Video Transcript
In this video, we’re talking about
specific heat capacity. As we’ll see, specific heat
capacity helps us understand how different materials respond to changes in their
internal energy level.
As we get started, say we have a
piece of a given material. For our purposes, the material
could be anything. It could be wood or stone or
plastic. Using this material, we want to run
an experiment. We wanna see what happens when we
put energy into the material in terms of its temperature. Since one of the easiest ways to
put energy into something is to heat it, let’s say we put a Bunsen burner underneath
this block. Over time, the effect is that heat
energy is transferred into the block.
Let’s say we let this process go on
for one minute. Then over this time, an amount of
energy — we’ll call 𝐸 — is input to the block. Over this one-minute interval,
we’re also measuring the temperature of the block. And what we find is that it goes
through a temperature change. We’ll call it Δ𝑇 that change in
temperature is equal to the final temperature of the block — that is, its
temperature after one minute of heating — minus its initial temperature, its
temperature before the heating started.
So what we’ve seen so far is that
if we put an energy 𝐸 into our block, then what we get out is a change in
temperature Δ𝑇. But let’s imagine that we let our
experiment continue on for another minute. At the end of the second minute,
two times 𝐸 joules of energy have been input into the block. And as we continue to keep track of
the block’s temperature, we find that after two 𝐸 joules of energy is input, the
temperature change is now two times Δ𝑇. That is, it’s twice the temperature
change that we found after one minute had elapsed.
What our measurements of this
heated block are showing us is that the energy that we input to the block is
proportional to the change in temperature that the block undergoes. That is, if we double the energy
input, then the temperature change of the block doubles. If we quadruple the energy input,
the temperature change quadruples, and so on. We can write out a relationship
that summarises this relationship or discovering. We’re finding that the energy input
to the block — we’ve called it 𝐸 — is directly proportional to the change in
temperature of the block.
Now an important caveat for this
relationship is that it’s only true if the state of matter of our block isn’t
changing. In other words, it started out
solid and it needs to remain as a solid in order for this relationship to hold. Assuming the block stays the way it
is though, it doesn’t melt and it doesn’t turn into a gas, then this relationship is
true. Looking at this expression, there
is another way we can write it, a mathematically equivalent way to say this.
Whenever we have an expression that
says that one value is directly proportional to a second value, then we can write
that that first value is equal to some constant value — we’ll call that constant
capital 𝐶 — multiplied by the second value. Even though these two expressions
look different, mathematically they say the same thing. 𝐸 being directly proportional to
Δ𝑇 means that 𝐸 is equal to some constant times Δ𝑇. Looking at this second form of the
equation, the question naturally comes up “what is this constant capital 𝐶?” It turns out that the name of this
constant is the heat capacity.
Here’s the definition of heat
capacity. It’s the amount of energy that’s
needed to raise a material’s temperature by one degree Celsius. So this makes sense. If we go back to our block, then we
would see the heat capacity of this block is how much energy we need to put into it
so that its temperature goes up by one degree Celsius. And an important thing to realise
about heat capacity is that it varies from material to material. The heat capacity of stone is
different from the heat capacity of brick, is different from the heat capacity of
wood, is different heat capacity of water, and so on.
So the energy that’s input into a
substance is equal to the heat capacity of that substance multiplied by its change
in temperature. Knowing this, let’s imagine we want
to expand our experiment. We want to make a comparison
between different materials as they’re heated. To that end, we are adding these
other materials of different shapes from our original block. And then, we start to heat each
one, adding energy to the material. Just like before, as we heat these
various materials we track their temperature.
But soon enough, we discover that
making this comparison between various pairs of these materials isn’t as easy as we
thought. That’s because in our experimental
setup we have more than one independent variable. One independent variable is the
material that each of our objects is made of. But another variable is the fact
that all of these objects have different masses. The mass of our blue block is not
the same as our orange cube, is not the same as our pink pyramid, is not the same as
our green cylinder. This means there’s another variable
that we want to account for if we want to establish a relationship between energy
input and temperature change.
So here’s what we decide to do. We modify all of our shapes so that
now they all have the same amount of mass. In each case, the material has a
mass of exactly one kilogram. At this point, now that the rate of
energy being put into our materials is equal across the materials and the mass of
the materials is all the same, now as we track temperature increase of these
materials over time, we can really compare one material to another. In order to make this comparison
between materials, for the first time, we’ve been careful to notice and keep track
of what the mass of each one is.
But this raises another
question. How does the mass of an individual
material, say our original blue block, affect the energy transfer into that block
and how its temperature changes? In other words, if we once again
put an amount of energy 𝐸 into this blue block and we measured a temperature change
Δ𝑇, how does the mass of the block affect that relationship? To find out, we try another
experiment. We tried doubling the mass of our
block from one to two kilograms and then we put the same amount of energy 𝐸 into
this block.
What we’ll find is that compared to
the temperature change Δ𝑇 that corresponded to a mass of the block of one kilogram,
the temperature change now is half of that: Δ𝑇 divided by two. Then trying something more, say
that we double the size of this block once again. Now, it’s four kilograms. If we once again add the same
amount of energy 𝐸 to the block, now the change in temperature isn’t Δ𝑇 over
two. It’s Δ𝑇 over four. Here’s what we’re seeing. When we start to account for the
mass of our material, we see that this mass and the change in temperature of the
material are related. The more mass there is to a
material, the smaller its temperature change given a fixed amount of energy
input.
This makes some intuitive
sense. Imagine heating a single ice cube
using some fixed amount of energy. Then, imagine applying that same
amount of energy to an entire iceberg. Which one — the ice cube or the
iceberg — would increase more in temperature given the same amount of energy
input? We know it will be the smaller mass
of ice, the ice cube. And here, we’re finding the same
thing that as the mass of a material increases, for a given fixed amount of energy
input, the change in temperature that that material experiences decreases.
This most recent experiment we’ve
performed means we can make a modification to our original proportionality
expression here that energy is proportional to the change in temperature of a
material. This expression is true. But notice that it has nothing to
do with a mass of this material. It’s basically saying for some mass
of material — we don’t know what that mass is, but we know that it’s fixed — the
energy we put into it is proportional to its increase in temperature. But now, we have more
information. We know that the mass of the
material when we vary it affects what the change in temperature of that material
is.
And even more than that, we can say
that there is a relationship between the mass and the change in temperature as well
as the energy input. That relationship is that the
energy we input into our material is proportional to the product of the material’s
mass and its change in temperature. We see that now we have this
product, we start to account for our ice cube-iceberg example. In the case of the ice cube, the
mass was very very small. And so that means for a given
amount of energy input, the change in temperature will be relatively high, whereas
for our iceberg, since the mass is very very high, for that same amount of energy
input, the overall change in temperature of the iceberg would be low.
Now just like before, there is a
second equivalent way we can express this mathematically. We can say that it means that the
energy input into our material is equal to some constant — we’ll call this one
lowercase 𝑐 — multiplied by 𝑚 times Δ𝑇. In the previous version of this
equation, we saw that constant was the heat capacity of the material. But now that our equation has
changed, now that it includes the mass of our material, this constant will have a
bit of a different aspect. Rather than our constant being the
heat capacity of the material, our constant now refers to something called the
specific heat capacity. And this is the amount of energy
needed to raise the temperature of one kilogram of a material by one degree
Celsius.
Notice the only difference between
heat capacity — what we had before — and specific heat capacity — what we have now —
is that when we’re talking about specific heat capacity, we’ve specified that the
material is one kilogram in mass. It’s this quantity the specific
heat capacity of a material that we often use to compare one material to another
one. One way to better understand
specific heat capacity is to think of it in terms of its units. When we look at this expression 𝐸
is equal to 𝑐 times 𝑚 times Δ𝑇, we know some of the units involved here.
For example, we know that the base
units of energy, the standard units, are joules. So we have some amount of joules is
equal to the unit of specific heat capacity, whatever that is, multiplied by a
mass. And we know the base unit of mass
is kilograms. And then, that’s multiplied by a
change in temperature. And a change in temperature will
have units of degrees Celsius. And looking at this equation, we
can now see what the units of specific heat capacity must be. Because this is an equation —
because units of joules is equal to the units of specific heat capacity times
kilograms times degrees Celsius — in order for this equation to be true, it must be
the case that specific heat capacity has units of joules per kilogram degrees
Celsius. And based on these units, we can
see how this expression agrees with our description of specific heat capacity.
Specific heat capacity is how much
energy measured in joules is needed to raise the temperature of a one-kilogram
object by one degree Celsius. So the great thing about knowing
the specific heat capacity for a given material, we often either look it up in a
table or are given that information in our problem statement, is that once we know
it based on other information that might be given we could solve for say the mass of
a sample or the change in temperature that undergoes or how much energy is input
into a material.
Before getting some practice doing
this, it’s helpful to know just as a side note that this equation for specific heat
capacity sometimes shows up in a different form. Occasionally, instead of
representing a change in temperature as Δ𝑇, it’s represented as a capital 𝜃
symbol. Which form of the equation we
encounter depends on which textbook we use. But just know that the two
equations are entirely equivalent and they say the same thing. It’s just that one uses Δ𝑇 to
represent change in temperature and the other uses capital 𝜃. Now that’s it. Let’s look at an example involving
specific heat capacity.
Asphalt concrete is used to surface
roads. When under direct sunlight for a
long time, it can get very hot. If 2500 kilograms of asphalt
increases in temperature from 18 degrees Celsius to 40 degrees Celsius, absorbing 50
megajoules of energy from sunlight, what is the specific heat capacity of asphalt
concrete? Give your answer to two significant
figures.
Okay, so say we have this huge slab
of asphalt. We’re told it has a mass of 2500
kilograms and that this mass is heating up in the sun. Thanks to all the energy it absorbs
from the sun, the asphalt goes from a temperature of 18 degrees Celsius to 40
degrees Celsius. That tells us that the change in
temperature undergoes or we can call Δ𝑇 is equal to 40 degrees Celsius minus 18
degrees Celsius or 22 degrees Celsius. We’re told that the amount of
energy the asphalt absorbs in order to effect this change in temperature is 50
megajoules. That’s 50000000 joules of
energy.
Based on all this information, we
want to solve for the specific heat capacity of asphalt concrete. At this point, we can recall that
specific heat capacity is the amount of energy needed to increase the temperature of
one kilogram of a given material by one degree Celsius. As a mathematical equation, we can
say that this amount of energy — what we can call 𝐸 — is equal to the mass of our
substance multiplied by the specific heat capacity multiplied by the change in
temperature of the substance, Δ𝑇.
Like we said, it’s the specific
heat capacity 𝑐 that we want to solve for in this case. So to do that, let’s rearrange this
equation algebraically. Let’s divide both sides of the
equation by the mass of our substance multiplied by its change in temperature. When we do that, we find the
specific heat capacity is equal to the energy added into our material divided by the
mass of that material multiplied by its change in temperature.
In our case, we know the mass of
our substance. It’s given as 2500 kilograms. We also know its change in
temperature. We calculated that to be 22 degrees
Celsius. And moreover, we know the amount of
energy input to our substance, 50000000 joules. So to calculate specific heat
capacity, it’s just a matter of substituting in these values into this
expression. When we do that, our mass of 2500
kilograms and our change in temperature of 22 degrees Celsius can go straight in as
is.
But our input energy, 50
megajoules, needs to change a bit. That’s because mega is a prefix
that means million. So to express our energy in base
units of joules, we need to convert it from 50000000 joules and write it as 50 times
10 to the six joules. With this conversion done, we can
see that the final units of our specific heat capacity will be joules per kilogram
degrees Celsius, just what we want. When we calculate this value to two
significant figures, we find a result of 910 joules per kilogram degrees
Celsius.
This answer means that in order to
heat one kilogram of asphalt concrete by one degree Celsius, we would need to input
910 joules to the concrete. That’s the specific heat capacity
of the asphalt.
Let’s take a moment now to
summarise what we’ve learnt about specific heat capacity.
In this lesson, we first saw that a
material’s heat capacity is how much energy it needs to go up in temperature by one
degree Celsius. But then as we experimented with
this idea, we saw there’s a way to extend it. By accounting for material mass, we
learn that a material’s specific heat capacity is how much energy one kilogram of
the material needs to go up in temperature by one degree Celsius.
Furthermore, we’ve learned that
specific heat capacity, often symbolised using a lowercase 𝑐, fits into a
mathematical equation. This equation tells us that if we
multiply the specific heat capacity of a given material multiplied by a particular
chunk of mass of that material times the change in temperature that mass has
undergone, then that overall product is equal to the total amount of energy input
into that particular mass of that particular material. And we also learned that another
way to write this expression is 𝐸 is equal to 𝑚 times 𝑐 times 𝜃.
Specific heat capacity is useful
for comparing materials and showing how they respond to changes in internal
energy.
