Thermodynamics:
Heat and Work
Work and energy
• Recall W = ΔKE
• Work can increase the
internal energy of a
substance
• Internal energy decreases
by giving off heat.
• The reverse can also happen
First Law of Thermo
• ΔU = Q – W,
• The change in internal
energy (ΔU) equals the heat
(Q) added to the system
minus the work (W) done by
the system.
First Law Explained
• Remember that heat is
kinetic energy and work is
a change in kinetic energy.
• To do work ON something
means to GIVE it energy (as
either heat or motion)
• To do work means to SPEND
energy.
Consider this
• What is meant by internal
energy again?
• Can internal energy be
negative?
• What does this mean for
the first law?
Question 1
• What happens to the
internal energy of water
vapor in the air that
condenses on the outside of
a cold glass of water? Is
work done or is heat
exchanged?
Answer
• It decreases
• Heat transfer
– Water condensating on a
glass doesn’t move it or break
it (no work)
Math Example
• 2500J of heat is added to a
system, and 1800J of work
is done ON the system.
What is the change in
internal energy of the
system?
Solution
• Use ΔU = Q – W,
• ΔU = 2500J – (-1800J) = 4300J
• NOTE: work is negative
because it is being done ON
the system.
Math Example 2
• 2500J of heat is added to a
system and 1800J of work
is done BY the system.
What is the change in
internal energy of the
system?
Solution
• Again use ΔU = Q – W,
• ΔU = 2500J – 1800J = 700J
• NOTE: work is positive here
because work is done BY
the system.
• On = negative
• By = positive
The system
• So far we have treated the
internal energy of a
substance or a combination
of substances as a single
quantity to or from which
energy is transferred.
• Such a substance or
combination is called a
system.
Examples of
Systems
• The hammer, nail, and
board
• The water and metal
• A hot air balloon
• A car engine
In real life
• Systems are almost never
completely isolated from
its surroundings.
• A system interacts with
its environment.
Work done on gasses
• Again recall,
Work = force x distance
• Recall pressure = force /
area and
volume = area x distance.
• These three facts lead us
to the following conclusion:
W = Fd = PΔV =
pressure x change in volume
Important!
• W = PΔV makes one very
important assumption. Can
you guess what that
assumption is?
• That temperature remains
constant.
Example
• A cylindrical piston with an
area of 0.025m2 is pushed
0.05m by a gas that exerts
a constant pressure of 9 x
105 Pa. How much work was
done by the gas on the
piston?
Solution
•
•
•
•
A = 0.025m2
d = 0.05m
P = 9 x 105 Pa = 9 x 105 N/m2
W = PΔV = PAd =
9 x105N/m2(0.025m2)(0.05m) =
1125J
Question 2
• Use the conservation of
energy to explain why the
temperature of a gas
increases when it is
compressed and decreases
when it expands.
Answer
• Compressing a gas takes
work. Doing work on the
gas increases its internal
energy in the form of heat.
• Expanding gasses are doing
work (either pushing other
gasses out of the way or
pushing the walls of its
container out) and losing
energy.
Thermodynamic
Processes
• In reality energy is
transferred as both heat
and work to some extent.
• However, in most cases one
type of energy transfer
dwarfs the other.
• Therefore, we can use ideal
processes to approximate
real life ones.
On strike
• When a gas changes
temperature but does not
change in volume no work is
done.
• Such a process is called a
constant volume process or
isovolumetric process
• Iso = same
• Volumetric = volume
Conceptual Example
• A bomb calorimeter is an
enclosure in which substances
undergo a combustion reaction.
• The gas released by the
reaction has nowhere to go. All
the energy is transferred as
heat.
• This device tells you how much
energy is released by the
reaction.
Staying cool
• When a gas expands slowly
enough it stays in
thermodynamic equilibrium
with the environment.
• As it expands it does work on
the environment but no net
heat is transferred.
• Such a process is called
isothermal
• Iso = same
• Thermal = temperature
Conceptual Example
• Picture a partially inflated
balloon.
• If the atmospheric pressure
decreases, like before a storm,
the gas will expand.
• As it expands it decreases in
internal energy and
temperature.
• However, heat transfers from
the air into the balloon, keeping
it a constant temperature.
Question 3
• In an isothermal process,
3700J of work is done by an
ideal gas. Is this enough
information to tell how
much heat has been added
to the system? If so, how
much?
Answer
• In an isothermal process
there is NO net heat
transfer.
• So, we do have enough info.
• There is 0 heat added.
Question 4
• Is it possible for the
temperature of a system
to remain constant even
though heat flows into or
out of it?
• Yes, as long as the system
is free to expand or
contract.
Rapid Expansion
• Isothermal processes must
happen slowly.
• If a gas expands rapidly its
temperature, pressure, and
internal energy decrease.
• If this happens in a closed
environment, no heat can be
transferred to or from the
environment, such a process is
called an adiabatic process
from a Greek word meaning
impassible
Question 5
• Explain why the
temperature of a gas
increases when it is
adiabatically compressed.
Answer
• Adiabatically means the
heat can’t get out.
• The trapped heat must be
used to increase the
internal energy of the gas,
which will increase its
temperature.
One Way Street
• In nature many things
happen spontaneously while
the reverse thing does not.
• Heat flow from hot to cold
• Falling objects creating
heat
• A falling glass breaking
2nd law of
thermodynamics
• There are several ways to
state this law but here is
the most common:
“Heat flows naturally from
a hot object to a cold
object; heat will not flow
spontaneously from a cold
object to a hot object” –
R.J.E. Clausius
2nd law and heat
engines
• Heat engine – any device that
changes thermal energy into
mechanical work.
• Examples: steam engines and
car engines
• Work can only be obtained when
heat is allowed to flow from a
high temperature to a low
temperature
How it works
• A heat input,
QH, at a high
temperature,
TH, is partly
transformed
into work, W,
and partly
exhausted as
heat, QL, at a
lower
temperature,
TL
Operating
Temperature
• TH and TL are called the
operating temperatures of
the engine and are
important in calculating
how efficient your engine
is.
Efficiency
• The efficiency, e, of any
heat engine is the ratio
comparing the work it does
to the heat input it uses.
• e = W/QH
• Conservation of energy
tells us:
QH = W + QL or W = QH – QL
• This changes e = W/QH to
(QH – QL) / QL or 1 – (QL / QH)
Example
• An automobile engine has an
efficiency of 20 percent
and produces an average of
23,000J of mechanical work
per second. How much heat
is discharged from this
engine per second?
Solution
•
•
•
•
•
W = 23,000J
e = 0.20
e = 1 – (QL / QH)
QL / QH = 1 – e = 0.80
Also QH = W/e = 23,000J /
0.20 = 1.15E5J
• QL = 0.80QL = (0.80)(1.15E5J) =
9.2E4J
• This is 92,000 watts.
Efficiency in terms
of Temp
Because heat is proportional to
temperature we can rewrite
our efficiency equation:
eideal = (TH – TL) / TH = 1 – (TL / TH)
We call this e the eideal because
this would be the efficiency of
a perfect engine, one that
takes in heat and uses all of it
for work.
D.N.E
• This leads us to our next
statement of the 2nd law:
• “No device is possible whose
sole effect is to transform
a given amount of heat
completely into work.” –
Kelvin – Planck
• Translation: a perfect
engine does not exist
Question 6
• Is it possible to cool down a
room on a hot summer day
by leaving the refrigerator
door open?
Answer
• No way, the refrigerator
must give off waste heat,
which would be released
into the room.
Question 7
• Which will give the greater
improvement in the
efficiency of a Carnot
engine, a 10oC increase in
the high-temperature
reservoir, or a 10oC
decrease in the lowtemperature reservoir?
Answer
• Start with the fraction
3/4.
• What is bigger, 2/4 or 3/5?
• 3/5 obviously,
• Because e = 1 – TL / TH,
• 1 – 2/4 is greater than
1 – 3/5. Dropping the lower
temp is better.
In other words
• The previous statement
(the Kelvin – Planck
statement) is commonly
explained this way:
• You can’t get something for
nothing (1st law)
• You can’t even break even
(2nd law)
Entropy
• Entropy: a measure of the
order or disorder of a system
• When dealing with entropy it is
the change in entropy that we
care about.
• ΔS = Q / T
• Where S is entropy, Q is still
heat, and T is the temperature
in Kelvin
Question 8
• The oceans contain a
tremendous amount of
thermal energy. Why, in
general, is it not possible
to put this energy to useful
work?
Answer
• The ocean also has a very large
entropy, every water molecule
is going every which way.
• Imagine a bunch of 5 year olds
running around in all
directions, it’s a lot of energy
but you couldn’t use it to push
a boulder unless they were all
going the same direction.
Example
• An ice cube of mass 60g is
taken from a storage
compartment at 0oC and
placed in a paper cup. After
a few minutes, exactly half
of the mass of the ice cube
has melted, becoming
water at 0oC. Find the
change in entropy of the
ice/water.
Solution
• Step 1: find Q
– The heat needed to melt 30g
of ice is found using the
latent heat of fusion.
– Q = mL = (30g)(79.7cal/g) =
2400cal = 2.4Cal
• Step 2: Find ΔS
– ΔS = Q / T = 2400cal / 273K =
+8.8 cal / K
But wait
• In the last example the
temperature stayed the same.
• What happens to the entropy
(or disorder) of the system if
the temperature changes?
• Answer: ΔStotal = ΔSH + ΔSC, where
ΔSH is the entropy at the higher
temp and ΔSC is the entropy at
the lower temp.
Example
• Find the change in entropy
when 50.0kg of water at
20.0oC is mixed with 50.0kg
of water at 24.0oC.
Solution
• Step 1: find Q
– Q = mc ΔT = (50.0kg)(1.00kcal
/kg*K)(2.0oC) = 100kcal
– Because equal masses of
water were mixed the final
temperature will be the
average of the 2, or 22oC
Solution
• Step 2: find ΔSH
ΔSH = -100kcal / 296K = 0.338kcal / K
– Note Q is negative because
the hot water gives off heat.
• Step 3: find ΔSC
ΔSC = 100kcal / 294K = 0.340kcal
• Step 4: add, ΔS = -0.338 + 0.340 =
0.002kcal/K
Notes
• Notice in the last example that
even though the entropy went
down for one part of the
system, the entropy for the
total system when up.
• ΔS > 0
• If the system is not isolated,
then the total entropy is the
entropy of the system plus the
entropy of the environment.
• ΔS = ΔSs + ΔSenv  0
Entropy and the 2nd
Law
• We have finally come to our
general statement of the 2nd
Law:
• The total entropy of any
system plus that of its
environment increases as a
result of any natural process.
• This can also be stated as:
Natural processes tend to move
toward a state of greater
disorder.
Information Theory
• It is accepted that a more
orderly system requires
more information to
describe it and a less
orderly system requires
less information.
Conceptual example
• Remember our falling rocks
creating heat when it hits the
ground?
• Why can’t rocks absorb heat
and jump up into the air?
• The random motion of thermal
energy is 1 piece of info.
• In a falling body all molecules
fall at the same speed as well
as randomly due to thermal
energy. (2 pieces)
Question 9
• What happens if you open a
container of chlorine gas?
Does the reverse ever
happen?
Answer
• The gas molecules will
spread out and fill the
room.
• No, the molecules will not
randomly all gather back
into the bottle.
Question 10
• Suppose you collect a lot of
papers scattered all over
the floor and put them in a
neat stack; does this
violate the second law of
thermodynamics? Explain.
Answer
• The work you did
organizing the papers
disturbed the air in the
room and created more
heat, both of which will
increase the entropy of the
system MORE than the
decrease in the entropy of
the papers.
Heat Death
• As a system goes from orderly to
disorderly the amount of work you
can get out of the system decreases.
• Well since the entropy of the
universe is always increasing, the
logical conclusion is that eventually
the universe will be a uniform
mixture at thermal equilibrium.
• No work can happen and no change
will occur, this is called the Heat
Death of the Universe (note: assumes
a finite universe)