostly correct. Enthalpy is a measure of the total heat content of a system. It's defined as the sum of internal energy and pressure times volume. If the reaction takes place in a constant pressure and volume system, then the change in that term (delta PV) would be zero, so enthalpy is equal to internal energy. if changes in pressure or volume are very small, we can approximate enthalpy to be the internal energy. The same can be said about changes (i.e. the change in enthalpy is approximately the change in internal energy, and so on).(1 vote)

## Video transcript

- [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.