[narrator] As we’re learning about the broader topic of thermodynamics let’s talk about
the quantity of internal energy.
The first thing that we’ve got to do in any thermodynamics problem, when we’re talking
about the movement of heat which is thermos and dynamics--thus where we’re coming from,
movement of heat, we have to carefully define in a system that we’re going to use as our
reference point.
The system is the only thing that we care about and so when energy goes into a system
it’s greater than zero.
If energy turns around and leaves a system we’re going to say that that energy change
is negative.
Okay, and so everything else other than the system is defined as the surroundings and
the surroundings encompass everything that is not within the system.
Defining our system is going to be very important.
We have to make sure we know what the system is in every single problem that we do.
Internal energy, or rather in our discussion we’re looking at the change in internal
energy, this capital delta (Δ) symbol means “change in”.
Really, any time you see that change in it means the final state minus the initial state.
That delta just adds final minus initial to any quantity that it precedes.
So, for thermodynamics one of the more important equations is being able to determine the change
in internal injury as the sum of heat energy change plus the sum of work done on, or by,
the system.
Let’s reconsider our system for a moment.
If heat energy goes in to a system we’re going to consider that change to be greater
than 0.
The number will be positive.
Likewise if heat leaves the change in heat energy will be less than 0.
If the system gets warmer we’ve added heat, that’s positive.
If the system loses energy then it’s going to be negative.
Work--if work is done on the system, by the surroundings, that’s going to be greater
than zero as well.
Then if the system does work--any time you do work you get tired because you lose energy.
The system is no different.
The system loses energy when it does work, so we’ll consider that to be a negative
change in energy.
This--being able to interpret this diagram is going to be very important.
Another thing we have to consider is the first law of thermodynamics.
That’s where we get the basis of this positive and negative from is that heat that is gained
or lost by the system will be equal to the negative of heat gained or lost by the surroundings.
What that means is energy cannot be created nor destroyed.
It’s, rather, transferred.
The heat energy can only be transferred from one to the other.
If the system has positive heat then the surroundings is going to have negative heat.
Likewise, if the system is losing heat, that’s a negative times a negative, that means the
surroundings gains heat.
This is if system gains the surroundings lose.
If the system loses heat energy the surroundings will gain energy because negative times a
negative gives us a positive.
The best thing to do is examples for this.
If we read a problem; 5.00 J of heat is transferred from the surroundings to the system while
the system does 5.00 J of work what is the net change in internal energy?
We have to decide, for our net change, whether q + w is either positive or negative for each
individual quantity.
If we go from surroundings into the system the system is going to gain energy.
q is going to be positive and work, the system does the work--suppose the system lifts something,
suppose the system has to push something a certain distance.
Force times distance is energy as well, the units for work and the formula for work, and
so that means the system, since it’s doing the work, it’s going to lose energy.
We can say negative for w.
We plug those numbers in.
We get +5 plus a negative, so minus, 5 as well.
We get 0 as the net change.
To keep our units the same 0 J of change.
What about if we have a problem where 500.
J of heat is transferred?
We’re not really telling us which way it’s going, directly.
Write our formula down.
Resulting in a temperature drop in the system so that tells us that heat is leaving the
system and going into the surroundings.
We have to define our heat and work transfer by the system, that’s why we define it carefully
so it’s our reference point.
We don’t care about the surroundings in these types of problems, other than their
relationship to the system.
-q is a temperature drop.
5.00 kJ of work is done on the system by the surroundings.
The surroundings are doing the work, that means energy is going into the system as a
quantity from work.
Let’s put everything together.
500.
J, the point there holds three significant figures in place.
It’s negative because we’re losing heat and then 5.00 kJ is going in from work.
Now, we’ve got to be careful here because we’ve got units that don’t match.
All right, so we know how to convert with conversion factors, at this point, so we need
to do that here.
We could say 500.
J. What do we want to get rid of?
We want to get rid of joules.
We want to get kilojoules.
What are two equal quantities?
So we’re only multiplying by one.
Well, there’s 1,000 joules in one kilojoule, that’s what the k means; is 1,000 times
a joule.
We end up with 0.500, to three significant figures and we convert to kilojoules.
We’ve got to be careful.
What does the problem ask for?
Does it ask for kilojoules?
Or does it ask for joules?
Well, it doesn’t say in this problem.
But, it might on a future problem.
Be careful and look for that.
-0.500 kJ + 5.0 kJ is going to be, to three significant figures--well now we have to be
careful now, though, because we’re doing addition and subtraction.
We can only go to the hundredths place because we only go to the hundredths place here.
It’s not like multiplication where we take all three significant figures.
1,000 has an unlimited number of significant figures because it’s a definition.
So, here, we get 4.50 kJ is the net change, and that’s a positive change in energy for
the system.