Paperwork
•
•
•
•
•
•
•
•
Assignments fo 18&19 due next week
Today 20.4-20.5
Problems for 18&19
Will finish chapter 20 Monday (Guest)
Tuesday – Lab3 & Quiz3
Wed – Problems for Ch. 20
Fri – Ch.21 (Guest)
Exam Following Friday
Engines = Cyclic Processes
•
•
•
•
•
•
•
Initial State = Final State
Q = DU + W
What does this imply about DU?
DU = UF – U0 = U – U = 0
Engine returns to starting state (p,V,T)
But Heat can enter system (Fuel, Energy In)
Work is Done (Energy Out)
– Q=0+W
Engine
Some Energy tapped to do work. Some Lost.
Not Path Independent means non-conservative!
Heat not a liquid, but think H20 Tower
QH
Work
QC
Perpetual Motion?
QH
Work
QC
Why must some heat (energy) be lost to cold side?
WaterWheel
• Converts Water Flow (H20 KE) to Usable
Energy (Wheel KE)
WaterWheel – on a H20 Tower
Tower Reservoir - Higher Graviational PE
Not all H20 Energy Used
Some flows to ground
Could convert all?
Ground - Lower Graviational PE
WaterWheel – on a H20 Tower
Tower Reservoir - Higher Graviational PE
Stop flow down just means higher “ground”
H20 KE less, also stops wheel from turning
Same Problem with heat
Need a “sink” to direct flow
Perpetual Motion?
QH
Work
QC
Why must some heat (energy) be lost to cold side?
No cold sink, no heat flow! [Warning heat is NOT a liquid]
What Happens if I Reverse W?
QH
Work
QC
Work Entering System instead of Work being done by system.
dQ = dU + W (Here, W <1)
What Happens if I Reverse W?
QH
Work
QC
Work Entering System instead of Work being done by system.
dQ = dU + W (Here, W <1) Can Make dQ negative (Flow backwards)
Makes cold side colder?
At least it removes heat energy
QC is heat taken out of cold side (what you want done)
W is work it takes to do this
QH is heat that goes into hot side (QH>QC)
Fridge exhaust - RT, Hotter or Colder?
QH
Work
|QH|+|W|=|QC|
QC
Refrigerator
Fridges have K, like efficiency:
K
QC
W

QC
QH  QC
QH
Work
QC
Efficiency
• If 100% efficient
– Perpetual Motion (patents everwhere…)
• Engines – Not all Heat converted to work
• Fridge – Need some work to force heat to
flow from cold object to hot object
• Statements = 2nd law of thermodynamics
– No Free Lunch?
?’s
Choose 1
• If you increase the temperature of an ideal gas?
–
–
–
–
its volume must increase
its pressure must increase
the speed of its molecules must increase
the average distance a molecule travels between
collisions must increase
– the average time between collisions of its molecules
must increase.
• Answer – speed. Implications of each?
Choose 1?
• Which of the following is FALSE?
– A dumbbell molecule like O2 is considered to have
eight degrees of freedom.
– The average kinetic energy of a molecule moving in
three dimensions is always 3kT/2.
– Rotational motion, as well as translational motion, can
contribute to the heat capacities of gases.
– The average kinetic energy associated with each
degree of freedom of a molecule is 1/2kT.
– Vibrational motion, as well as translational motion,
can contribute to the heat capacities of gases.
• 5 not 8. Remainder?
Choose 1
• Q = DU + W. Which of the following statements
is FALSE?
– W is the work done by the system, and not on the
system.
– Q can be positive or negative.
– This is called the law of conservation of energy.
– This is called the First Law of Thermodynamics.
– It follows that since Q and W are path dependent,
then DU must also be path dependent.
• Last One – Internal Energy depends only on
Temp for ideal gas (+ KE, PE’s, etc… for others)
Choose Any
• Which of the following are true statements about an ideal
gas? (What is isochoric?)
– During an isochoric process, the change in the internal energy of
the gas is exactly equal to the amount of heat that goes into (or
out of) the gas.
– During an isothermal process, no heat enters or leaves the gas.
– During an isochoric process, no work is done on or by the gas.
– During an isothermal process, the temperature of the gas does
not change.
– During an adiabatic process, the temperature of the gas does not
change.
• Only 1st one. What about 3rd one?
Choose any?
• Which of the following is an accurate statement?
– A typical gasoline engine has an efficiency of about 2%.
– An important distinction between the Diesel cycle and the Otto
cycle is that for the Diesel cycle high efficiencies may be
obtained with low compression ratios.
– An important distinction between the Diesel cycle and the Otto
cycle is that there is no fuel in the cylinder at the beginning of the
compression stroke and no spark plug is used.
– Because a Diesel engine requires no fuel ignition system, Diesel
engines tend to be lighter and easier to start than a comparable
gasoline engine.
– The efficiency of the Otto cycle does not depend on the
compression ratio.
• Middle: Implications of others?
Friday
• Return Quiz 1
• Cover one problem each from 18,19,20
– Or at least 18&19