Using the

1
Using the Clicker

• If you have a clicker now, and did not do this
  last time, please enter your ID in your clicker.
• First, turn on your clicker by sliding the power
  switch, on the left, up. Next, store your student
  number in the clicker. You only have to do this
  once.
• Press the button to enter the setup menu.
• Press the up arrow button to get to ID
• Press the big green arrow key
• Press the T button, then the up arrow to get a U
• Enter the rest of your BU ID.
• Press the big green arrow key.

2
Post-test on WebCT Assignment 12

• Dont forget to submit the post-test!
• Also, Assignment 12 on WebAssign is optional
  (although you will be responsible for this
  material on the final exam), but you can use your
  score on the assignment to replace your lowest
  assignment score (out of 20 points) of the
  semester.
• Exam location (Monday Dec. 17th from 6-9 pm)
• 8 am and 6 pm classes MOR 101
• 2 pm class SCI 107

3
Worksheet

• A thermodynamic system undergoes a three-step
  process. An adiabatic expansion takes it from
  state 1 to state 2 heat is added at constant
  pressure to move the system to state 3 and an
  isothermal compression returns the system to
  state 1. The system consists of a diatomic ideal
  gas with CV 5R/2.
• The number of moles is chosen so nR 100 J/K.
• The following information is known about states 2
  and 3.
• Pressure P2 P3 100 kPa
• Volume V3 0.5 m3
• What is the temperature
• of the system in state 3?

4
Apply the ideal gas law

• The system does 20000 J of work in the constant
  pressure process that takes it from state 2 to
  state 3. What is the volume and temperature of
  the system in state 2?

5
The temperature in state 2
What is the temperature of the system in state 2?
1. 200 K 2. 300 K 3. 500 K 4. 700 K 5.
None of the above
6
Finding work

• The system does 20000 J of work in the constant
  pressure process that takes it from state 2 to
  state 3. What is the volume and temperature of
  the system in state 2?
• For constant pressure, we can use
• Finding volume
• Finding temperature (use the ideal gas law, or
  )

7
Complete the table

• For the same system, complete the table. The
  total work done by the system in the cycle is
  19400 J.
• First fill in all the terms that are zero.
• Each row satisfies the First Law of
  Thermodynamics.
• Also remember that

Process Q ?Eint W
1 to 2
2 to 3 20000 J
3 to 1
Entire cycle -19400 J
8
Complete the table

• For the same system, complete the table. The
  total work done by the system in the cycle is
  19400 J.
• Q is zero for an adiabatic process.
• The change in internal energy is zero for an
  isothermal process, and is always zero for a
  complete cycle.

Process Q ?Eint W
1 to 2 0
2 to 3 20000 J
3 to 1 0
Entire cycle 0 -19400 J
9
Complete the table

• For the same system, complete the table. The
  total work done by the system in the cycle is
  19400 J.
• Even the last row has to satisfy the first law

Process Q ?Eint W
1 to 2 0
2 to 3 20000 J
3 to 1 0
Entire cycle -19400 J 0 -19400 J
10
Complete the table

• For the same system, complete the table. The
  total work done by the system in the cycle is
  19400 J.
• Find the change in internal energy for the 2 ? 3
  process.

Process Q ?Eint W
1 to 2 0
2 to 3 50000 J 20000 J
3 to 1 0
Entire cycle -19400 J 0 -19400 J
11
Complete the table

• For the same system, complete the table. The
  total work done by the system in the cycle is
  19400 J.
• Rows have to obey the first law.
• Columns have to sum to the value for the entire
  cycle.

Process Q ?Eint W
1 to 2 0 -50000 J
2 to 3 70000 J 50000 J 20000 J
3 to 1 0
Entire cycle -19400 J 0 -19400 J
12
Complete the table

• For the same system, complete the table. The
  total work done by the system in the cycle is
  19400 J.
• Rows have to obey the first law.
• Columns have to sum to the value for the entire
  cycle.

Process Q ?Eint W
1 to 2 0 -50000 J 50000 J
2 to 3 70000 J 50000 J 20000 J
3 to 1 -89400 J 0 -89400 J
Entire cycle -19400 J 0 -19400 J
13
A heat engine

• A heat engine is a device that uses heat to do
  work. A gasoline-powered car engine is a good
  example.
• To be useful, the engine must go through cycles,
  with work being done every cycle. Two
  temperatures are required. The higher temperature
  causes the system to expand, doing work, and the
  lower temperature re-sets the engine so another
  cycle can begin. In a full cycle, three things
  happen
• Heat QH is added at a relatively high temperature
  TH. Some of this energy is used to do work W.
  The rest is removed as heat QL at a lower
  temperature TL.
• For the cycle QH W QL (all positive
  quantities)

14
Efficiency

• In general, efficiency is the ratio of the work
  done divided by the heat needed to do the work.
• The net work done in one cycle
• is the area enclosed by the cycle
• on the P-V diagram.

15
Carnots principle

• Sadi Carnot (1796 1832), a French engineer,
  discovered an interesting result that is a
  consequence of the Second Law of Thermodynamics.
• Even in an ideal situation, the efficiency of a
  heat engine is limited by the temperatures
  between which the engine operates. 100
  efficiency is not possible, and most engines,
  even in ideal cases, achieve much less than 100
  efficiency.
• Carnots principle
• Ideal (Carnot) efficiency

16
A refrigerator
If you had a refrigerator in a closed,
well-insulated room and you left the fridge door
open for a long time, what would happen to the
temperature in the room? 1. It would increase
2. It would decrease 3. It would stay the same
17
Heat engines running backwards

• Refrigerators and air conditioners are heat
  engines that run backward. Work is done on the
  system to pump some heat QL from a low
  temperature region TL. An amount of heat QH QL
  W is then removed from the system at a higher
  temperature TH.
• (a) Represents the cylinder in a car engine
  while (b) represents a refrigerator.

18
A heat pump

• If you heat your home using electric heat, 1000 J
  of electrical energy can be transformed into 1000
  J of heat. An alternate way of heating is to use
  a heat pump, which extracts heat from a
  lower-temperature region (outside the house) and
  transfers it to the higher-temperature region
  (inside the house). Let's say the work done in
  the process is 1000 J, and the temperatures are
  TH 27C 300 K and TL -13 C 260 K. What
  is the maximum amount of heat that can be
  transferred into the house?
• Something less than 1000 J
• 1000 J
• Something more than 1000 J

19
A heat pump

• The best we can do is determined by the Carnot
  relationship.
• Using this in the energy equation gives
• For our numerical example this gives
• This is why heat pumps are much better than
  electric heaters. Instead of 1000 J of work going
  to 1000 J of heat we have 1000 J of work
  producing 7500 J of heat.

20
Entropy

• Entropy is in some sense a measure of disorder.
• The symbol for entropy is S, and the units are
  J/K.
• A container of ideal gas has an entropy value,
  just as it has a pressure, a volume, and a
  temperature. Unlike P, V, and T, which are quite
  easy to measure, the entropy of a system is
  difficult to calculate.
• On the other, a change in entropy is easy to
  determine.

21
Change in entropy

• Entropy changes whenever there is a transfer of
  heat. The change in entropy is the heat added
  divided by the temperature at which the transfer
  took place.
• If the heat transfer takes place at a single
  temperature, the change in entropy is simply
• isothermal process  
• If the heat transfer takes place over a range of
  temperatures then, as long as ?T is small
  compared to the absolute temperature T, the
  change in entropy is approximately

22
The Second Law of Thermodynamics

• The entropy of a closed system is constant for
  reversible processes and increases for
  irreversible processes. Entropy never decreases
  (for a closed system).
• For a closed system,

23
Dropping a glass
You drop a glass of milk and the glass smashes
into 27 pieces and milk spills all over the
floor. If you videotaped this and ran the film
backwards, it would be obvious to you that the
film was running backwards. Why? The process
violates 1. The Law of Conservation of Energy
2. The Law of Conservation of Momentum 3. The
Second Law of Thermodynamics 4. All of the above
24
Entropy times arrow

• In the process of smashing the glass of milk,
  both energy and momentum are conserved. However,
  the entropy is increased The direction of time is
  the direction of increasing entropy.

Reversible and irreversible processes
In an irreversible process, the entropy of a
closed system increases. In a reversible
process, the entropy stays the same.
25
Reversible or not?
You have two styrofoam containers of water. Each
holds 1 kg of water. In one the water temperature
is 17C, while in the other it is 37C. The
colder water is then poured into the warmer
water, and the system is allowed to come to
equilibrium. Is this process reversible or
irreversible? 1. Reversible 2. Irreversible
26
Irreversible!

• The container of water will not spontaneously
  separate into two parts that differ in
  temperature by 20, so this process is
  irreversible. Lets calculate the change in
  entropy.
• Find the heat
• Use the change in entropy equation, using an
  average temperature of 22C 295 K for the
  cooler water and 32C 305 K for the warmer
  water.

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Whiteboard