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## AP®︎/College Chemistry

### Course: AP®︎/College Chemistry > Unit 10

Lesson 1: Internal energy- First Law of Thermodynamics introduction
- More on internal energy
- Calculating internal energy and work example
- Heat and temperature
- Specific heat and latent heat of fusion and vaporization
- Chilling water problem
- Pressure-volume work
- Work from expansion

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# More on internal energy

Getting more intuition of internal energy, heat, and work. Examples of using the first law to calculate work. Created by Sal Khan.

## Want to join the conversation?

- What is the exact difference between enthalpy and heat? If there is a difference..(50 votes)
- Intuitively, enthalpy can be said to be the internal energy of a system as well as the energy it takes that system to establish itself in differentiation or displacement from its environment. For example, when considering the enthalpy of a puddle, it is the amount of internal energy in the puddle, as well as the amount of energy required for that puddle to establish its pressure and volume in relation to the air above it and the earth around and beneath it.(11 votes)

- I really get confused what is meant by work done by system and work done to the system ? i wanna a clear example illustrating that(13 votes)
- A diagram will help illustrate each version of the equation.

Imagine a container drawn as a box like in the video.

The container represents the system.

The boundaries of the container represent the boundaries of the system.

Write "U" inside the container. .This is the system's internal energy.

Draw an arrow starting from outside the system pointing toward the the system and crossing the system boundary. Label it "Q".

This is the heat added to the system.

Draw an arrow starting from inside the system pointing out and crossing the system boundary and label it as "W".

This is the work done by the system across the boundary to external systems.

So the system starts with an initial internal energy U.

Heat Q is added to the system, thus increasing the internal energy U.

Work W is "taken" from the system or done by the system, thus decreasing the internal energy U.

The change in U is just:

a) the heat Q added to (crossing the boundaries of) the system; and

b) the work W done by ("taken from" and crossing the boundaries of) the system.

Change the direction of the arrow representing Q, and now heat is "taken from" the system.

Change the direction of the arrow representing W, and now work is done on the system.

In short, always draw a picture.

A picture really is worth a thousand words.(18 votes)

- Im kind of stuck but will you answer me this question....?

What will be the internal energy of hot tea in a thermos flask after it is vigorously shaked?

what i think is:

since, Del U= q-w

or, Del U= Q - P(Del V)

now, since there is no change in volume so workdone will be zero.

now whats left is, U = Q

and thermos flask is nearer to an adiabatic(isolated) system so Q will also be zero because no heat can flow in or out.therefore,

Del U= 0

which means there is no change in internal energy that is remains constant.

Is this correct?(9 votes)- Assume the temperature of the tea was T1 before you shake it. The internal energy before you shake it will depend (mostly) on the mass of the tea and the temperature. When you shake it you inject a lot more kinetic energy into the tea, but initially, that kinetic energy is not molecular level, random, kinetic energy. It is represented by complex and chaotic and turbulent flow patterns in the tea that will subside when you set the flask down. Until it subsides, it is not part of what is called the internal energy. Until it subsides, it is kinetic energy of macroscopic motion. Once it subsides, it will have become randomized molecular motion and will add to the internal energy. This will be manifested in an increase in the temperature to T2 > T1. The longer and more vigorously you shake it, the greater the final value of T2 and the greater the increase in the internal energy.(22 votes)

- my text book says that there is no intermolecular force between gas molecules in an ideal gas and therefore the potential energy of the molecules is 0 . why is that ? whats causes a molecule to have potential energy ?(8 votes)
- Well, potential energy comes from interaction with a charge, be it mass (gravity), electric charge (electric potential), weak or strong nuclear forces... by reducing intermolecular interaction to 0, and recognizing that the mass of a gas molecule is so tiny as to render gravity negligble (by mgh), the molecules have no potential energy.(11 votes)

- Are heat and work the only two ways to add/subtract energy of a system? Can't there be any more?(11 votes)
- nice question

doing work on a system or exchanging heat. yes, they seem o me to be the only ways,

looking at the second law of thermodynamics i think confirms your idea(3 votes)

- If your question involves Cal, do you need to convert that to joules first?(5 votes)
- what do you think? How do we determine what units we need to use in a particular problem?(2 votes)

- Is heat at molecular level work?(4 votes)
- Hello Ayudh,

You are very close. Heat is a form of energy. By doing work on an object you can increase its temperature and consequently its energy.

For example, rubbing your hands together would increase the temperature of your hands. Your muscles are doing the work to move your hands against friction causing them to heat up.

On a related note, know that heat can be converted to other forms of energy. For example, the heat from burning cola can be converted into electricity...

Regards,

APD(3 votes)

- why is it that work done by the system is negative in a chemistry text book and positive in a physics textbook?(5 votes)
- It's probably because the aforementioned examples are different kinds of work. I'm not sure, really.(1 vote)

- At6:43Sal said that change in interal energy=Q+W is the definition for internal energy can you give me a more precise definition for it?(2 votes)
- Internal energy is all the energy contained in an object, including both kinetic energy and potential energy.(6 votes)

- is Q considered a rate?(2 votes)
- Q is not a rate. It is just a way of transfering energy.(1 vote)

## Video transcript

In the last video we defined
internal energy as literally all the energy that's
in a system. That's kind of the most
inclusive version, at least in my head. So that's my system, it's
some type of container. And I have a bunch of
particles in here. It's literally the sum of
the kinetic energies of all these particles. If they had potential
energies, we'd throw that in there. If they have electrical
potential, we'd throw that in there. If they have bonds with each
other, the energies in that bond, we would throw
it in there. It's everything. It's all inclusive. And, I told you in the last
video it's unintuitive that U stands for internal energy. But I kind of think that, you
know, U, it contains the universe of energy. You know, that's just for
me to memorize it. Don't think too deeply into
what I just said. But this is internal energy. If you show me a system, it
has some internal energy. It's a function of its state. I don't care how it got to that
state, but if you told me a system at a certain state,
maybe with a certain pressure, or a certain volume, or at a
certain temperature-- although if you give me pressure and
volume that should be enough-- I can tell you what its
internal energy is. Especially if you tell me the
type of molecule I have, and things like that. Now we also said in the last
video that because internal energy is all the energy in
the system, it can't be randomly created or destroyed. It can just be transformed
from one form to another. So if I have a change in
internal energy, it can only be due to-- well, it can be
due to more than what I'm describing. But in our simple world where
all of the energy in a system-- and maybe we're dealing
with gases, because that's normally what you deal
with in a first-year chemistry course-- it's going to be the
heat that could be added to the system, plus the work
done on the system. And like I said in the last
video, sometimes they'll say, instead of plus the work done
on the system, sometimes they'll say, minus the work
that the system does. Either way. And here I just want to make
another side discussion here, because I decided to write
it without the little deltas here. And the reason why I did that
is if you were to write this equation, which you will see,
you'll see it in textbooks, teachers will do it-- nothing
inherently criminal about that-- but I just do that
because it clears in my mind what heat and what work are
relative to internal energy. If I were to write delta U is
equal to heat delta Q plus delta W to me this implies that
at some point I had some amount of heat in my system,
and then I have a different amount of heat in my system,
and I took the difference between the two and I got
a change in heat. So this implies that heat is
somehow an inherent macrostate of the system, and that's
not the case. I can tell you what the
internal energy of this system is. I can tell you its pressure. I can tell you its volume. I can tell you its
temperature. These are all macrostates
of the system. I don't know how it got to this
situation, but I can tell you about it. I cannot tell you what the
heat is of this system. And that might be a little
unintuitive, because if I ask you, hey I have a cup of coffee,
what's the heat of it? You might say, oh, it's, you
know, you might give me the temperature. Because in our everyday language
we use things like heat and temperature
interchangeably. But in thermodynamics, heat
is a transfer of energy. A way to think about it
is, if internal energy is your bank account. So you could say change
in bank account. And it really is like
the energy bank account of a system. If you have a change in a bank
account, you had some deposits or withdrawals. And heat and work are really
just deposits or withdrawals into our energy bank account. Heat is one kind. Maybe heat is like
a wire transfer. So you could say it's wire
transfers to your account. Transfers to your account
plus check deposits. Now it makes a lot of sense to
say, you could say, what is my value of my bank account? Or you could say what is my
change in my bank account? You take two snapshots of
your bank account at two different times. But would it make sense to say--
you know, I could say I wire transferred $10, right? So I could say, this statement
would be plus $10. And I could say that I
wrote $20 in checks, minus $20 in checks. In which case, my change
in my bank account would be minus $10. Now would it make sense
for me to say change in wire transfer? That implies that when I started
off maybe my bank account had $100 in it
and now it has $90. When my bank account had $100
in it, did it have any wire transfer amount? No, wire transfer was how money
was deposited or taken away from my account. Likewise, it didn't have a
checking deposit account, so I can't really-- it just seems
weird to me to say change in wire transfer is $10, or change
in check deposits is $20 or minus $20. Would you say, I made a
$10 wire transfer and I paid $20 in checks. So I had a net change in
my bank account of $10. Likewise, I say how much work
was done to me or I did, which is essentially a deposit or
withdrawal of energy. Or I could say much heat was
given to me or how much heat was released, which is another
way of depositing or withdrawing energy from my
energy bank account. So that's why I like
to stick to this. And I like to stay
away from this. And just like I said, you can't
say how much heat is in the system. So someone will say, oh, how
much heat is in this? You cannot say that. There's no heat state variable
for that system. You have internal energy. The closest thing to heat,
we'll talk about it in a future video, is enthalpy. Enthalpy is essentially a way of
measuring how much heat is in a system, but we won't talk
about that just now. And you can't say how much
work is in a system. You can't say, oh I have x
amount of work in a system. The system can do work or have
work done to it, but there's not a certain amount of work,
because that energy in the system could be all used for
work, it could all be used for heat, it could all be used for
a ton of different things. So you can't say those things. And that's why I don't like
treating them like state variables, or state functions. So with that said, this
is our definition. Let's do a couple of
simple problems, just to give you intuition. And I really want to make
you comfortable. My real goal is to make you
comfortable with, when to know to use plus or minus
on the work. And the best way to do it is not
to memorize a formula, but just to kind of think about
what's happening. So let's say that I have some
system here and, I don't know, it's a balloon. And let's say that I have no
change in internal energy. Internal energy is 0. And for our purposes we can kind
of view it as the kinetic energy of the particles
haven't changed. And let's say by expanding a
bit, by my balloon expanding a bit-- I did some work. I'll do this in more detail
in the next video. So the system does 10
joules of work. My question is, how much heat
was added or taken away from the system? So the way I can think about
it-- you don't even have to write the formula down, or you
can write it-- you could say, look, the internal energy, the
amount of energy in the system hasn't changed. The system did 10
joules of work. So that's energy going
out of the system. It did 10 joules. So 10 joules went out
in the form of work. If the internal energy did not
change, then essentially 10 joules of energy had to
go into the system. 10 joules had to go
into the system. If it didn't, the internal
energy would have gone down by the amount of work we had. And the only way, if is this is
the net work, the only way that we're talking about, that
we can add energy outside of work, is through heat. So 10 joules of heat must
have been added. So we can write 10 joules
of heat added to system. Now let's look at that from
the point of view of the actual formula. If we have delta U is equal to
heat added, plus work done to the system, then we would
say, OK, this was 0. It's equal to the heat
added to the system. Remember, in this way we're
saying this is the heat done, or the work done,
to the system. W is work done. Now, the system did work
to something else. It didn't have work
done to it. So if this is work done to the
system, and it did work, then this is going to
be a minus 10. Minus 10 joules. And then you solve both
sides of this. You add 10 to both sides and
you get 10 is equal to Q, which is exactly what
we got up here. But this can get confusing
sometimes, because you're like, oh, is this heat
that the system did? Is this heat that was added to
the system or taken away? The convention tends to be
that this is heat added. But then sometimes
it's confusing. Is this the work done to the
system or work done by? And that's why I like just
doing it this way. If the system does work
it loses energy. If the system has work done
to it, it gains energy. So let's do another problem. I mean, I could have done this
exact same thing using the other formula that you'll
sometimes see. Delta U is equal to Q minus the
work that the system does by the system-- Work
done by the system. And in this case, once
again, change in internal energy was 0. That is equal to heat
added to the system, minus the work done. So minus-- I told you at the
beginning of the problem that the system did 10 joules of
work-- so minus the work done. Minus 10 joules. We get the same situation
up here from two different formulas. And we got 10 is equal to Q. Either way, the heat added to
the system is 10 joules. Let's do one more. Let's say that, I don't know,
5 joules of heat taken away from system. And let's say that 1 joule
of work done on system. So maybe we're compressing the
balloon on the system. What is our change in
internal energy? Let's just figure
out our change. So the way I think of it is 5
joules of heat taken away from the system, that's going
to reduce our internal energy by minus 5. And if 1 joule work is done onto
the system, we're putting energy into it so that'll
be plus 1. So minus 5, plus 1, is
going to be minus 4. Or enter our change in internal energy is minus 4 joules. Now we could have done that a
little bit more formally with the formula, change in internal
energy is equal to heat added to the system, plus
work done on the system. So it's equal to heat
added to the system. We had 5 joules heat taken away,
so this is minus 5, plus work done on the system. We have 1 joule of work done
on the system, and there we get, once again, minus 4. Now, I could have written that
same formula the other way. I could have said change in
internal energy is equal to heat added to the system minus
the work that the system does. I want to do this both ways,
because I don't want you to get confused, either way you
see it on problems or in school, or maybe you take one
class that does it one way and one class that does
it the other. If you use it this way,
what was the heat added to the system? We had heat taken away,
so it's minus 5. So minus the work that the
system does, how much work does the system do? Well, the system had 1 joule
of work done to it. So it did itself minus
1 joules of work. So it's minus minus 1. I want to be clear, when I use
this formula, this is work done by system, done by. This is work done to. The system had work
done to it. So it had minus 1 joules of work
done by the system, so these become pluses, and you
get back to a minus 4. Do whatever's intuitive
for you. For me, this tends to be the
most intuitive, where you don't even use a formula. We just think, if I'm doing
work, I'm using up energy. If I get work done to me, I'm
being handed over energy. If I have heat taken away from
me, I'm giving away energy. If I have heat added to me,
I'm getting energy. See you in the next video.