- [Instructor] Calorimetry refers to the measurement
of heat flow.
And a device that's used to measure heat flow
is called a calorimeter.
An easy way to make a calorimeter is to use two coffee cups.
So at the base here, we have one coffee cup,
and then we can also use another coffee cup
as a loose fitting lid.
And since this top coffee cup is loose fitting,
our calorimeter is exposed to the constant pressure
of the atmosphere.
Therefore, we could use this coffee cup calorimeter
for constant-pressure calorimetry.
Other components of our calorimeter include some water
and then we also have a stir bar to stir up the water,
and a thermometer to measure the temperature change
of the water.
Let's say we have 150.0 grams of water
at an initial temperature of 25.0 degrees Celsius.
Next, let's take a block of copper, 120.0 grams of it,
and let's heat up that block of copper
to 100.0 degrees Celsius.
Once the copper has reached that temperature,
we add the copper block to our calorimeter.
Here, we can see the copper block has been added
to the calorimeter, and since the copper
is at a higher temperature than the water,
heat flows from the copper block to the water,
and therefore the temperature of the water will increase,
which we will see on the thermometer.
So we'll see the temperature increase on the thermometer.
Heat is transferred from the copper block to the water
until thermal equilibrium has been reached.
And we know when thermal equilibrium has been reached
by looking at the thermometer
and measuring the highest temperature that's reached.
Let's say the final temperature is equal
to 30.0 degrees Celsius.
So at thermal equilibrium, both the pieces of copper,
both the copper block and the water
are at the same final temperature.
Next, let's calculate the heat gained by the water
by using the equation q is equal to mc delta T.
So q is what we're trying to calculate,
the heat gained by the water.
M is the mass of the water, which is 150.0 grams.
So we can write in 150.0 grams.
C is the specific heat of water, which is 4.18 joules
per gram degrees Celsius.
And delta T is the change in the temperature,
which would be the final temperature,
so Tf minus the initial temperature Ti.
The final temperature of the water is 30.0 degrees Celsius,
and the initial temperature of the water
was 25.0 degrees Celsius.
So 30.0 minus 25.0 is equal to 5.0 degrees Celsius.
So we can write that in.
And next, we look at units and see what cancels out here.
So the grams cancel out, degrees Celsius cancels out,
and we're left with joules as our unit.
So q is equal to, when we go to two significant figures,
this is positive 3.1 times 10 to the third joules.
The positive sign means that this was the energy gained
by the water.
Next, let's do the same calculation for copper.
So we're trying to find q.
The mass of the copper was 120.0 grams.
So we can plug that in.
The specific heat of copper is .39 joules
per gram degrees Celsius.
And let's think about the change
in the temperature of the copper.
The final temperature of the copper
was 30.0 degrees Celsius,
and the initial temperature of the copper
was 100.0 degrees Celsius.
So the change in the temperature would be 30.0 minus 100.0,
which of course is negative 70.0.
So let's plug in negative 70.0 degrees Celsius.
Once again, we see what cancels for our units.
Grams will cancel, degrees Celsius will cancel,
and our answer will be in joules.
So q is equal to, using two significant figures,
negative 3.3 times 10 to the third joules.
And the negative sign,
so this negative sign means this is the energy
that was lost by the copper.
Next, let's look at these two numbers that we got
from our calculations.
Let's think about the magnitude of these two numbers.
If our coffee cup calorimeter were a perfect insulator,
the magnitude of these two numbers would be the same.
So it could be something like 3.3 times 10
to the third joules for both of them.
But since these two numbers are not the same, right?
We can see that we've lost more heat from the copper
than we've gained in terms of energy for the water,
which means we could have lost some of the energy
to the environment.
So not all of the heat was transferred directly
to the water.
Some of it could have escaped our coffee cup calorimeter.
Next, let's think about calorimetry for a chemical reaction.
So before we do that,
let's review some terms for thermodynamics.
So the system is the part of the universe
that we are studying.
So in the case of a chemical reaction,
the reactants and the products make up the system.
The surroundings are everything else,
which would include the water in the calorimeter,
the coffee cup itself, the thermometer,
the environment outside,
so the surroundings are everything else.
And finally, the universe would be the system
plus the surroundings.
So the reactants and the products make up the system.
So that's what the S stands for here in our calorimeter.
That's our system.
And let's say we run a reaction,
and in the reaction, heat is given off.
So in that case,
heat would flow from the system to the surroundings,
and so the temperature of the water would increase.
So we would see that as the temperature increases
on the thermometer.
Next, we could calculate the heat gained by the water
by using our q is equal to mc delta T equation.
And let's say q is equal to positive 1.0 times 10
to the second joules.
The positive sign means that the water gained energy.
If we assume a perfect transfer of heat
from the system to the surroundings,
if the surroundings gained positive 1.0 times 10
to the second joules,
that means the system must have lost negative 1.0 times 10
to the second joules.
So the same magnitude, but we changed the sign here,
because if we're talking about
the energy lost by the system,
it's the same in magnitude, but opposite in sign.
Next, remember that our lid over here is loose fitting,
which makes this constant pressure calorimetry,
and therefore this heat that was transferred is the heat
that's transferred at constant pressure.
So we can write a subscript p in here, so qp.
The heat transfer at a constant pressure is the definition
for the change in the enthalpy delta H,
so we can write that qp is equal to delta H.
And when delta H is negative,
we're talking about an exothermic reaction.
So when a reaction is exothermic,
heat is transferred from the system to the surroundings,
and therefore we see an increase
in the temperature of the water.
Finally, let's think about an endothermic reaction.
In an endothermic reaction,
heat is transferred from the surroundings to the system.
So here we can show heat flowing
from the surroundings to the system.
Since energy is leaving the surroundings,
the temperature of the water will decrease
for an endothermic reaction.
And since heat is being transferred to the system,
we can go ahead and write heat over here,
we can go ahead and write heat on the reactants side,
and delta H would be positive for an endothermic reaction.
So for an endothermic reaction,
energy is transferred from the surroundings to the system,
and therefore the temperature of the water will decrease.