Enthalpy of formation

WEBVTT
Kind: captions
Language: en

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- [Instructor] Enthalpy of a formation

00:00:01.890 --> 00:00:03.830
refers to the change in enthalpy

00:00:03.830 --> 00:00:06.400
for the formation of one mole

00:00:06.400 --> 00:00:09.320
of a substance from the most stable form

00:00:09.320 --> 00:00:11.870
of its constituent elements.

00:00:11.870 --> 00:00:13.380
Change in enthalpy is symbolized

00:00:13.380 --> 00:00:17.060
by delta H and the f stands for formation.

00:00:17.060 --> 00:00:20.670
And the superscript
nought refers to the fact

00:00:20.670 --> 00:00:23.440
that everything is under
standard state conditions,

00:00:23.440 --> 00:00:25.600
which refers to atmospheric pressure

00:00:25.600 --> 00:00:28.830
of one atmosphere and
a specified temperature

00:00:28.830 --> 00:00:33.240
that is usually 25 degrees Celsius.

00:00:33.240 --> 00:00:36.490
So when we're thinking about
standard enthalpy of formation,

00:00:36.490 --> 00:00:38.100
we're thinking about the elements

00:00:38.100 --> 00:00:40.730
and the state that they exist
under standard conditions.

00:00:40.730 --> 00:00:43.880
So the elements have to be
in their standard states.

00:00:43.880 --> 00:00:48.420
So let's think about forming
one mole of carbon dioxide.

00:00:48.420 --> 00:00:51.370
So carbon dioxide is
composed of the elements

00:00:51.370 --> 00:00:53.680
carbon and oxygen.

00:00:53.680 --> 00:00:55.610
And under standard conditions,

00:00:55.610 --> 00:01:00.200
the most stable form
of carbon is graphite.

00:01:00.200 --> 00:01:02.710
So we're gonna write
carbon in the solid state

00:01:02.710 --> 00:01:05.590
and we're gonna write graphite over here.

00:01:05.590 --> 00:01:08.900
And next, when you think
about the most stable form

00:01:08.900 --> 00:01:11.910
of oxygen under standard conditions.

00:01:11.910 --> 00:01:15.030
And so at one atmosphere,
so atmospheric pressure

00:01:15.030 --> 00:01:17.590
and room temperature
of 25 degrees Celsius,

00:01:17.590 --> 00:01:20.700
the most stable form of
oxygen is oxygen gas.

00:01:20.700 --> 00:01:23.403
So we can go ahead and write in here O2.

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And since we're forming
one mole of carbon dioxide

00:01:27.200 --> 00:01:29.920
from the elements that
make up carbon dioxide

00:01:29.920 --> 00:01:33.620
in their most stable form
under standard conditions,

00:01:33.620 --> 00:01:35.590
the change in enthalpy for this

00:01:35.590 --> 00:01:38.420
would be the standard
enthalpy of formation.

00:01:38.420 --> 00:01:39.910
So we have our subscript f

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and our superscript nought
indicate standard conditions.

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The change in enthalpy for the formation

00:01:46.370 --> 00:01:51.247
of one mole of CO2 is equal
to negative 393.5 kilojoules

00:01:54.160 --> 00:01:56.673
per one mole of carbon dioxide.

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Let's look at some more
equations showing the formation

00:01:59.590 --> 00:02:01.210
of one mole of a substance.

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For example, let's look at the equation

00:02:02.930 --> 00:02:06.030
showing the formation
of one mole of water.

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So water is composed
of hydrogen and oxygen

00:02:08.490 --> 00:02:10.720
and the most stable forms
of those two elements

00:02:10.720 --> 00:02:14.170
under standard conditions are
hydrogen gas and oxygen gas.

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And for the coefficients
to make one mole of water,

00:02:17.590 --> 00:02:21.010
we need a 1/2 as our
coefficient in front of O2.

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The standard change in
enthalpy of formation

00:02:23.960 --> 00:02:26.270
for the formation of one mole of water

00:02:26.270 --> 00:02:30.720
is negative 285.8 kilojoules per mole.

00:02:30.720 --> 00:02:32.850
We can do the same thing for
the formation of one mole

00:02:32.850 --> 00:02:34.800
of methane CH4.

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We already know that the most stable form

00:02:36.640 --> 00:02:38.210
of carbon is graphite

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and the most stable form of
hydrogen is hydrogen gas.

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And the standard change
in enthalpy of formation

00:02:44.810 --> 00:02:46.720
for the formation of one mole of methane

00:02:46.720 --> 00:02:50.770
is equal to negative
74.8 kilojoules per mole.

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Next, let's think about
forming one mole of oxygen gas.

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Well, we're forming the oxygen gas

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from the most stable form of oxygen

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under standard conditions,

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which is also diatomic oxygen gas, O2.

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So we're not changing anything
we're going from O2 to O2.

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And since there's no change,
there's no change in enthalpy.

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Therefore, the standard enthalpy

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of formation is equal to zero.

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And this is true for the most
stable form of any element.

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The standard enthalpy of formation

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of the most stable form
of any element is zero

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since you'd be making it from itself.

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Standard enthalpies of formation
and kilojoules per mole

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are often found in the
appendices of many textbooks.

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And if you look in the
appendix of a textbook,

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you'll see the standard
enthalpy of formation

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for diatomic oxygen gas,
O2, is equal to zero.

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Ozone, which is O3, also exists
under standard conditions.

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However, it's not the
most stable form of oxygen

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under standard conditions and therefore,

00:03:49.930 --> 00:03:54.930
its standard enthalpy formation
is not zero, it's 142.3.

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Graphite is the most stable form of carbon

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under standard conditions.

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Therefore, it has a standard enthalpy

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of formation of zero, but of course,

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diamond also exists
under standard conditions

00:04:06.500 --> 00:04:07.860
but it's not the most stable form.

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So its standard enthalpy
formation is not zero,

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it's 1.88 kilojoules per mole.

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Enthalpies of formation
can be used to calculate

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the change in enthalpy
for a chemical reaction.

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We can do this by using
the following equation.

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The standard change in enthalpy
for a chemical reaction

00:04:26.160 --> 00:04:31.160
is equal to the sum of the
standard enthalpies of formation

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of the products minus the sum
of the standard enthalpies

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of formation of the reactants.

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Let's say our goal is to
find the standard change

00:04:39.810 --> 00:04:42.960
in enthalpy for the
following chemical reaction.

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So we have one mole of methane

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reacting with two moles of oxygen

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to form one mole of carbon
dioxide and two moles of water.

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The first thing we need to do is sum

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all the standard enthalpies
of formation of the products.

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So if we look at our
two products over here

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and we'll start with one
mole of carbon dioxide.

00:05:02.760 --> 00:05:04.460
So let's go ahead and
write this down here.

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We have one mole of carbon dioxide

00:05:07.120 --> 00:05:10.760
and the standard molar
enthalpy of carbon dioxide

00:05:10.760 --> 00:05:14.340
we've already seen as
negative 393.5 kilojoules

00:05:14.340 --> 00:05:16.590
per mole of carbon dioxide.

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So we're gonna multiply
one mole of carbon dioxide

00:05:18.980 --> 00:05:23.980
by negative 393.5 kilojoules
per mole of carbon dioxide.

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Our other product is two moles of water.

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So we're going to add
this to the other ones.

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We have two moles of H2O.

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And the standard enthalpy
of formation of H2O

00:05:40.040 --> 00:05:42.460
is negative 285.8.

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So we're gonna multiply this

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by negative 285.8 kilojoules per mole.

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So moles cancel out and we
get negative 393.5 kilojoules.

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And then for the other one,
moles cancel out again.

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And this would be plus
negative 571.6 kilojoules,

00:06:07.340 --> 00:06:12.340
which is equal to
negative 965.1 kilojoules.

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So that's the sum of all

00:06:15.550 --> 00:06:20.540
of the standard enthalpies
of formation of our products.

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Next, we need to sum
the standard enthalpies

00:06:23.170 --> 00:06:24.920
of formation of our reactants.

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So the two reactants that we
have are methane and oxygen

00:06:28.930 --> 00:06:31.730
and we have one mole of methane.

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So let's go ahead and write that in here.

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So we have one mole of methane.

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The standard molar enthalpy
of formation of methane

00:06:38.570 --> 00:06:42.260
is negative 74.8 kilojoules per mole.

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So we're multiplying one mole

00:06:43.430 --> 00:06:48.430
by negative 74.8 kilojoules per mole.

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Our other reactant is oxygen.

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And we know that diatomic oxygen gas

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has a standard enthalpy
of formation of zero.

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So we could go ahead and write this in

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just to show it.

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So we have two moles of oxygen

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but we're multiplying that number by zero.

00:07:07.300 --> 00:07:12.300
Some moles cancel and give
us negative 74.8 kilojoules.

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And we're adding zero to that.

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So negative 74.8 kilojoules

00:07:18.300 --> 00:07:22.980
is the sum of all the standard
enthalpies of formation

00:07:22.980 --> 00:07:24.547
of our reactants.

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So to find the standard change
in enthalpy for our reaction,

00:07:28.020 --> 00:07:30.820
we take the summation of
the enthalpies of formation

00:07:30.820 --> 00:07:35.040
of our products, which was
negative 965.1 kilojoules.

00:07:35.040 --> 00:07:36.940
And from that, we subtract the sum

00:07:36.940 --> 00:07:39.160
of the standard enthalpies of
formation of the reactants,

00:07:39.160 --> 00:07:43.220
which we found was
negative 74.8 kilojoules.

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So negative 965.1 minus negative 74.8

00:07:47.700 --> 00:07:52.700
is equal to negative 890.3 kilojoules.

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For the unit, sometimes
you see kilojoules,

00:07:56.130 --> 00:07:58.610
sometimes you see kilojoules per mole,

00:07:58.610 --> 00:08:01.830
and sometimes you see
kilojoules per mole of reaction.

00:08:01.830 --> 00:08:04.080
And what kilojoules per
mole a reaction means

00:08:04.080 --> 00:08:06.480
is how the balanced equation is written.

00:08:06.480 --> 00:08:07.530
For this balanced equation,

00:08:07.530 --> 00:08:10.790
we're showing the combustion
of one mole of methane.

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So combusting one mole of methane

00:08:12.570 --> 00:08:17.510
releases 890.3 kilojoules of energy.

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So that's what kilojoules
per mole of reaction

00:08:19.670 --> 00:08:20.573
is referring to.

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Let's go back to the step

00:08:22.730 --> 00:08:26.090
where we summed the standard
enthalpies of formation

00:08:26.090 --> 00:08:28.880
of the products to see how we
could actually get kilojoules

00:08:28.880 --> 00:08:32.460
per mole of reaction as our units.

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To do this, we need to
use a conversion factor.

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For how the equation is written,

00:08:36.500 --> 00:08:39.340
we're producing one
mole of carbon dioxide.

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So we can use as a conversion factor,

00:08:41.510 --> 00:08:46.510
there's one mole of carbon
dioxide per one mole of reaction.

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We can do the same thing
for our other product,

00:08:51.800 --> 00:08:52.930
which is water.

00:08:52.930 --> 00:08:54.500
For how the equation is written,

00:08:54.500 --> 00:08:56.790
we're forming two moles of water.

00:08:56.790 --> 00:09:01.440
So our conversion factor can
be there are two moles of water

00:09:01.440 --> 00:09:04.803
for every one mole of reaction.

00:09:06.190 --> 00:09:09.030
Next, moles of carbon dioxide cancels out

00:09:09.030 --> 00:09:11.660
and moles of water cancel out.

00:09:11.660 --> 00:09:14.710
And this gives us kilojoules
per mole of reaction

00:09:14.710 --> 00:09:16.350
as our units.

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It's a little more time-consuming

00:09:17.730 --> 00:09:19.310
to write out all the units this way.

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So often, it's faster
to do it the first way

00:09:21.840 --> 00:09:23.933
and add in these units at the end.


Worked example: Measuring enthalpy of reaction using coffee-cup calorimetry

WEBVTT
Kind: captions
Language: en

00:00:00.680 --> 00:00:02.600
- [Instructor] A constant
pressure calorimeter

00:00:02.600 --> 00:00:04.730
can be used to find the change in entropy

00:00:04.730 --> 00:00:06.980
for a chemical reaction.

00:00:06.980 --> 00:00:08.390
Let's look at the chemical reaction

00:00:08.390 --> 00:00:10.990
between an aqueous
solution of silver nitrate

00:00:10.990 --> 00:00:13.210
and aqueous solution of sodium chloride

00:00:13.210 --> 00:00:15.470
to form a precipitate of silver chloride

00:00:15.470 --> 00:00:18.490
and an aqueous solution of sodium nitrate.

00:00:18.490 --> 00:00:20.880
Let's say we have 25.0 milliliters

00:00:20.880 --> 00:00:25.490
of a 0.100 molar solution
of silver nitrate

00:00:25.490 --> 00:00:30.080
and 25.0 milliliters of
a 0.100 molar solution

00:00:30.080 --> 00:00:31.760
of sodium chloride.

00:00:31.760 --> 00:00:34.010
Both solutions are
initially at a temperature

00:00:34.010 --> 00:00:38.710
of 25.000 degrees Celsius.

00:00:38.710 --> 00:00:41.900
Next, we add our two
solutions to our calorimeter

00:00:41.900 --> 00:00:44.270
which has made of two coffee cups.

00:00:44.270 --> 00:00:47.240
And since the top coffee
cup is loose fitting,

00:00:47.240 --> 00:00:49.630
this reaction is under
the constant pressure

00:00:49.630 --> 00:00:50.800
of the atmosphere.

00:00:50.800 --> 00:00:54.080
So this is constant pressure calorimetry.

00:00:54.080 --> 00:00:57.570
After the two aqueous solutions
mix, the reaction occurs

00:00:57.570 --> 00:01:00.490
and we watch the thermometer
in the calorimeter.

00:01:00.490 --> 00:01:04.240
In this case, the temperature
of the solution increases

00:01:04.240 --> 00:01:05.860
and the final temperature,

00:01:05.860 --> 00:01:07.700
the highest one reached in our experiment

00:01:07.700 --> 00:01:12.100
is 25.781 degrees Celsius.

00:01:12.100 --> 00:01:14.130
So the change in the
temperature of the solution

00:01:14.130 --> 00:01:15.670
would be the final temperature

00:01:15.670 --> 00:01:17.220
minus the initial temperature,

00:01:17.220 --> 00:01:22.217
which is 25.781 minus 25.0

00:01:24.510 --> 00:01:29.510
which is equal to positive
0.781 degrees Celsius.

00:01:32.350 --> 00:01:36.170
The total volume of solution
would be 25 plus 25,

00:01:36.170 --> 00:01:41.170
which is equal to 50.0
milliliters of solution.

00:01:41.800 --> 00:01:43.640
If we assume the density of the solution

00:01:43.640 --> 00:01:45.840
is one gram per milliliter,

00:01:45.840 --> 00:01:50.840
50.0 milliliters is equal to 50.0 grams.

00:01:52.410 --> 00:01:55.580
Next we need to solve for
the heat gained by the water

00:01:55.580 --> 00:01:59.600
and we can use the Q is
equal to MC delta T equation.

00:01:59.600 --> 00:02:01.190
So we're solving for the heat,

00:02:01.190 --> 00:02:03.180
which is symbolized by Q,

00:02:03.180 --> 00:02:05.000
M is the mass of our solution

00:02:05.000 --> 00:02:07.530
which we saw was 50.0 grams,

00:02:07.530 --> 00:02:09.990
So we can plug that in.

00:02:09.990 --> 00:02:12.610
We can assume that the
specific heat of the solution

00:02:12.610 --> 00:02:15.240
is the same as the specific heat of water,

00:02:15.240 --> 00:02:20.050
which is 4.18 joules
per gram degrees Celsius

00:02:20.050 --> 00:02:22.170
and the change in the
temperature of the solution

00:02:22.170 --> 00:02:24.940
was 0.781 degrees Celsius.

00:02:24.940 --> 00:02:27.593
So we can plug that in as well.

00:02:29.630 --> 00:02:33.900
Grams cancels out, degrees
Celsius cancels out

00:02:33.900 --> 00:02:38.900
and we find that Q is equal to

00:02:39.040 --> 00:02:44.040
positive 1.63 times 10
to the second joules.

00:02:47.170 --> 00:02:51.030
The positive sign means that
energy was gained by the water.

00:02:51.030 --> 00:02:52.680
But let's think about the distinction

00:02:52.680 --> 00:02:56.520
between system and surroundings.

00:02:56.520 --> 00:02:58.740
The system consists of the reactants

00:02:58.740 --> 00:03:01.320
and products for our particular reaction

00:03:01.320 --> 00:03:03.320
and the surroundings are everything else

00:03:03.320 --> 00:03:05.880
which includes the water.

00:03:05.880 --> 00:03:06.770
So in this case,

00:03:06.770 --> 00:03:09.660
since the temperature of the
surroundings increased, right?

00:03:09.660 --> 00:03:11.830
We saw an increase in the temperature,

00:03:11.830 --> 00:03:13.110
that means that heat flowed

00:03:13.110 --> 00:03:15.880
from the system to the surroundings

00:03:15.880 --> 00:03:18.190
and so the surroundings
increased in energy

00:03:18.190 --> 00:03:20.823
and that's what we see with
this positive sign here.

00:03:21.940 --> 00:03:24.440
If we assume a perfect transfer of heat

00:03:24.440 --> 00:03:26.490
from the system to the surroundings,

00:03:26.490 --> 00:03:28.530
if the surroundings gained energy,

00:03:28.530 --> 00:03:31.210
that means the system lost energy.

00:03:31.210 --> 00:03:32.450
So if we're thinking about the heat

00:03:32.450 --> 00:03:34.830
transferred for the reaction,

00:03:34.830 --> 00:03:36.260
it's the same in magnitude,

00:03:36.260 --> 00:03:40.160
1.63 times 10 to the second joules.

00:03:40.160 --> 00:03:43.040
However, we need to put
a negative sign in here

00:03:43.040 --> 00:03:47.223
which indicates that energy
was given off by the reaction.

00:03:48.130 --> 00:03:50.590
The heat that's transferred
under constant pressure

00:03:50.590 --> 00:03:52.730
is equal to the change in the entropy

00:03:52.730 --> 00:03:55.230
of the reaction, delta H.

00:03:55.230 --> 00:03:57.950
However, let's find the
change in entropy, delta H

00:03:57.950 --> 00:04:01.270
in terms of kilojoules per mole

00:04:01.270 --> 00:04:05.120
of silver chloride for our units.

00:04:05.120 --> 00:04:07.010
Since silver chloride
is one of our products,

00:04:07.010 --> 00:04:09.770
we first need to find
moles of our reactants

00:04:09.770 --> 00:04:12.650
and we're gonna do that
using the molarity equation

00:04:12.650 --> 00:04:13.770
which says that molarity

00:04:13.770 --> 00:04:16.540
is equal to moles divided by liters.

00:04:16.540 --> 00:04:19.290
For our silver nitrate solution,

00:04:19.290 --> 00:04:23.900
the concentration was 0.100 molar

00:04:23.900 --> 00:04:26.610
and trying to solve for moles so that's X.

00:04:26.610 --> 00:04:28.560
The volume of our silver nitrate solution

00:04:28.560 --> 00:04:30.710
was 25.0 milliliters

00:04:30.710 --> 00:04:35.310
which is 0.0250 liters.

00:04:35.310 --> 00:04:38.690
So we solve for X and we get 0.00250.

00:04:42.580 --> 00:04:46.140
So that's how many moles of silver nitrate

00:04:46.140 --> 00:04:48.010
that we started with

00:04:48.010 --> 00:04:50.100
and it's the exact same calculation

00:04:50.100 --> 00:04:52.230
for sodium chloride as well.

00:04:52.230 --> 00:04:56.913
So this is also how many moles
of sodium chloride we have.

00:04:57.850 --> 00:05:00.800
Next, we go back to our
balanced chemical equation

00:05:00.800 --> 00:05:03.290
and we can see we have coefficients of one

00:05:03.290 --> 00:05:04.610
in front of silver nitrate,

00:05:04.610 --> 00:05:05.820
in front of sodium chloride

00:05:05.820 --> 00:05:08.390
and in front of silver chloride.

00:05:08.390 --> 00:05:11.967
Therefore, we're also going
to produce 0.00250 moles

00:05:14.930 --> 00:05:16.710
of silver chloride.

00:05:16.710 --> 00:05:18.530
Next we're gonna calculate the change

00:05:18.530 --> 00:05:21.870
in the entropy, delta H, for our reaction.

00:05:21.870 --> 00:05:23.400
The heat that was transferred

00:05:23.400 --> 00:05:28.400
was negative 1.63 times
10 to the second joules

00:05:30.060 --> 00:05:31.650
and we're gonna divide that

00:05:31.650 --> 00:05:33.690
by the moles of silver chloride

00:05:33.690 --> 00:05:38.690
which was 0.00250 moles
of silver chlorides.

00:05:41.890 --> 00:05:45.447
This is equal to negative 65,200 joules

00:05:48.350 --> 00:05:51.050
per mole of silver chloride

00:05:51.050 --> 00:05:53.720
and we could convert that into kilojoules.

00:05:53.720 --> 00:05:58.720
And so this is equal to
negative 65.2 kilojoules

00:05:59.550 --> 00:06:02.983
per mole of silver chloride.

00:06:04.930 --> 00:06:08.030
We could stop right here and
give this as our final answer,

00:06:08.030 --> 00:06:09.020
but let's keep going

00:06:09.020 --> 00:06:13.200
and convert two kilojoules
per mole of reaction.

00:06:13.200 --> 00:06:14.560
First, let's rewrite this.

00:06:14.560 --> 00:06:19.210
We have negative 65.2 kilojoules

00:06:19.210 --> 00:06:23.590
per mole of silver chlorides

00:06:23.590 --> 00:06:26.540
and what it means by
kilojoules per mole of reaction

00:06:26.540 --> 00:06:28.590
is how the reaction is written

00:06:28.590 --> 00:06:31.660
in the balanced equation down here.

00:06:31.660 --> 00:06:33.440
So if we look at the balanced equation,

00:06:33.440 --> 00:06:37.010
there's one mole of silver chloride

00:06:37.010 --> 00:06:38.860
for how the reaction is written.

00:06:38.860 --> 00:06:41.370
So we can write a conversion factor

00:06:41.370 --> 00:06:45.340
of one mole of silver chloride

00:06:45.340 --> 00:06:49.520
per one mole of reaction

00:06:49.520 --> 00:06:52.490
and writing it this way
for the conversion factor,

00:06:52.490 --> 00:06:55.850
the moles of silver
chloride would cancel out

00:06:55.850 --> 00:07:00.850
and give us negative 65.2
kilojoules per mole of reaction.

00:07:06.100 --> 00:07:08.993
So this could also be our final answer.

00:07:11.540 --> 00:07:12.740
Finally, the negative sign

00:07:12.740 --> 00:07:15.350
means that we have an exothermic reaction.

00:07:15.350 --> 00:07:17.730
The reaction gave off energy

00:07:17.730 --> 00:07:19.720
and this value when you do

00:07:19.720 --> 00:07:21.750
a constant pressure calorimetry experiment

00:07:21.750 --> 00:07:24.520
is often a little bit
lower than the actual value

00:07:24.520 --> 00:07:27.940
because in reality there's
not always a perfect transfer

00:07:27.940 --> 00:07:31.610
of heat from the reaction to the water.

00:07:31.610 --> 00:07:35.063
Often some of the energy
is lost to the environment.


Enthalpy of reaction

WEBVTT
Kind: captions
Language: en

00:00:00.660 --> 00:00:01.580
- [Instructor] The change in enthalpy

00:00:01.580 --> 00:00:03.500
for a chemical reaction delta H,

00:00:03.500 --> 00:00:06.240
we could even write delta
H of reaction in here

00:00:06.240 --> 00:00:08.230
is equal to the heat transferred

00:00:08.230 --> 00:00:11.100
during a chemical reaction
at constant pressure.

00:00:11.100 --> 00:00:13.963
So delta H is equal to qp.

00:00:14.930 --> 00:00:17.100
Let's say we are performing
a chemical reaction,

00:00:17.100 --> 00:00:20.710
an aqueous solution under
constant atmospheric pressure.

00:00:20.710 --> 00:00:23.060
The reactants and products
of that chemical reaction

00:00:23.060 --> 00:00:25.270
make up the system and
everything else makes

00:00:25.270 --> 00:00:27.550
up the surroundings.

00:00:27.550 --> 00:00:30.800
When heat flows from the
surroundings to the system,

00:00:30.800 --> 00:00:33.500
the system or the reaction absorbs heat

00:00:33.500 --> 00:00:35.690
and therefore the change in enthalpy

00:00:35.690 --> 00:00:38.010
is positive for the reaction.

00:00:38.010 --> 00:00:41.320
This is called an endothermic reaction.

00:00:41.320 --> 00:00:44.450
If heat flows from the
system to the surroundings,

00:00:44.450 --> 00:00:46.730
the reaction gave off energy.

00:00:46.730 --> 00:00:48.290
Therefore the change in enthalpy

00:00:48.290 --> 00:00:50.030
for the reaction is negative

00:00:50.030 --> 00:00:52.853
and this is called an exothermic reaction.

00:00:53.790 --> 00:00:57.050
As an example of a reaction,
let's look at the decomposition

00:00:57.050 --> 00:01:01.930
of hydrogen peroxide to form
liquid water and oxygen gas.

00:01:01.930 --> 00:01:04.470
The change in the
enthalpy for this reaction

00:01:04.470 --> 00:01:07.540
is equal to negative 196 kilojoules.

00:01:07.540 --> 00:01:11.330
The negative sign means
the reaction is exothermic.

00:01:11.330 --> 00:01:14.420
And for the units, sometimes
you might see kilojoules.

00:01:14.420 --> 00:01:18.320
Sometimes you might see
kilojoules per mole,

00:01:18.320 --> 00:01:22.553
and sometimes you might see
kilojoules per mole of reaction.

00:01:23.540 --> 00:01:25.150
What kilojoules per mole of reaction

00:01:25.150 --> 00:01:28.900
is referring to is how
the equation is written.

00:01:28.900 --> 00:01:31.120
So if we look at this balanced equation,

00:01:31.120 --> 00:01:33.980
there's a two as a coefficient
in front of hydrogen peroxide

00:01:33.980 --> 00:01:36.290
and therefore two moles
of hydrogen peroxide

00:01:36.290 --> 00:01:39.470
are decomposing to form two moles of water

00:01:39.470 --> 00:01:42.080
and one mole of oxygen gas.

00:01:42.080 --> 00:01:45.770
So when two moles of
hydrogen peroxide decompose,

00:01:45.770 --> 00:01:50.770
196 kilojoules of energy are given off.

00:01:50.810 --> 00:01:54.120
Next, let's calculate
how much heat is released

00:01:54.120 --> 00:01:58.580
when 5.00 grams of hydrogen
peroxide decomposes

00:01:58.580 --> 00:02:00.400
at a constant pressure.

00:02:00.400 --> 00:02:02.530
The first step is to
find out how many moles

00:02:02.530 --> 00:02:04.680
of hydrogen peroxide that we have.

00:02:04.680 --> 00:02:07.020
So we take the mass of hydrogen peroxide

00:02:07.020 --> 00:02:08.490
which is five grams

00:02:08.490 --> 00:02:12.040
and we divide that by the
molar mass of hydrogen peroxide

00:02:12.040 --> 00:02:15.530
which is 34.0 grams per mole.

00:02:15.530 --> 00:02:18.713
Grams cancels out and this gives us 0.147

00:02:21.570 --> 00:02:25.033
moles of hydrogen peroxide.

00:02:26.120 --> 00:02:29.770
Next, we take our negative 196 kilojoules

00:02:29.770 --> 00:02:32.430
per mole of reaction

00:02:32.430 --> 00:02:36.540
and we're gonna multiply
this by a conversion factor.

00:02:36.540 --> 00:02:39.130
When we look at the balanced
equation for how it's written,

00:02:39.130 --> 00:02:42.590
there are two moles of hydrogen peroxide.

00:02:42.590 --> 00:02:44.720
So for our conversion factor

00:02:44.720 --> 00:02:49.480
for every one mole of
reaction as it is written,

00:02:49.480 --> 00:02:53.590
there are two moles of hydrogen peroxide.

00:02:53.590 --> 00:02:55.743
So two moles of H2O2.

00:02:58.450 --> 00:03:01.230
Now the of reaction will cancel out

00:03:01.230 --> 00:03:06.230
and this gives us negative 98.0 kilojoules

00:03:07.730 --> 00:03:12.033
per one mole of H2O2.

00:03:13.220 --> 00:03:15.560
So two moles of hydrogen peroxide

00:03:15.560 --> 00:03:18.810
would give off 196 kilojoules of energy.

00:03:18.810 --> 00:03:21.350
And one mole of hydrogen
peroxide would give

00:03:21.350 --> 00:03:25.793
off half that amount or
98.0 kilojoules of energy.

00:03:26.690 --> 00:03:30.520
Next, we take our 0.147
moles of hydrogen peroxide.

00:03:30.520 --> 00:03:31.353
So let me just go ahead

00:03:31.353 --> 00:03:32.970
and write this down here really quickly.

00:03:32.970 --> 00:03:37.390
So we have 0.147 moles of H202.

00:03:38.710 --> 00:03:41.730
And remember, we're trying to calculate,

00:03:41.730 --> 00:03:44.930
we're trying to calculate
the amount of heat

00:03:44.930 --> 00:03:46.410
that was released.

00:03:46.410 --> 00:03:51.153
So next we multiply that
by negative 98.0 kilojoules

00:03:52.760 --> 00:03:57.760
per mole of H202, and moles
of H2O2 will cancel out

00:04:00.910 --> 00:04:03.280
and this gives us our final answer.

00:04:03.280 --> 00:04:07.100
So the heat that was
released when 5.00 grams

00:04:07.100 --> 00:04:10.440
of hydrogen peroxide decompose
at constant pressure,

00:04:10.440 --> 00:04:15.430
this turns out to be equal
to negative 14.4 kilojoules.


Heating curve for water

WEBVTT
Kind: captions
Language: en

00:00:00.220 --> 00:00:02.700
- [Instructor] Let's look at
the heating curve for water.

00:00:02.700 --> 00:00:05.360
A heating curve has
temperature on the y-axis.

00:00:05.360 --> 00:00:07.730
In this case, we have
it in degrees Celsius.

00:00:07.730 --> 00:00:10.550
And heat added on the x-axis,

00:00:10.550 --> 00:00:12.780
let's say it's in kilojoules.

00:00:12.780 --> 00:00:15.740
Let's say we have 18.0 grams of ice

00:00:15.740 --> 00:00:18.090
and our goal is to
calculate the total heat

00:00:18.090 --> 00:00:21.460
necessary to convert that 18 grams of ice

00:00:21.460 --> 00:00:26.460
at -25 degrees Celsius to
steam at 125 degrees Celsius.

00:00:28.880 --> 00:00:31.890
So we're starting with
ice at -25 degrees Celsius

00:00:31.890 --> 00:00:34.010
and first we need to heat up the ice

00:00:34.010 --> 00:00:37.770
to zero degrees Celsius, which
we know is the melting point.

00:00:37.770 --> 00:00:38.950
So on our heating curve,

00:00:38.950 --> 00:00:43.110
we're going from point A to point B.

00:00:43.110 --> 00:00:44.690
To calculate the heat necessary,

00:00:44.690 --> 00:00:49.690
we need to use the equation
Q is equal to mc delta T,

00:00:50.350 --> 00:00:52.000
where q is the heat added,

00:00:52.000 --> 00:00:53.850
m is the mass of the ice.

00:00:53.850 --> 00:00:55.760
c is the specific heat of ice

00:00:55.760 --> 00:00:57.990
and delta T is the change in temperature,

00:00:57.990 --> 00:00:59.370
which is the final temperature

00:00:59.370 --> 00:01:01.520
minus the initial temperature.

00:01:01.520 --> 00:01:03.560
So we're trying to calculate q.

00:01:03.560 --> 00:01:08.147
We know the mass of our ice is 18.0 grams.

00:01:08.147 --> 00:01:10.740
The specific heat of ice

00:01:10.740 --> 00:01:15.740
is 2.03 joules per gram degrees Celsius.

00:01:18.420 --> 00:01:21.350
And for the change in temperature,
it's final minus initial.

00:01:21.350 --> 00:01:24.898
So the final temperature
would be zero degrees Celsius,

00:01:24.898 --> 00:01:27.030
initial is -25.

00:01:27.030 --> 00:01:32.030
So zero minus -25 gives
us +25 degrees Celsius.

00:01:34.730 --> 00:01:38.130
So grams will cancel out,
degrees Celsius cancels out.

00:01:38.130 --> 00:01:43.130
And this gives us q is
equal to 9.1 times 10

00:01:44.580 --> 00:01:49.070
to the second joules to
two significant figures

00:01:49.070 --> 00:01:54.070
or we could also write 0.91 kilojoules.

00:01:55.660 --> 00:01:57.620
Now that the ice is at
zero degrees Celsius,

00:01:57.620 --> 00:01:58.850
we know what's going to melt.

00:01:58.850 --> 00:02:00.520
So we're gonna go from point B

00:02:00.520 --> 00:02:03.260
on the heating curve to point C.

00:02:03.260 --> 00:02:07.020
And to calculate how much heat
is necessary to melt the ice,

00:02:07.020 --> 00:02:10.450
we need to know the heat of fusion of ice,

00:02:10.450 --> 00:02:14.200
which is equal to 6.01
kilojoules per mole.

00:02:18.213 --> 00:02:21.690
So we need to figure out how
many moles of ice we have.

00:02:21.690 --> 00:02:24.890
After starting with 18.0 grams,

00:02:24.890 --> 00:02:27.050
we divide by the molar mass of H2O

00:02:27.050 --> 00:02:31.590
which is 18.0 grams per mole.

00:02:31.590 --> 00:02:35.370
And the grams will cancel
and give us one mole.

00:02:35.370 --> 00:02:39.440
So we have 1.00 moles of ice

00:02:39.440 --> 00:02:43.520
and we multiply that by
6.01 kilojoules per mole

00:02:43.520 --> 00:02:48.040
and the moles cancel out
and give us 6.01 kilojoules.

00:02:50.940 --> 00:02:53.790
Now that all the ice is
melted, we have liquid water.

00:02:53.790 --> 00:02:54.940
And so on our heating curve,

00:02:54.940 --> 00:02:56.370
we're gonna heat that liquid water

00:02:56.370 --> 00:02:59.940
from zero degrees Celsius to 100 Celsius

00:02:59.940 --> 00:03:01.820
which is the boiling point of water.

00:03:01.820 --> 00:03:04.340
So we're going from point C to point D

00:03:04.340 --> 00:03:05.840
on the heating curve.

00:03:05.840 --> 00:03:07.430
To calculate the heat added,

00:03:07.430 --> 00:03:12.400
we use the Q is equal to
mc delta T equation again.

00:03:12.400 --> 00:03:14.880
So we're solving for Q.

00:03:14.880 --> 00:03:18.450
The mass is still 18.0 grams

00:03:18.450 --> 00:03:20.060
but the specific heat now,

00:03:20.060 --> 00:03:21.250
since we have liquid water,

00:03:21.250 --> 00:03:24.020
we need to use the specific
heat of liquid water,

00:03:24.020 --> 00:03:29.020
which is 4.18 joules per
gram degrees Celsius.

00:03:29.720 --> 00:03:31.320
And for the change in temperature,

00:03:31.320 --> 00:03:34.489
the final temperature is 100.

00:03:34.489 --> 00:03:39.489
So 100 minus zero gives
us +100 degrees Celsius.

00:03:42.870 --> 00:03:46.440
So grams cancel out,
degrees Celsius cancels out

00:03:46.440 --> 00:03:50.447
and we find that Q is
equal to 7.52 times 10

00:03:53.180 --> 00:03:57.113
to the third joules, let me
just correct three there,

00:03:57.113 --> 00:04:00.470
7.52 times 10 to the third joules,

00:04:00.470 --> 00:04:03.177
which is equal to 7.52 kilojoules.

00:04:07.170 --> 00:04:09.170
Once we reached a point
D in the heating curve,

00:04:09.170 --> 00:04:10.630
we're at the boiling point of water.

00:04:10.630 --> 00:04:12.680
So the heat that we add now is gonna go

00:04:12.680 --> 00:04:15.760
into turning the liquid
water into gaseous water.

00:04:15.760 --> 00:04:18.730
So going from point D to point E,

00:04:18.730 --> 00:04:19.820
we're doing a phase change.

00:04:19.820 --> 00:04:23.240
We need to know the heat
of vaporization of water,

00:04:23.240 --> 00:04:28.240
and that's equal to 40.7
kilojoules per mole.

00:04:31.030 --> 00:04:35.600
And we already know we
have one mole of H2O.

00:04:35.600 --> 00:04:40.040
So one mole times 40.7
moles, the moles cancel

00:04:40.040 --> 00:04:45.040
and it takes 40.7 kilojoules of energy

00:04:45.740 --> 00:04:50.350
to convert the liquid water
in to gaseous water or steam.

00:04:50.350 --> 00:04:52.130
Next we're gonna heat the gaseous water

00:04:52.130 --> 00:04:55.480
from 100 degrees Celsius
to 125 degrees Celsius.

00:04:55.480 --> 00:04:59.050
So we're going from point E to
point F on the heating curve.

00:04:59.050 --> 00:05:01.280
And to figure out how
much heat we need to add,

00:05:01.280 --> 00:05:06.010
we use the Q is equal to mc
delta T equation one more time.

00:05:06.010 --> 00:05:10.700
So we're solving for Q and
we still have 18.0 grams.

00:05:10.700 --> 00:05:13.990
This time we need to use
these specific heat of steam,

00:05:13.990 --> 00:05:18.990
which is 1.84 joules
per gram degree Celsius.

00:05:20.570 --> 00:05:24.380
The change in temperature
would be 125 minus 100

00:05:24.380 --> 00:05:28.780
or +25 degrees Celsius.

00:05:28.780 --> 00:05:31.540
So grams cancel, units cancel out

00:05:31.540 --> 00:05:36.540
and we get Q is equal to 8.3
times 10 to the second joules

00:05:40.130 --> 00:05:42.020
to two significant figures,

00:05:42.020 --> 00:05:46.823
which is equal to 0.83 kilojoules.

00:05:48.760 --> 00:05:50.700
Finally, we need to add everything up.

00:05:50.700 --> 00:05:53.960
So going from point A to
point B in the heating curve.

00:05:53.960 --> 00:05:57.060
So think about just the X
axis this time, all right?

00:05:57.060 --> 00:05:59.460
So going from point A to point B,

00:05:59.460 --> 00:06:04.460
we calculated that to be
equal to 0.91 kilojoules.

00:06:05.370 --> 00:06:08.492
And then from point B to point C,

00:06:08.492 --> 00:06:13.480
we calculated that to be 6.01 kilojoules.

00:06:13.480 --> 00:06:18.480
From C to D, so this
distance here was 7.52.

00:06:21.463 --> 00:06:24.990
From D to E, this was the big one here.

00:06:24.990 --> 00:06:29.047
This was equal to 40.7 kilojoules.

00:06:30.650 --> 00:06:32.760
And finally from E to F we calculated

00:06:32.760 --> 00:06:35.487
this was equal to 0.83 kilojoules.

00:06:39.010 --> 00:06:40.240
And when you add everything up

00:06:40.240 --> 00:06:44.870
this is equal to 56.0 kilojoules.

00:06:44.870 --> 00:06:47.360
So that's how much energy it takes

00:06:47.360 --> 00:06:52.360
to convert 18.0 grams of
ice at -25 degrees Celsius

00:06:53.450 --> 00:06:58.270
to gaseous water at 125 degrees Celsius.

00:06:58.270 --> 00:06:59.970
Next, let's think about the slopes

00:06:59.970 --> 00:07:02.540
of the different lines
on our heating curve.

00:07:02.540 --> 00:07:06.770
So let's look at the
line going from B to C

00:07:06.770 --> 00:07:11.770
and also the line going
from point D to point E.

00:07:12.170 --> 00:07:14.110
Both of these lines
represent phase changes,

00:07:14.110 --> 00:07:16.280
going from point B to point C

00:07:16.280 --> 00:07:18.420
was going from a solid to a liquid

00:07:18.420 --> 00:07:20.370
and going from point D to E

00:07:20.370 --> 00:07:22.938
was going from a liquid to a gas.

00:07:22.938 --> 00:07:26.270
And since the slope of both
of these lines is zero,

00:07:26.270 --> 00:07:29.930
that means as you add heat on the x-axis,

00:07:29.930 --> 00:07:32.207
the temperature doesn't change.

00:07:32.207 --> 00:07:34.070
So during a phase change,

00:07:34.070 --> 00:07:36.490
all the energy goes into disrupting

00:07:36.490 --> 00:07:39.660
the intermolecular forces that are present

00:07:39.660 --> 00:07:42.500
and they don't go into
increasing the temperature.

00:07:42.500 --> 00:07:44.410
So there is no increase in temperature

00:07:44.410 --> 00:07:46.190
during a phase change.

00:07:46.190 --> 00:07:48.740
Think about going from point D to point E,

00:07:48.740 --> 00:07:52.160
this was converting our liquid
water into gaseous water.

00:07:52.160 --> 00:07:54.470
So as the heat is being added,

00:07:54.470 --> 00:07:59.100
all that energy goes into
breaking the intermolecular forces

00:07:59.100 --> 00:08:01.470
between water molecules and pulling apart

00:08:01.470 --> 00:08:02.790
those liquid water molecules

00:08:02.790 --> 00:08:06.400
and turning them into
gaseous water molecules.

00:08:06.400 --> 00:08:09.190
So it's only after all of
the liquid water molecules

00:08:09.190 --> 00:08:11.720
are converted into
gaseous water molecules,

00:08:11.720 --> 00:08:14.000
that's when we see the
temperature increase again.

00:08:14.000 --> 00:08:16.730
So talking about from point E to point F,

00:08:16.730 --> 00:08:19.160
everything is now in the gaseous state

00:08:19.160 --> 00:08:22.050
and then we see the
increase in temperature.

00:08:22.050 --> 00:08:25.700
Finally, let's compare the
slope of the line from A to B

00:08:25.700 --> 00:08:29.284
to the slope of the line from C to D.

00:08:29.284 --> 00:08:31.080
If we look at it, the slope of the line

00:08:31.080 --> 00:08:34.330
from A to B is a little bit steeper

00:08:34.330 --> 00:08:37.600
than the slope of the line from C to D.

00:08:37.600 --> 00:08:39.090
The reason for the different slopes

00:08:39.090 --> 00:08:41.671
has to do with the
different specific heats.

00:08:41.671 --> 00:08:45.270
From A to B, we used the
specific heat for ice

00:08:45.270 --> 00:08:49.060
which is 2.03 joules per
gram degrees Celsius.

00:08:49.060 --> 00:08:51.840
From C to D in our calculation,

00:08:51.840 --> 00:08:53.910
we used the specific heat for water

00:08:53.910 --> 00:08:57.800
which is 4.1 joules per
gram degrees Celsius.

00:08:57.800 --> 00:09:00.160
The higher the value
for the specific heat,

00:09:00.160 --> 00:09:02.410
the more energy it takes to raise

00:09:02.410 --> 00:09:06.030
the temperature of a
substance by a certain amount.

00:09:06.030 --> 00:09:07.660
So if we think about comparing these two,

00:09:07.660 --> 00:09:09.290
let's say we try to raise the temperature

00:09:09.290 --> 00:09:11.850
of ice by 25 degrees Celsius.

00:09:11.850 --> 00:09:15.460
So lets think about this
distance here on the y-axis.

00:09:15.460 --> 00:09:18.560
We would have to put in only
a small amount of energy

00:09:18.560 --> 00:09:20.700
to get ice to increase its temperature

00:09:20.700 --> 00:09:23.200
by 25 degrees Celsius.

00:09:23.200 --> 00:09:26.160
We think about that same
temperature change on liquid water.

00:09:26.160 --> 00:09:28.820
So if we tried to increase the
temperature of liquid water

00:09:28.820 --> 00:09:31.690
by that same amount, 25 degrees,

00:09:31.690 --> 00:09:34.280
we would have to put in more energy.

00:09:34.280 --> 00:09:37.440
So on the x-axis, we have
to put in more energy

00:09:37.440 --> 00:09:40.020
to accomplish the same
change in temperature.

00:09:40.020 --> 00:09:44.820
And that's because liquid water
has a higher specific heat.

00:09:44.820 --> 00:09:47.100
Since it might be a little bit
hard to see on that diagram,

00:09:47.100 --> 00:09:50.870
let's think about putting some
heat into a substance here.

00:09:50.870 --> 00:09:52.590
So I'm gonna draw a horizontal line,

00:09:52.590 --> 00:09:54.050
and then we're trying to accomplish

00:09:54.050 --> 00:09:55.420
a certain temperature change.

00:09:55.420 --> 00:09:57.200
So I'll draw a vertical line.

00:09:57.200 --> 00:10:02.060
Those two give me a line with a slope.

00:10:02.060 --> 00:10:03.790
So let's say we're trying to accomplish

00:10:03.790 --> 00:10:06.060
the same change in temperature.

00:10:06.060 --> 00:10:09.182
So I'll draw this Y
distance the same as before

00:10:09.182 --> 00:10:11.450
but we have a higher specific heat.

00:10:11.450 --> 00:10:14.190
So it takes more energy.

00:10:14.190 --> 00:10:17.370
Therefore this X distance
is going to increase.

00:10:17.370 --> 00:10:19.740
And when we increase the X distance,

00:10:19.740 --> 00:10:23.230
we see that the slope decreases.

00:10:23.230 --> 00:10:26.780
So the greater the value
for the specific heat,

00:10:26.780 --> 00:10:30.033
the lower the slope on the heating curve.


Enthalpy and phase changes

WEBVTT
Kind: captions
Language: en

00:00:00.030 --> 00:00:01.720
- [Instructor] Let's say
that we have some solid water

00:00:01.720 --> 00:00:04.860
or ice and we want to melt the ice

00:00:04.860 --> 00:00:09.170
and turn the solid
water into liquid water.

00:00:09.170 --> 00:00:10.900
This phase change of solid water

00:00:10.900 --> 00:00:13.010
to liquid water is called melting

00:00:13.010 --> 00:00:16.460
and it takes positive 6.01 kilojoules

00:00:16.460 --> 00:00:20.010
per one mole to melt ice.

00:00:20.010 --> 00:00:23.970
This change in enthalpy
is symbolized by delta H

00:00:23.970 --> 00:00:27.090
with a subscript fus,
which stands for fusion.

00:00:27.090 --> 00:00:29.913
So this is called the heat of fusion.

00:00:31.020 --> 00:00:32.580
Next let's think about the phase change

00:00:32.580 --> 00:00:35.630
of converting liquid
water into gaseous water.

00:00:35.630 --> 00:00:38.040
This phase change is called vaporization

00:00:38.040 --> 00:00:40.810
and it also takes energy
to convert liquid water

00:00:40.810 --> 00:00:42.860
into gaseous water.

00:00:42.860 --> 00:00:46.970
Specifically for water
it takes 40.7 kilojoules

00:00:46.970 --> 00:00:50.700
per one mole of liquid
water to vaporize it.

00:00:50.700 --> 00:00:54.490
And so this change in energy
is called the enthalpy

00:00:54.490 --> 00:00:59.140
of vaporization or simply
the heat of vaporization.

00:00:59.140 --> 00:01:01.600
Let's go back and think
about the structure of ice.

00:01:01.600 --> 00:01:05.790
Ice has water molecules in a
repeating crystal structure

00:01:05.790 --> 00:01:10.270
and the water molecules are
held together by hydrogen bonds.

00:01:10.270 --> 00:01:12.690
So between these two water molecules here,

00:01:12.690 --> 00:01:16.810
when we add energy, we
increase the freedom of motion,

00:01:16.810 --> 00:01:19.300
so over here is a picture of liquid water.

00:01:19.300 --> 00:01:22.100
So this is still held
together by hydrogen bonds.

00:01:22.100 --> 00:01:23.850
These water molecules
are still held together

00:01:23.850 --> 00:01:26.940
by hydrogen bonds but we no
longer have a crystal structure.

00:01:26.940 --> 00:01:29.490
So we have increased freedom of motion

00:01:29.490 --> 00:01:33.690
and it takes energy to disrupt
that crystal structure.

00:01:33.690 --> 00:01:35.680
And next, let's think about
converting liquid water

00:01:35.680 --> 00:01:38.320
into gaseous water or steam.

00:01:38.320 --> 00:01:39.790
When water is in the gaseous state,

00:01:39.790 --> 00:01:42.240
there are no more intermolecular forces

00:01:42.240 --> 00:01:43.450
between the molecules.

00:01:43.450 --> 00:01:45.450
There's nothing holding them together.

00:01:45.450 --> 00:01:48.170
And so it takes a lot
of energy to pull these

00:01:48.170 --> 00:01:49.600
two water molecules apart.

00:01:49.600 --> 00:01:52.610
It takes a lot of energy to
overcome these hydrogen bonds.

00:01:52.610 --> 00:01:55.950
And that's the reason why
we have such a large value

00:01:55.950 --> 00:01:58.730
for the heat of vaporization.

00:01:58.730 --> 00:02:01.703
So it takes a lot more
energy to completely pull

00:02:01.703 --> 00:02:05.190
these molecules apart
than it did to simply

00:02:05.190 --> 00:02:06.500
increase the freedom of motion.

00:02:06.500 --> 00:02:11.500
So 40.7 is a much bigger number than 6.01.

00:02:12.060 --> 00:02:15.070
If it takes positive
40.7 kilojoules per mole

00:02:15.070 --> 00:02:18.030
of energy to go from the liquid
state to the gaseous state.

00:02:18.030 --> 00:02:19.090
If we go in reverse

00:02:19.090 --> 00:02:21.730
from the gaseous state
back to the liquid state

00:02:21.730 --> 00:02:24.390
that same amount of energy is given off.

00:02:24.390 --> 00:02:28.930
So we can write 40.7 kilojoules per mole.

00:02:28.930 --> 00:02:31.110
However, since the energy is given off,

00:02:31.110 --> 00:02:33.870
we need to include a negative sign,

00:02:33.870 --> 00:02:35.130
going from the gaseous state

00:02:35.130 --> 00:02:37.900
to the liquid state is
called condensation.

00:02:37.900 --> 00:02:39.410
So we could call this value

00:02:39.410 --> 00:02:42.260
of negative 40.7 kilojoules per mole,

00:02:42.260 --> 00:02:44.453
the heat of condensation.

00:02:45.470 --> 00:02:48.900
And if it takes positive 6.01
kilojoules per mole to go

00:02:48.900 --> 00:02:50.970
from the solid state to the liquid state.

00:02:50.970 --> 00:02:52.080
If we go in reverse

00:02:52.080 --> 00:02:54.880
from the liquid state
back to the solid state

00:02:54.880 --> 00:02:59.880
we would give off 6.01
kilojoules per mole of energy.

00:03:00.690 --> 00:03:03.020
And so we need to write
a negative sign here

00:03:03.020 --> 00:03:05.250
to indicate the energy is given off.

00:03:05.250 --> 00:03:08.770
When we go from a liquid to
a solid, that's freezing.

00:03:08.770 --> 00:03:12.543
So this value is called the
heat of freezing for water.


Constant-volume calorimetry

WEBVTT
Kind: captions
Language: en

00:00:00.690 --> 00:00:03.170
- [Instructor] Calorimetry
refers to the measurement

00:00:03.170 --> 00:00:04.370
of heat flow.

00:00:04.370 --> 00:00:07.060
And there are many different
types of calorimeters.

00:00:07.060 --> 00:00:10.970
In this case, we're looking at
a constant volume calorimeter

00:00:10.970 --> 00:00:14.460
which is also called a bomb calorimeter.

00:00:14.460 --> 00:00:17.190
Let's look at how a
bomb calorimeter works.

00:00:17.190 --> 00:00:20.490
First, the sample to
be combusted is placed

00:00:20.490 --> 00:00:23.220
in a container that has some oxygen.

00:00:23.220 --> 00:00:25.750
And then there's some
ignition wires that go into

00:00:25.750 --> 00:00:29.690
this little container here,
and the sample is ignited

00:00:29.690 --> 00:00:33.440
and heat is given off by
the combustion reaction.

00:00:33.440 --> 00:00:37.460
So heat is being
transferred from our sample

00:00:37.460 --> 00:00:41.460
to the water in our containers.

00:00:41.460 --> 00:00:43.530
Let me go ahead and
draw some water in here.

00:00:43.530 --> 00:00:45.600
So imagine we have some water,

00:00:45.600 --> 00:00:49.560
and it's important to know
that the container is rigid.

00:00:49.560 --> 00:00:54.560
So the walls are very,
very solid and cannot move.

00:00:55.350 --> 00:00:57.550
There's also something to stir the water

00:00:57.550 --> 00:00:59.400
and since heat is being transferred

00:00:59.400 --> 00:01:02.110
from the combustion reaction to the water,

00:01:02.110 --> 00:01:04.480
the temperature of the water will increase

00:01:04.480 --> 00:01:07.980
which we can see on the thermometer.

00:01:07.980 --> 00:01:11.270
Now that we understand how
a bomb calorimeter works,

00:01:11.270 --> 00:01:12.610
let's think about that heat

00:01:12.610 --> 00:01:15.420
that's being transferred
from the combustion reaction

00:01:15.420 --> 00:01:19.980
to the water, so that heat is q.

00:01:19.980 --> 00:01:22.780
Let's go back to the first
law of thermodynamics,

00:01:22.780 --> 00:01:25.220
which says that the change
in the internal energy

00:01:25.220 --> 00:01:28.210
of the system is equal to q plus w,

00:01:28.210 --> 00:01:30.890
where q is the heat that's transferred

00:01:30.890 --> 00:01:33.840
and w is the work done.

00:01:33.840 --> 00:01:37.020
Let's say we do this combustion
reaction in a container

00:01:37.020 --> 00:01:39.070
with a movable piston.

00:01:39.070 --> 00:01:41.830
And the combustion reaction is performed

00:01:41.830 --> 00:01:45.080
under the constant
pressure of the atmosphere.

00:01:45.080 --> 00:01:47.971
So this time, when we do
the combustion reaction,

00:01:47.971 --> 00:01:50.590
we will transfer some heat.

00:01:50.590 --> 00:01:53.640
So heat is being transferred
from the combustion reaction

00:01:53.640 --> 00:01:57.060
and we would also produce some
gases, which would push up

00:01:57.060 --> 00:02:00.200
on the piston and so the
piston would move up,

00:02:00.200 --> 00:02:01.960
and since the piston's moving,

00:02:01.960 --> 00:02:06.130
work is being done by
the combustion reaction.

00:02:06.130 --> 00:02:08.960
In this case, the heat
that's transferred q

00:02:08.960 --> 00:02:11.090
is done under constant pressure,

00:02:11.090 --> 00:02:14.935
and so we can write qp
here and by definition,

00:02:14.935 --> 00:02:17.395
the heat that's transferred
at constant pressure,

00:02:17.395 --> 00:02:20.895
that's the change in the enthalpy delta H.

00:02:23.150 --> 00:02:25.399
So for this example with the container

00:02:25.399 --> 00:02:26.918
with the movable piston,

00:02:26.918 --> 00:02:29.385
when we did our combustion reaction,

00:02:29.385 --> 00:02:31.706
the heat that's transferred
at constant pressure is equal

00:02:31.706 --> 00:02:35.401
to the enthalpy delta H, the
change in enthalpy delta H,

00:02:35.401 --> 00:02:40.012
and as the gases expand and
pushing the piston work is done.

00:02:40.012 --> 00:02:42.236
Let's compare the example
with the movable piston

00:02:42.236 --> 00:02:44.293
to our bomb calorimeter.

00:02:44.293 --> 00:02:46.513
Our bomb calorimeter has rigid walls

00:02:46.513 --> 00:02:49.013
and therefore no work can be done.

00:02:49.013 --> 00:02:51.846
So the work done is equal to zero.

00:02:52.751 --> 00:02:55.841
When we plugged that into the
first law of thermodynamics,

00:02:55.841 --> 00:03:00.038
we find that the change in
the internal energy delta E

00:03:00.038 --> 00:03:02.955
is equal to the heat transferred q.

00:03:03.861 --> 00:03:06.617
And since this is a constant volume

00:03:06.617 --> 00:03:10.369
calorimeter right, the walls are rigid.

00:03:10.369 --> 00:03:12.536
We can write q sub v here.

00:03:13.406 --> 00:03:17.665
So this heat that's transferred
from our combustion reaction

00:03:17.665 --> 00:03:20.782
in this case is not equal to
the change in the enthalpy.

00:03:20.782 --> 00:03:25.782
It's equal to the change in
the internal energy delta E.

00:03:26.209 --> 00:03:27.280
So the heat that's transferred

00:03:27.280 --> 00:03:30.210
at constant pressure
is equal to the change

00:03:30.210 --> 00:03:32.180
in the enthalpy delta H,

00:03:32.180 --> 00:03:35.140
while the heat that's
transferred at constant volume

00:03:35.140 --> 00:03:38.453
is equal to the change in
the internal energy delta E.

00:03:39.330 --> 00:03:42.100
To do a constant volume
calorimetry problem,

00:03:42.100 --> 00:03:44.650
we need to know the heat
capacity of the calorimeter

00:03:44.650 --> 00:03:49.090
which is symbolized by
C with a subscript cal.

00:03:49.090 --> 00:03:51.190
To find the heat capacity
of the calorimeter,

00:03:51.190 --> 00:03:54.380
we need to combust something
that we know the exact amount

00:03:54.380 --> 00:03:55.250
of heat for them.

00:03:55.250 --> 00:03:58.510
For example, if you
combust exactly one gram

00:03:58.510 --> 00:04:00.170
of benzoic acid,

00:04:00.170 --> 00:04:05.170
you'll get 26.38 kilojoules
released of energy.

00:04:05.500 --> 00:04:10.220
So let's say we have a 0.2350
gram sample of benzoic acid.

00:04:10.220 --> 00:04:13.080
And we put that in our calorimeter

00:04:13.080 --> 00:04:16.400
and we go ahead and
combust the benzoic acid.

00:04:16.400 --> 00:04:18.610
And we find that the temperature increases

00:04:18.610 --> 00:04:22.793
by positive 1.642 degrees Celsius.

00:04:23.730 --> 00:04:25.870
To find the heat capacity
for the calorimeter,

00:04:25.870 --> 00:04:27.210
first we take our known amount

00:04:27.210 --> 00:04:32.210
which is 26.38 kilojoules per gram

00:04:32.820 --> 00:04:35.820
and we multiply that by
how much benzoic acid

00:04:35.820 --> 00:04:40.820
we used in our calorimeter
which was 0.2350 grams.

00:04:41.430 --> 00:04:43.570
And so the grams will cancel out

00:04:43.570 --> 00:04:48.000
and this is equal to 6.199 kilojoules.

00:04:50.580 --> 00:04:53.090
Next we divide this by
our temperature change

00:04:53.090 --> 00:04:56.467
which was positive 1.642 degrees Celsius.

00:04:59.070 --> 00:05:01.240
And this gives us the heat capacity

00:05:01.240 --> 00:05:06.240
of our calorimeter, which turns
out to be 3.3775 kilojoules

00:05:11.590 --> 00:05:14.083
per degree Celsius.

00:05:15.060 --> 00:05:16.960
Now that we know the heat capacity

00:05:16.960 --> 00:05:19.340
for our specific calorimeter,

00:05:19.340 --> 00:05:21.610
we can use this value
to calculate the heat

00:05:21.610 --> 00:05:24.500
of combustion for another substance.

00:05:24.500 --> 00:05:27.230
So the heat of combustion
for another substance

00:05:27.230 --> 00:05:30.030
or just q would be equal to the negative

00:05:30.030 --> 00:05:32.650
of the heat capacity of the calorimeter

00:05:32.650 --> 00:05:35.740
times the change in the
temperature of the water

00:05:35.740 --> 00:05:37.113
in that calorimeter.

00:05:38.320 --> 00:05:40.030
Let's say our goal is
to calculate the heat

00:05:40.030 --> 00:05:43.760
of combustion of caffeine
in kilojoules per mole.

00:05:43.760 --> 00:05:46.650
So we take 0.265 grams of caffeine.

00:05:46.650 --> 00:05:50.100
We put that in our
calorimeter, we combust it

00:05:50.100 --> 00:05:52.510
and we find the temperature
of the water increases

00:05:52.510 --> 00:05:57.510
by positive 1.525 degrees Celsius.

00:05:57.510 --> 00:06:00.530
So to calculate q, q is
equal to the negative

00:06:00.530 --> 00:06:02.570
of the heat capacity of the calorimeter,

00:06:02.570 --> 00:06:07.570
which is 3.775 kilojoules
per degree Celsius.

00:06:09.270 --> 00:06:11.580
And we multiply that by
the temperature change

00:06:11.580 --> 00:06:15.880
which has 1.525 degrees Celsius.

00:06:15.880 --> 00:06:18.400
So degrees Celsius cancels out

00:06:18.400 --> 00:06:23.400
and this gives us
negative 5.757 kilojoules.

00:06:26.830 --> 00:06:28.323
And technically this is the heat transfer

00:06:28.323 --> 00:06:32.640
to a constant volume so we could
even write q sub v in here,

00:06:32.640 --> 00:06:35.130
and remember this is equal to the change

00:06:35.130 --> 00:06:37.970
in the internal energy of our system.

00:06:37.970 --> 00:06:41.690
So this is qv is equal to delta E.

00:06:41.690 --> 00:06:43.670
Since our goal is to find
the heat of combustion

00:06:43.670 --> 00:06:46.270
of caffeine in kilojoules per mole,

00:06:46.270 --> 00:06:48.660
next we need to take
our grams of caffeine,

00:06:48.660 --> 00:06:51.390
which is 0.265 grams, and divide that

00:06:51.390 --> 00:06:53.200
by the molar mass of caffeine.

00:06:53.200 --> 00:06:57.740
And so grams will cancel out
and give us moles of caffeine.

00:06:57.740 --> 00:07:00.450
So this calculation is equal to 1.36

00:07:01.930 --> 00:07:06.930
times 10 to the negative
third moles of caffeine.

00:07:08.800 --> 00:07:10.810
Now, all we have to do is divide our heat,

00:07:10.810 --> 00:07:15.810
which is negative 5.757
kilojoules by our moles

00:07:17.340 --> 00:07:22.340
of 1.36 times 10 to the
negative third moles,

00:07:23.620 --> 00:07:26.830
to give us a final value of negative 4.23

00:07:28.289 --> 00:07:33.289
times 10 to the third kilojoules per mole

00:07:33.740 --> 00:07:36.713
with a negative sign,
meaning heat is given off.

00:07:37.800 --> 00:07:39.490
So we can say that this is the value,

00:07:39.490 --> 00:07:41.890
this is the change in the internal energy

00:07:41.890 --> 00:07:44.920
for our reaction in kilojoules per mole.

00:07:44.920 --> 00:07:48.180
And often the change in
enthalpy is about the same

00:07:48.180 --> 00:07:50.250
as the change in the internal energy.

00:07:50.250 --> 00:07:52.640
So we can say that this
is approximately equal

00:07:52.640 --> 00:07:57.023
to the change in the enthalpy
for the reaction as well.


Worked example: Measuring the energy content of foods using soda-can calorimetry

WEBVTT
Kind: captions
Language: en

00:00:00.252 --> 00:00:01.450
- [Instructor] Calorimetry refers

00:00:01.450 --> 00:00:04.040
to the measurement of heat flow.

00:00:04.040 --> 00:00:05.550
And in this worked example,

00:00:05.550 --> 00:00:07.850
we're going to burn a marshmallow

00:00:07.850 --> 00:00:11.820
and find the energy
content of the marshmallow.

00:00:11.820 --> 00:00:15.320
First, let's look at the setup
for our soda can calorimeter.

00:00:15.320 --> 00:00:17.800
So our soda can has some water in it.

00:00:17.800 --> 00:00:19.970
So here's the water in our soda can.

00:00:19.970 --> 00:00:23.720
And then we also have a
thermometer in the soda can

00:00:23.720 --> 00:00:26.950
to measure the change in the
temperature of the water.

00:00:26.950 --> 00:00:29.560
If you take a stir rod
and you put the stir rod

00:00:29.560 --> 00:00:31.840
through the tab on the soda can,

00:00:31.840 --> 00:00:35.290
you can attach the soda can to a stand.

00:00:35.290 --> 00:00:37.480
Next, we can put our marshmallow on a pin

00:00:37.480 --> 00:00:39.650
that's attached to a piece of cork.

00:00:39.650 --> 00:00:40.900
Before we start the experiment,

00:00:40.900 --> 00:00:43.360
we need to take the
mass of the marshmallow

00:00:43.360 --> 00:00:44.470
with the cork and the pin,

00:00:44.470 --> 00:00:47.553
so we'll call that the initial mass.

00:00:47.553 --> 00:00:51.464
And we also need the initial
temperature of the water.

00:00:51.464 --> 00:00:53.297
So we'll call that Ti.

00:00:54.990 --> 00:00:57.210
Next, we'd light the marshmallow on fire.

00:00:57.210 --> 00:01:00.190
As the marshmallow burns heat is given off

00:01:00.190 --> 00:01:04.290
and that heat is transferred
to the water in the soda can.

00:01:04.290 --> 00:01:05.910
Therefore the water and the soda can

00:01:05.910 --> 00:01:07.520
will increase in temperature,

00:01:07.520 --> 00:01:10.370
which we can see on the thermometer.

00:01:10.370 --> 00:01:12.170
After the marshmallow
burns for a little while,

00:01:12.170 --> 00:01:14.614
we can stop the burning process.

00:01:14.614 --> 00:01:17.700
And once we stop that, we
wanna look at the thermometer

00:01:17.700 --> 00:01:20.576
for the maximum temperature reached.

00:01:20.576 --> 00:01:22.870
And when we find that maximum temperature

00:01:22.870 --> 00:01:24.230
we can go ahead and record it.

00:01:24.230 --> 00:01:26.230
So we have our final temperature

00:01:26.230 --> 00:01:28.350
and once the marshmallow
with the cork and the pin

00:01:28.350 --> 00:01:31.313
cools down we can find our final mass.

00:01:33.430 --> 00:01:34.790
Let's say the initial mass

00:01:34.790 --> 00:01:39.130
of our marshmallow pin cork was 6.08 grams

00:01:39.130 --> 00:01:42.371
and the final mass was 6.00 grams.

00:01:42.371 --> 00:01:45.000
The initial temperature of
the water in the soda can

00:01:45.000 --> 00:01:48.180
was 25.0 degrees Celsius
and the final temperature

00:01:48.180 --> 00:01:52.080
was 30.0 degrees Celsius.

00:01:52.080 --> 00:01:57.080
Also, let's say that we
started with 50.0 grams

00:01:57.240 --> 00:01:59.253
of water in the soda can.

00:02:00.360 --> 00:02:03.830
Let's calculate the heat
gained by the water.

00:02:03.830 --> 00:02:07.850
To do that we can use the Q is
equal to mc delta T equation

00:02:07.850 --> 00:02:11.527
where Q is equal to the heat transferred,

00:02:11.527 --> 00:02:16.527
m is the mass of the
water which is 50.0 grams,

00:02:17.410 --> 00:02:20.160
c is the specific heat of water

00:02:20.160 --> 00:02:25.160
which is 4.18 joules per
gram degrees Celsius,

00:02:27.153 --> 00:02:29.710
and delta T is the change in temperature

00:02:29.710 --> 00:02:31.780
which should be the final temperature

00:02:31.780 --> 00:02:35.640
minus the initial temperature of the water

00:02:35.640 --> 00:02:40.640
which is 30.0 minus 25.0

00:02:43.160 --> 00:02:48.160
which is equal to 5.0 degrees Celsius.

00:02:50.300 --> 00:02:53.240
Grams cancels out, degrees
Celsius cancels out

00:02:53.240 --> 00:02:57.593
and we find that Q is
equal to +1.0 times 10

00:03:00.973 --> 00:03:02.983
to the third joules.

00:03:02.983 --> 00:03:05.816
That's to two significant figures.

00:03:07.338 --> 00:03:09.304
The positive sign here means

00:03:09.304 --> 00:03:11.839
that heat was gained by the water

00:03:11.839 --> 00:03:15.154
which is why we saw an
increase in the temperature.

00:03:15.154 --> 00:03:17.215
If we assume a perfect transfer of heat,

00:03:17.215 --> 00:03:19.394
so all the heat that was
given off by the burning

00:03:19.394 --> 00:03:21.572
of the marshmallow was
transferred to the water,

00:03:21.572 --> 00:03:25.342
if we think about Q for the
burning of the marshmallow,

00:03:25.342 --> 00:03:27.592
it should be equal in magnitudes.

00:03:27.592 --> 00:03:30.341
So we can write Q is equal to,

00:03:30.341 --> 00:03:31.817
this time we're gonna write a negative

00:03:31.817 --> 00:03:34.886
since heat was given off by
the burning of the marshmallow,

00:03:34.886 --> 00:03:37.636
1.0 times 10 to the third joules.

00:03:41.930 --> 00:03:43.940
So assuming a perfect transfer of heat,

00:03:43.940 --> 00:03:46.800
the magnitude of these
two numbers is equal.

00:03:46.800 --> 00:03:49.260
However, there's no way that
all of the heat was transferred

00:03:49.260 --> 00:03:51.320
from the combustion of the marshmallow

00:03:51.320 --> 00:03:53.954
to the soda can with our simple setup.

00:03:53.954 --> 00:03:56.200
So definitely not all
of it was transferred.

00:03:56.200 --> 00:03:58.120
For example, some of
it could have been lost

00:03:58.120 --> 00:03:59.743
simply to the environment.

00:04:00.820 --> 00:04:02.670
And since the soda can is open

00:04:02.670 --> 00:04:06.390
to the atmospheric pressure
of the environment,

00:04:06.390 --> 00:04:08.960
so I'll go ahead and write
atmospheric pressure in here,

00:04:08.960 --> 00:04:11.330
this soda can calorimeter is an example

00:04:11.330 --> 00:04:14.210
of constant pressure calorimetry.

00:04:14.210 --> 00:04:16.070
And since this is the heat

00:04:16.070 --> 00:04:18.360
that's transferred
under constant pressure,

00:04:18.360 --> 00:04:20.970
I can go ahead and write
QP here to indicate

00:04:20.970 --> 00:04:23.510
the heat transferred
under constant pressure.

00:04:23.510 --> 00:04:27.863
That's the definition of the
change in enthalpy delta H.

00:04:30.920 --> 00:04:34.240
So burning the marshmallow gave off energy

00:04:34.240 --> 00:04:36.850
which is an exothermic reaction,

00:04:36.850 --> 00:04:40.383
therefore, the sign for
delta H is negative.

00:04:41.470 --> 00:04:46.240
Finally, let's relate this soda
can calorimetry experiments

00:04:46.240 --> 00:04:49.010
to calories in everyday life.

00:04:49.010 --> 00:04:51.250
And so let's find the energy content

00:04:51.250 --> 00:04:54.870
of the marshmallow in calories per gram.

00:04:54.870 --> 00:04:58.370
A food calorie has a capital C

00:04:58.370 --> 00:05:02.100
and in chemistry, there's
also a unit of energy

00:05:02.100 --> 00:05:05.260
with calorie with a lowercase C.

00:05:05.260 --> 00:05:07.925
So one food calorie with a capital C

00:05:07.925 --> 00:05:12.925
is equal to one kilocalorie
or 1000 calories

00:05:13.260 --> 00:05:14.533
with a lowercase C.

00:05:15.830 --> 00:05:17.510
When we burned the marshmallow,

00:05:17.510 --> 00:05:20.360
we started off with a mass of 6.08 grams

00:05:20.360 --> 00:05:22.660
for the marshmallow pin cork.

00:05:22.660 --> 00:05:25.780
And the final mass was 6.00 grams,

00:05:25.780 --> 00:05:29.473
which means we burned 0.08
grams of marshmallows.

00:05:30.595 --> 00:05:31.610
And when we burned the marshmallow,

00:05:31.610 --> 00:05:32.830
we found there was a transfer

00:05:32.830 --> 00:05:37.830
of 1.0 times 10 to the
third joules of energy.

00:05:38.460 --> 00:05:40.790
So first, let's convert that

00:05:40.790 --> 00:05:44.190
into calories with a lowercase C.

00:05:44.190 --> 00:05:46.950
So if we multiply by the conversion factor

00:05:46.950 --> 00:05:51.560
of there is one calorie with a lowercase C

00:05:51.560 --> 00:05:54.507
for every 4.184 joules.

00:05:57.530 --> 00:05:59.440
Joules will cancel out

00:05:59.440 --> 00:06:04.440
and this gives us 239
calories with a lowercase C.

00:06:08.040 --> 00:06:12.680
Next, let's convert 239
calories into food calories.

00:06:12.680 --> 00:06:16.760
So first let's take 239 calories,

00:06:16.760 --> 00:06:20.150
and we can multiply by
the conversion factor of,

00:06:20.150 --> 00:06:24.430
there are 1000 calories
with the lowercase C

00:06:24.430 --> 00:06:26.768
for every one kilocalorie.

00:06:26.768 --> 00:06:31.768
And that's gonna give
us 0.239 kilocalories.

00:06:33.142 --> 00:06:35.780
And going back up here to our chart,

00:06:35.780 --> 00:06:40.020
remember, one kilocalorie
is equal to one food calorie

00:06:40.020 --> 00:06:41.920
with a capital C.

00:06:41.920 --> 00:06:46.250
Therefore we have 0.239 kilocalories

00:06:46.250 --> 00:06:51.250
or 0.239 Calories with a capital
C which is a food calorie.

00:06:54.340 --> 00:06:56.270
To find the answer
content of the marshmallow

00:06:56.270 --> 00:06:59.170
in calories per gram,
we just need to divide

00:06:59.170 --> 00:07:02.850
our food calories by how many
grams of marshmallows we use

00:07:02.850 --> 00:07:06.060
which was 0.08 grams in our combustion.

00:07:06.060 --> 00:07:09.410
So dividing by 0.08 grams

00:07:09.410 --> 00:07:14.240
gives us approximately
three calories per gram

00:07:14.240 --> 00:07:15.840
of marshmallows.

00:07:15.840 --> 00:07:16.700
And that's useful

00:07:16.700 --> 00:07:20.940
because let's say we had one
serving size of marshmallows

00:07:20.940 --> 00:07:23.630
which is about 30 grams.

00:07:23.630 --> 00:07:25.688
So therefore, if we
know the energy content

00:07:25.688 --> 00:07:27.891
is three calories per gram,

00:07:27.891 --> 00:07:31.690
we can simply multiply
our four marshmallows

00:07:31.690 --> 00:07:36.660
approximately by three calories per gram

00:07:36.660 --> 00:07:38.660
and grams would cancel out.

00:07:38.660 --> 00:07:41.210
And we would find that
those four marshmallows

00:07:41.210 --> 00:07:45.023
that we wanted to eat are
about 90 food calories.


Constant-pressure calorimetry

WEBVTT
Kind: captions
Language: en

00:00:00.080 --> 00:00:02.400
- [Instructor] Calorimetry
refers to the measurement

00:00:02.400 --> 00:00:03.460
of heat flow.

00:00:03.460 --> 00:00:05.790
And a device that's used
to measure heat flow

00:00:05.790 --> 00:00:07.863
is called a calorimeter.

00:00:08.870 --> 00:00:13.690
An easy way to make a calorimeter
is to use two coffee cups.

00:00:13.690 --> 00:00:16.010
So at the base here,
we have one coffee cup,

00:00:16.010 --> 00:00:18.580
and then we can also
use another coffee cup

00:00:18.580 --> 00:00:20.800
as a loose fitting lid.

00:00:20.800 --> 00:00:23.800
And since this top coffee
cup is loose fitting,

00:00:23.800 --> 00:00:26.760
our calorimeter is exposed
to the constant pressure

00:00:26.760 --> 00:00:28.280
of the atmosphere.

00:00:28.280 --> 00:00:31.090
Therefore, we could use
this coffee cup calorimeter

00:00:31.090 --> 00:00:33.783
for constant-pressure calorimetry.

00:00:34.710 --> 00:00:38.720
Other components of our
calorimeter include some water

00:00:39.720 --> 00:00:43.060
and then we also have a stir
bar to stir up the water,

00:00:43.060 --> 00:00:46.530
and a thermometer to measure
the temperature change

00:00:46.530 --> 00:00:47.453
of the water.

00:00:48.670 --> 00:00:51.810
Let's say we have 150.0 grams of water

00:00:51.810 --> 00:00:56.650
at an initial temperature
of 25.0 degrees Celsius.

00:00:56.650 --> 00:01:01.650
Next, let's take a block of
copper, 120.0 grams of it,

00:01:01.710 --> 00:01:03.530
and let's heat up that block of copper

00:01:03.530 --> 00:01:07.500
to 100.0 degrees Celsius.

00:01:07.500 --> 00:01:09.860
Once the copper has
reached that temperature,

00:01:09.860 --> 00:01:14.063
we add the copper block
to our calorimeter.

00:01:15.320 --> 00:01:17.530
Here, we can see the
copper block has been added

00:01:17.530 --> 00:01:20.220
to the calorimeter, and since the copper

00:01:20.220 --> 00:01:22.660
is at a higher temperature than the water,

00:01:22.660 --> 00:01:26.680
heat flows from the
copper block to the water,

00:01:26.680 --> 00:01:30.010
and therefore the temperature
of the water will increase,

00:01:30.010 --> 00:01:32.860
which we will see on the thermometer.

00:01:32.860 --> 00:01:35.873
So we'll see the temperature
increase on the thermometer.

00:01:37.070 --> 00:01:40.210
Heat is transferred from the
copper block to the water

00:01:40.210 --> 00:01:43.150
until thermal equilibrium
has been reached.

00:01:43.150 --> 00:01:45.540
And we know when thermal
equilibrium has been reached

00:01:45.540 --> 00:01:46.800
by looking at the thermometer

00:01:46.800 --> 00:01:49.730
and measuring the highest
temperature that's reached.

00:01:49.730 --> 00:01:52.720
Let's say the final temperature is equal

00:01:52.720 --> 00:01:57.560
to 30.0 degrees Celsius.

00:01:57.560 --> 00:02:01.170
So at thermal equilibrium,
both the pieces of copper,

00:02:01.170 --> 00:02:03.050
both the copper block and the water

00:02:03.050 --> 00:02:05.463
are at the same final temperature.

00:02:06.500 --> 00:02:08.840
Next, let's calculate the
heat gained by the water

00:02:08.840 --> 00:02:12.660
by using the equation q
is equal to mc delta T.

00:02:12.660 --> 00:02:15.290
So q is what we're trying to calculate,

00:02:15.290 --> 00:02:17.010
the heat gained by the water.

00:02:17.010 --> 00:02:21.310
M is the mass of the water,
which is 150.0 grams.

00:02:21.310 --> 00:02:25.310
So we can write in 150.0 grams.

00:02:25.310 --> 00:02:30.310
C is the specific heat of
water, which is 4.18 joules

00:02:31.450 --> 00:02:35.230
per gram degrees Celsius.

00:02:35.230 --> 00:02:38.150
And delta T is the change
in the temperature,

00:02:38.150 --> 00:02:40.080
which would be the final temperature,

00:02:40.080 --> 00:02:43.183
so Tf minus the initial temperature Ti.

00:02:44.140 --> 00:02:48.760
The final temperature of the
water is 30.0 degrees Celsius,

00:02:48.760 --> 00:02:50.630
and the initial temperature of the water

00:02:50.630 --> 00:02:53.540
was 25.0 degrees Celsius.

00:02:53.540 --> 00:02:58.540
So 30.0 minus 25.0 is equal
to 5.0 degrees Celsius.

00:03:01.230 --> 00:03:03.050
So we can write that in.

00:03:03.050 --> 00:03:07.170
And next, we look at units
and see what cancels out here.

00:03:07.170 --> 00:03:12.170
So the grams cancel out,
degrees Celsius cancels out,

00:03:12.370 --> 00:03:14.740
and we're left with joules as our unit.

00:03:14.740 --> 00:03:19.740
So q is equal to, when we go
to two significant figures,

00:03:20.680 --> 00:03:25.680
this is positive 3.1 times
10 to the third joules.

00:03:28.200 --> 00:03:33.200
The positive sign means that
this was the energy gained

00:03:33.410 --> 00:03:34.583
by the water.

00:03:35.460 --> 00:03:38.270
Next, let's do the same
calculation for copper.

00:03:38.270 --> 00:03:40.970
So we're trying to find q.

00:03:40.970 --> 00:03:45.150
The mass of the copper was 120.0 grams.

00:03:45.150 --> 00:03:47.760
So we can plug that in.

00:03:47.760 --> 00:03:52.760
The specific heat of copper is .39 joules

00:03:53.580 --> 00:03:57.320
per gram degrees Celsius.

00:03:57.320 --> 00:03:58.690
And let's think about the change

00:03:58.690 --> 00:04:00.980
in the temperature of the copper.

00:04:00.980 --> 00:04:02.710
The final temperature of the copper

00:04:02.710 --> 00:04:05.590
was 30.0 degrees Celsius,

00:04:05.590 --> 00:04:07.870
and the initial temperature of the copper

00:04:07.870 --> 00:04:11.480
was 100.0 degrees Celsius.

00:04:11.480 --> 00:04:15.813
So the change in the temperature
would be 30.0 minus 100.0,

00:04:17.980 --> 00:04:21.830
which of course is negative 70.0.

00:04:21.830 --> 00:04:26.830
So let's plug in negative
70.0 degrees Celsius.

00:04:28.700 --> 00:04:32.030
Once again, we see what
cancels for our units.

00:04:32.030 --> 00:04:35.370
Grams will cancel, degrees
Celsius will cancel,

00:04:35.370 --> 00:04:38.060
and our answer will be in joules.

00:04:38.060 --> 00:04:43.060
So q is equal to, using
two significant figures,

00:04:45.977 --> 00:04:49.290
negative 3.3 times 10 to the third joules.

00:04:53.100 --> 00:04:54.800
And the negative sign,

00:04:54.800 --> 00:04:57.860
so this negative sign
means this is the energy

00:04:57.860 --> 00:05:00.353
that was lost by the copper.

00:05:02.090 --> 00:05:04.040
Next, let's look at these
two numbers that we got

00:05:04.040 --> 00:05:05.350
from our calculations.

00:05:05.350 --> 00:05:09.020
Let's think about the
magnitude of these two numbers.

00:05:09.020 --> 00:05:12.480
If our coffee cup calorimeter
were a perfect insulator,

00:05:12.480 --> 00:05:15.580
the magnitude of these two
numbers would be the same.

00:05:15.580 --> 00:05:17.670
So it could be something like 3.3 times 10

00:05:17.670 --> 00:05:20.760
to the third joules for both of them.

00:05:20.760 --> 00:05:23.720
But since these two numbers
are not the same, right?

00:05:23.720 --> 00:05:27.330
We can see that we've lost
more heat from the copper

00:05:27.330 --> 00:05:30.010
than we've gained in terms
of energy for the water,

00:05:30.010 --> 00:05:32.980
which means we could have
lost some of the energy

00:05:32.980 --> 00:05:34.930
to the environment.

00:05:34.930 --> 00:05:38.810
So not all of the heat
was transferred directly

00:05:38.810 --> 00:05:39.870
to the water.

00:05:39.870 --> 00:05:43.313
Some of it could have escaped
our coffee cup calorimeter.

00:05:44.530 --> 00:05:48.920
Next, let's think about calorimetry
for a chemical reaction.

00:05:48.920 --> 00:05:50.010
So before we do that,

00:05:50.010 --> 00:05:53.110
let's review some terms
for thermodynamics.

00:05:53.110 --> 00:05:56.170
So the system is the part of the universe

00:05:56.170 --> 00:05:58.070
that we are studying.

00:05:58.070 --> 00:06:00.090
So in the case of a chemical reaction,

00:06:00.090 --> 00:06:03.960
the reactants and the
products make up the system.

00:06:03.960 --> 00:06:06.483
The surroundings are everything else,

00:06:07.680 --> 00:06:11.410
which would include the
water in the calorimeter,

00:06:11.410 --> 00:06:14.050
the coffee cup itself, the thermometer,

00:06:14.050 --> 00:06:15.720
the environment outside,

00:06:15.720 --> 00:06:17.460
so the surroundings are everything else.

00:06:17.460 --> 00:06:20.120
And finally, the universe
would be the system

00:06:20.120 --> 00:06:22.400
plus the surroundings.

00:06:22.400 --> 00:06:24.870
So the reactants and the
products make up the system.

00:06:24.870 --> 00:06:28.420
So that's what the S stands
for here in our calorimeter.

00:06:28.420 --> 00:06:29.690
That's our system.

00:06:29.690 --> 00:06:31.710
And let's say we run a reaction,

00:06:31.710 --> 00:06:35.880
and in the reaction, heat is given off.

00:06:35.880 --> 00:06:37.070
So in that case,

00:06:37.070 --> 00:06:40.740
heat would flow from the
system to the surroundings,

00:06:40.740 --> 00:06:43.940
and so the temperature of
the water would increase.

00:06:43.940 --> 00:06:47.500
So we would see that as
the temperature increases

00:06:47.500 --> 00:06:49.210
on the thermometer.

00:06:49.210 --> 00:06:51.640
Next, we could calculate
the heat gained by the water

00:06:51.640 --> 00:06:56.640
by using our q is equal
to mc delta T equation.

00:06:57.240 --> 00:07:02.240
And let's say q is equal
to positive 1.0 times 10

00:07:03.400 --> 00:07:06.180
to the second joules.

00:07:06.180 --> 00:07:11.180
The positive sign means that
the water gained energy.

00:07:12.770 --> 00:07:15.910
If we assume a perfect transfer of heat

00:07:15.910 --> 00:07:17.880
from the system to the surroundings,

00:07:17.880 --> 00:07:21.340
if the surroundings gained
positive 1.0 times 10

00:07:21.340 --> 00:07:22.630
to the second joules,

00:07:22.630 --> 00:07:27.533
that means the system must
have lost negative 1.0 times 10

00:07:29.210 --> 00:07:31.130
to the second joules.

00:07:31.130 --> 00:07:34.390
So the same magnitude, but
we changed the sign here,

00:07:34.390 --> 00:07:36.360
because if we're talking about

00:07:36.360 --> 00:07:38.550
the energy lost by the system,

00:07:38.550 --> 00:07:41.833
it's the same in magnitude,
but opposite in sign.

00:07:42.690 --> 00:07:46.370
Next, remember that our lid
over here is loose fitting,

00:07:46.370 --> 00:07:50.030
which makes this constant
pressure calorimetry,

00:07:50.030 --> 00:07:53.240
and therefore this heat that
was transferred is the heat

00:07:53.240 --> 00:07:55.150
that's transferred at constant pressure.

00:07:55.150 --> 00:07:59.580
So we can write a
subscript p in here, so qp.

00:07:59.580 --> 00:08:02.700
The heat transfer at a constant
pressure is the definition

00:08:02.700 --> 00:08:05.840
for the change in the enthalpy delta H,

00:08:05.840 --> 00:08:10.800
so we can write that
qp is equal to delta H.

00:08:11.820 --> 00:08:15.140
And when delta H is negative,

00:08:15.140 --> 00:08:18.383
we're talking about an
exothermic reaction.

00:08:22.940 --> 00:08:24.810
So when a reaction is exothermic,

00:08:24.810 --> 00:08:27.800
heat is transferred from the
system to the surroundings,

00:08:27.800 --> 00:08:29.680
and therefore we see an increase

00:08:29.680 --> 00:08:32.190
in the temperature of the water.

00:08:32.190 --> 00:08:35.280
Finally, let's think about
an endothermic reaction.

00:08:35.280 --> 00:08:36.710
In an endothermic reaction,

00:08:36.710 --> 00:08:39.940
heat is transferred from the
surroundings to the system.

00:08:39.940 --> 00:08:41.630
So here we can show heat flowing

00:08:41.630 --> 00:08:44.200
from the surroundings to the system.

00:08:44.200 --> 00:08:48.090
Since energy is leaving the surroundings,

00:08:48.090 --> 00:08:51.640
the temperature of the water will decrease

00:08:51.640 --> 00:08:54.430
for an endothermic reaction.

00:08:54.430 --> 00:08:57.640
And since heat is being
transferred to the system,

00:08:57.640 --> 00:09:00.860
we can go ahead and write heat over here,

00:09:00.860 --> 00:09:03.990
we can go ahead and write
heat on the reactants side,

00:09:03.990 --> 00:09:08.990
and delta H would be positive
for an endothermic reaction.

00:09:09.960 --> 00:09:11.910
So for an endothermic reaction,

00:09:11.910 --> 00:09:15.560
energy is transferred from the
surroundings to the system,

00:09:15.560 --> 00:09:19.123
and therefore the temperature
of the water will decrease.


Heat capacity

WEBVTT
Kind: captions
Language: en

00:00:01.200 --> 00:00:03.430
- [Instructor] The heat capacity
of an object is the amount

00:00:03.430 --> 00:00:05.640
of heat necessary to raise the temperature

00:00:05.640 --> 00:00:10.230
of the object by one degree
Celsius or one Kelvin.

00:00:10.230 --> 00:00:12.820
The specific heat capacity, which is often

00:00:12.820 --> 00:00:16.310
just called specific
heat is the heat capacity

00:00:16.310 --> 00:00:18.640
of one gram of a substance

00:00:18.640 --> 00:00:21.890
while the molar heat
capacity is the heat capacity

00:00:21.890 --> 00:00:24.313
for one mole of a substance.

00:00:25.170 --> 00:00:28.630
We symbolize specific
heat with a capital C

00:00:28.630 --> 00:00:31.000
and a subscript s for specific,

00:00:31.000 --> 00:00:34.250
and molar heat capacity
is symbolized by capital C

00:00:34.250 --> 00:00:36.430
with a subscript m.

00:00:36.430 --> 00:00:38.510
First let's look at specific heat.

00:00:38.510 --> 00:00:40.390
The specific heat of water is equal

00:00:40.390 --> 00:00:43.897
to 4.18 joules per gram degrees Celsius.

00:00:43.897 --> 00:00:48.750
And what this means is if we
have one gram of liquid water,

00:00:48.750 --> 00:00:50.030
and let's say the initial temperature

00:00:50.030 --> 00:00:53.540
is 14.5 degrees Celsius,

00:00:53.540 --> 00:00:58.380
it takes positive 4.18 joules

00:00:58.380 --> 00:01:00.880
of energy to increase the temperature

00:01:00.880 --> 00:01:05.640
of that one gram of water
by one degree Celsius.

00:01:05.640 --> 00:01:07.760
Therefore the final
temperature of the water

00:01:07.760 --> 00:01:12.760
would be 15.5 degrees Celsius
after we add 4.18 joules.

00:01:14.950 --> 00:01:17.360
Next let's calculate
the molar heat capacity

00:01:17.360 --> 00:01:20.160
of water from the specific heat.

00:01:20.160 --> 00:01:22.030
If we multiply the specific heat

00:01:22.030 --> 00:01:24.710
of water by the molar mass of water

00:01:24.710 --> 00:01:29.710
which is 18.0 grams per mole,

00:01:31.000 --> 00:01:36.000
the grams will cancel out
and that gives us 75.2 joules

00:01:39.149 --> 00:01:43.560
per mole degree Celsius.

00:01:43.560 --> 00:01:48.560
And so this is the molar
heat capacity of water.

00:01:49.250 --> 00:01:52.190
Let's say we had 18.0 grams of water.

00:01:52.190 --> 00:01:54.340
If we divide by the molar mass of water

00:01:54.340 --> 00:01:59.210
which is 18.0 grams per
mole, the grams cancel

00:01:59.210 --> 00:02:03.770
and that gives us one
mole of liquid water.

00:02:03.770 --> 00:02:06.300
So one mole of H2O.

00:02:06.300 --> 00:02:09.364
Using the molar heat capacity of water,

00:02:09.364 --> 00:02:14.364
it would take positive 75.2 joules

00:02:14.650 --> 00:02:17.470
of energy to increase the temperature

00:02:17.470 --> 00:02:22.470
of that 18.0 grams of water
by one degree Celsius.

00:02:22.890 --> 00:02:25.640
Next let's calculate how
much heat is necessary

00:02:25.640 --> 00:02:30.000
to warm 250 grams of water
from an initial temperature

00:02:30.000 --> 00:02:32.840
of 22 degrees Celsius
to a final temperature

00:02:32.840 --> 00:02:35.590
of 98 degrees Celsius.

00:02:35.590 --> 00:02:38.830
Using the units for specific heat,

00:02:38.830 --> 00:02:42.420
which are joules per gram degree Celsius.

00:02:42.420 --> 00:02:45.580
We can rewrite the specific heat is equal

00:02:45.580 --> 00:02:50.120
to joules is the quantity
of heat that's transferred.

00:02:50.120 --> 00:02:52.580
So we could just write q for that.

00:02:52.580 --> 00:02:55.690
Grams is the mass of the substance

00:02:55.690 --> 00:02:57.920
and degree Celsius is talking

00:02:57.920 --> 00:03:00.520
about the change in temperature delta T.

00:03:00.520 --> 00:03:04.360
So if we multiply both sides by m delta T,

00:03:04.360 --> 00:03:07.010
we arrive at the following equation

00:03:07.010 --> 00:03:12.010
which is q is equal to mC delta T.

00:03:12.500 --> 00:03:15.160
And we can use this equation
to calculate the heat

00:03:15.160 --> 00:03:17.370
transferred for different substances

00:03:17.370 --> 00:03:20.030
with different specific heats.

00:03:20.030 --> 00:03:23.220
However, right now we're only
interested in our liquid water

00:03:23.220 --> 00:03:26.160
and how much heat it takes
to increase the temperature

00:03:26.160 --> 00:03:29.880
of our water from 22 degrees
Celsius to a final temperature

00:03:29.880 --> 00:03:32.440
of 98 degrees Celsius.

00:03:32.440 --> 00:03:34.370
To find the change in temperature,

00:03:34.370 --> 00:03:36.860
that's equal to the final temperature

00:03:36.860 --> 00:03:39.610
minus the initial temperature

00:03:39.610 --> 00:03:42.480
which would be 98 degrees Celsius

00:03:42.480 --> 00:03:47.480
minus 22 which is equal
to 76 degrees Celsius.

00:03:49.470 --> 00:03:52.210
Next, we can plug everything
into our equation.

00:03:52.210 --> 00:03:54.410
Q is what we're trying to find.

00:03:54.410 --> 00:03:57.887
M is the mass of the
substance, which is 250 grams.

00:03:59.810 --> 00:04:02.410
C is the specific heat of water

00:04:02.410 --> 00:04:06.070
which is 4.18 joules per
gram degrees Celsius,

00:04:06.070 --> 00:04:10.490
and delta T we've just
found is 76 degrees Celsius.

00:04:10.490 --> 00:04:13.830
So let's plug everything
into our equation.

00:04:13.830 --> 00:04:18.830
Q would be equal to,
the mass is 250 grams.

00:04:19.410 --> 00:04:24.410
The specific heat of water is 4.18 joules

00:04:24.880 --> 00:04:28.900
per gram degree Celsius.

00:04:28.900 --> 00:04:33.570
And the change of temperature
is 76 degrees Celsius.

00:04:33.570 --> 00:04:37.720
So looking at that, we can
see that grams will cancel out

00:04:37.720 --> 00:04:42.030
and degrees Celsius will
cancel out and give us,

00:04:42.030 --> 00:04:47.030
q is equal to 79,420 joules
or to two significant figures,

00:04:51.910 --> 00:04:56.910
q is equal to 7.9 times
10 to the fourth joules.

00:04:59.266 --> 00:05:02.810
So 7.9 times 10 to the
fourth joules of energy

00:05:02.810 --> 00:05:05.260
has to be transferred to the water

00:05:05.260 --> 00:05:08.170
to increase the temperature of the water

00:05:08.170 --> 00:05:13.110
from 22 degrees Celsius
to 98 degrees Celsius.

00:05:13.110 --> 00:05:15.670
The specific heat can vary
slightly with temperature.

00:05:15.670 --> 00:05:18.590
So the temperature is often
specified when you're looking

00:05:18.590 --> 00:05:21.180
at a table for specific heats.

00:05:21.180 --> 00:05:22.930
For example, in the left column

00:05:22.930 --> 00:05:24.250
we have different substances,

00:05:24.250 --> 00:05:26.220
on the right column we
have their specific heats

00:05:26.220 --> 00:05:29.090
at 298 Kelvin.

00:05:29.090 --> 00:05:31.030
So we could use for our units

00:05:31.030 --> 00:05:35.220
for specific heat joules
per gram degrees Celsius,

00:05:35.220 --> 00:05:38.920
or we could use joules per gram Kelvin.

00:05:38.920 --> 00:05:43.920
For liquid water, the specific
heat is 4.18 at 298 Kelvin.

00:05:45.180 --> 00:05:50.180
For aluminum, solid aluminum,
the specific heat is 0.90.

00:05:50.430 --> 00:05:54.227
And for solid iron the
specific heat is 0.45

00:05:55.520 --> 00:05:58.310
joules per gram Kelvin.

00:05:58.310 --> 00:06:01.490
Let's compare the two
metals on our table here.

00:06:01.490 --> 00:06:04.780
Let's compare a solid
aluminum and solid iron.

00:06:04.780 --> 00:06:08.053
So we're gonna add 1.0 times 10

00:06:08.053 --> 00:06:13.053
to the second joules of
energy to both metals

00:06:13.560 --> 00:06:17.760
and see what happens in terms
of change in temperature.

00:06:17.760 --> 00:06:20.240
First, let's do the
calculation for aluminum.

00:06:20.240 --> 00:06:22.360
We're doing Q is equal to mC delta T

00:06:22.360 --> 00:06:27.360
and we're adding 1.0 times
10 to the second joules.

00:06:28.140 --> 00:06:32.970
And let's say we had 10
grams of both of our metals.

00:06:32.970 --> 00:06:36.870
So this would be 10.0 grams of aluminum.

00:06:36.870 --> 00:06:39.910
And then we multiply that by
this specific heat of aluminum,

00:06:39.910 --> 00:06:41.565
which is 0.90.

00:06:41.565 --> 00:06:46.565
So 0.90 joules per gram
Kelvin times delta T.

00:06:49.620 --> 00:06:51.500
When we do the math for this,

00:06:51.500 --> 00:06:55.010
the joules will cancel out,
the grams will cancel out

00:06:55.010 --> 00:06:58.100
and we would find that
delta T would be equal

00:06:58.100 --> 00:07:01.800
to 11 Kelvin or 11 degrees Celsius.

00:07:01.800 --> 00:07:05.040
It doesn't really matter
which units you're using here

00:07:05.040 --> 00:07:07.230
for the specific heat.

00:07:07.230 --> 00:07:10.630
Next let's do the same
calculation for iron.

00:07:10.630 --> 00:07:13.380
So we're adding the same amount of heat.

00:07:13.380 --> 00:07:16.410
So 1.0 times 10 to the
second joules of energy.

00:07:16.410 --> 00:07:20.450
So we can plug that in, 1.0 times 10

00:07:20.450 --> 00:07:22.370
to the second joules.

00:07:22.370 --> 00:07:24.090
We're dealing with the same mass

00:07:24.090 --> 00:07:27.710
so we have 10.0 grams of iron

00:07:27.710 --> 00:07:30.280
but this time we're using
these specific heat of iron

00:07:30.280 --> 00:07:35.280
which is 0.45 joules per
gram Kelvin times delta T.

00:07:39.450 --> 00:07:43.380
So once again, joules cancels
out, grams cancels out

00:07:43.380 --> 00:07:48.380
and we get that delta T
is equal to 22 Kelvin,

00:07:49.080 --> 00:07:52.890
or 22 degrees Celsius.

00:07:52.890 --> 00:07:55.380
What we can learn from
doing these two calculations

00:07:55.380 --> 00:07:58.270
is we had the same amount of heat added

00:07:58.270 --> 00:08:00.300
to our two substances

00:08:00.300 --> 00:08:01.980
with the mass of the two substances

00:08:01.980 --> 00:08:05.850
was the same, the difference
was their specific heats.

00:08:05.850 --> 00:08:10.850
So iron has a lower
specific heat than aluminum.

00:08:10.850 --> 00:08:13.840
And since iron has a lower specific heat,

00:08:13.840 --> 00:08:16.540
it's easier to change the
temperature of the iron.

00:08:16.540 --> 00:08:18.870
So the lower the value
for the specific heat,

00:08:18.870 --> 00:08:21.850
the higher the change in the temperature,

00:08:21.850 --> 00:08:24.440
or you could also say the higher the value

00:08:24.440 --> 00:08:25.860
for the specific heat,

00:08:25.860 --> 00:08:28.230
the smaller the change in the temperature,

00:08:28.230 --> 00:08:30.423
and going back to our chart,

00:08:31.651 --> 00:08:35.150
water, liquid water has a
relatively high specific heat

00:08:35.150 --> 00:08:36.960
which means the temperature of water

00:08:36.960 --> 00:08:39.953
is relatively resistant to change.


Heat transfer and thermal equilibrium

WEBVTT
Kind: captions
Language: en

00:00:00.450 --> 00:00:03.360
- [Instructor] Let's say we
have two samples of helium gas.

00:00:03.360 --> 00:00:05.730
One sample of helium
gas is at temperature T1

00:00:06.910 --> 00:00:09.973
and the other sample of helium
gas is at temperature T2.

00:00:10.840 --> 00:00:13.600
If T2 is greater than T1,

00:00:13.600 --> 00:00:15.080
that means on average,

00:00:15.080 --> 00:00:18.780
the particles of helium
gas in the second box

00:00:18.780 --> 00:00:22.550
are moving faster than the
particles of helium gas

00:00:22.550 --> 00:00:24.170
in the first box.

00:00:24.170 --> 00:00:26.020
We can tell the particles
in the second box

00:00:26.020 --> 00:00:28.320
are moving faster because on average,

00:00:28.320 --> 00:00:32.460
the length of these arrows
indicating the velocity vector

00:00:32.460 --> 00:00:34.620
in the box and the right are longer

00:00:34.620 --> 00:00:38.130
than the length of the
arrows in the first box.

00:00:38.130 --> 00:00:40.660
The equation for kinetic energy

00:00:40.660 --> 00:00:44.420
is equal to 1/2 mv squared,

00:00:44.420 --> 00:00:46.450
where m is the mass of a particle

00:00:46.450 --> 00:00:49.330
and v is the velocity of a particle.

00:00:49.330 --> 00:00:51.990
Since the gas particles
in the box on the right

00:00:51.990 --> 00:00:55.660
are on average traveling faster
and have higher velocities,

00:00:55.660 --> 00:00:58.170
the average kinetic energy

00:00:58.170 --> 00:01:01.130
of the particles in the box on the right

00:01:01.130 --> 00:01:03.800
is higher than the average kinetic energy

00:01:03.800 --> 00:01:06.980
for the particles in the box on the left.

00:01:06.980 --> 00:01:08.880
And so the average kinetic energy

00:01:08.880 --> 00:01:11.160
is proportional to the temperature.

00:01:11.160 --> 00:01:12.760
The higher the temperature,

00:01:12.760 --> 00:01:16.993
the higher the average kinetic
energy of the particles.

00:01:18.090 --> 00:01:21.180
Instead of gases, let's
look at two metal blocks

00:01:21.180 --> 00:01:23.050
made of the same material.

00:01:23.050 --> 00:01:25.670
So one of the metal blocks
is at temperature T1

00:01:25.670 --> 00:01:28.890
and the other metal block
is at temperature T2.

00:01:28.890 --> 00:01:32.800
Let's say the temperature
T2 is greater than T1.

00:01:32.800 --> 00:01:36.230
What that means is the
particles in the metal box

00:01:36.230 --> 00:01:39.170
on the right are on average moving faster

00:01:39.170 --> 00:01:42.530
than the particles in the
metal box on the left.

00:01:42.530 --> 00:01:44.970
And that means that the
average kinetic energy

00:01:44.970 --> 00:01:47.620
of the particles in the
box on the right is greater

00:01:47.620 --> 00:01:48.980
than the average kinetic energy

00:01:48.980 --> 00:01:52.393
in the particles in the box on the left.

00:01:53.340 --> 00:01:55.680
Also notice, right now
are two pieces of metal

00:01:55.680 --> 00:01:57.150
are not touching each other.

00:01:57.150 --> 00:01:59.450
So there's a little bit
of space between them.

00:02:00.450 --> 00:02:01.980
Next, we bring the two objects

00:02:01.980 --> 00:02:04.040
and we put them in
contact with each other.

00:02:04.040 --> 00:02:07.410
So now there are collisions
between the particles that touch

00:02:07.410 --> 00:02:10.170
and this results in the transfer of energy

00:02:10.170 --> 00:02:13.550
from the hotter object
to the cooler object.

00:02:13.550 --> 00:02:17.310
So heat flows from the object
at the higher temperature

00:02:17.310 --> 00:02:20.480
to the object at the lower temperature.

00:02:20.480 --> 00:02:24.410
And so the hot atoms in the
metal object on the right

00:02:24.410 --> 00:02:25.860
start to move a little slower,

00:02:25.860 --> 00:02:28.580
whereas the colder atoms
in the metal on the left

00:02:28.580 --> 00:02:30.710
start to move a little bit faster.

00:02:30.710 --> 00:02:33.810
And the transfer of energy continues

00:02:33.810 --> 00:02:38.350
until both objects have
the same final temperature.

00:02:38.350 --> 00:02:42.653
And we say that thermal
equilibrium has been reached.

00:02:43.620 --> 00:02:45.640
So here we have our two objects,

00:02:45.640 --> 00:02:50.070
and the two objects are at
the same final temperature,

00:02:50.070 --> 00:02:53.870
which means we have reached
thermal equilibrium.

00:02:53.870 --> 00:02:55.750
And since we've reached
thermal equilibrium,

00:02:55.750 --> 00:02:58.530
there's no more flow of heat
between the two objects,

00:02:58.530 --> 00:03:02.000
and since both objects are at
the same final temperature,

00:03:02.000 --> 00:03:04.190
that means the average kinetic energy

00:03:04.190 --> 00:03:07.663
of the particles in both
objects is now the same.