M i i f d ’ l
Physics 231 Lecture 36
• Main points of today’s lecture:• Refrigerators:
Ccoef of performance Q=
• Reversible Refrigerator i.e. ideal or Carnot refrigerator
coef . of performance =W
g
T is in Kelvin
C
h C
coef . of performanc T eT
=T−
T is in Kelvin.
• Entropy:reversibleQ
SΔ
• Review:
reversibleQS
TΔ =
Final Exam
A fi l ti i h d l d f ll ti f Ph i 231• A common final exam time is scheduled for all sections of Physics 231 • Time: Wednesday December 14, from 8-10 pm. • Location for section 002 : BPS 1410 (our regular lecture room). • This information can also be found on our course schedule page
• An alternate exam time will be scheduled for students who have• An alternate exam time will be scheduled for students who have conflicts with the regular time. – Two students have confirmed conflicts with me and will take the
exam thenexam then.
– You must contact me by email and obtain permission from me to take the exam at the alternate time. If you fail to do this, you will b b d f ki h l fi lbe barred from taking the alternate final exam.
– RCPD students should arrange to take their final exam at Bessey.• Alternate time: Tuesday December 13, from 1-3 pm• Location: BPS 1410 (this room)
Heat pumps and refrigerators• Heat engines can run in reverse
– Send in energy– Energy is extracted from the cold reservoir– Energy is transferred to the hot reservoir– Energy is transferred to the hot reservoir
• This process means the heat engine is running as a heat pump– A refrigerator is a common type of heat
pump– An air conditioner is another example of a
heat pump– In the south, people often use heat pumps
to heat homes
heat removed per cycle
• A standard measure of the performance of a refrigerator is its “coefficient of performance”
Ccoef . of performance Q =W
heat removed per cycle
work required to remove it
A Carnot refrigerator is a Carnot engine run in reverse
• A Carnot refrigerator maintains the food inside it at 276 K while theA Carnot refrigerator maintains the food inside it at 276 K while the temperature of the kitchen is 298 K. The refrigerator removes 3.00x 104 J of heat from the food. How much heat is delivered to the kitchen?
A C t h t i k th t
ccarnot
h
As a Carnot heat engine, we know thatQ
e 1Q
= − c
h
T1
T= − c c
h h
Q TQ T
=hQ hT h hQ
c
The Carnot engine run in reverse takes mechanical energy W to move Q from the inside of the refrigerator
d d i Q i h ki hhand deposit Q in the kitchen.
c c
h h
Q TQ T
= 4 4hh c
T 298Q Q 3x10 J 3.23x10 JT 276
= = =h hQ T cT 276
Note:
C C C
h c h C
Q Q TCarnot coefficient of performance=W Q Q T T
= =− −
Checking Understanding: Increasing Efficiency of a Heat PumppWhich of the following changes would allow your refrigerator to use less energy to run? (1) Increasing the temperature inside the refrigerator; (2) increasing the temperature of the kitchen; (3)refrigerator; (2) increasing the temperature of the kitchen; (3) decreasing the temperature inside the refrigerator; (4) decreasing the temperature of the kitchen.
A. All of the aboveB. 1 and 4C 2 d 3C. 2 and 3
Assume the temperature dependence is similar to that of a carnot refrigerator, where
h h
C c
Q TQ T
= c
h c
TCarnot Coefficent of performance =
T T−
Slide 11-31
This is largest when Th and TC are closer to each other in temperature.
Heat pump example• In warmer climates, it is now common to heat a room with a heat pump. It is
ti ll i diti hi h l th t id i d d it h t Q Qessentially an air conditioner, which cools the outside air and deposits heat Qh=Qc+Win inside your room. If your heat pump has a coefficient of performance that is 80% that of a reversible heat pump, how much power would be required to supply 1 kW of heat power to your room when the inside temperature is 250C and the1 kW of heat power to your room when the inside temperature is 25 C and the outside is 15oC.
The inside of your room c
in
QCoeff . of Perf. =
W
c
h c
T=0.8
T T−273 150.8 22.825 15
+= =−
Q 22 8 W =c inQ 22.8 W =
h c inQ Q W= + in in in22.8W W 23.8W= + =
Q 1kW t 23 8W= Δ = 23 8P t= Δ
Th i tdThe
h inQ 1kW t 23.8W= Δ = 23.8P t= Δ
1kW t 23.8P t Δ = Δ
The air outdoorscompressor of the heat pump
1kWP 42W23.8
= =
TQS reversible =Δ
T
• Entropy can only be calculated from a reversible path, and must be done that that way even if the system actually follows an irreversible path– To calculate the entropy for an irreversible process model it as a– To calculate the entropy for an irreversible process, model it as a
reversible process• When heat energy is absorbed, Q is positive and entropy increases
Wh h t i ll d Q i ti d t d• When heat energy is expelled, Q is negative and entropy decreases• In an adiabatic process Q=0 and entropy remains the same.• S ∝ ln(probability). • A disordered state with energy and matter spread out everywhere is more
probable than having all of the energy stored in an organized way that can be used to do work.
Example• The surface of the Sun is approximately at 5700 K, and the
temperature of Earth’s surface is approximately 290 K. What entropy change occurs when 1000 J of energy is transferred by heat from the Sun to Earth?
sun
The sun loses Q of heat and therefore decreases its entropy by the amountQS
T−Δ = sun
sunTThe earth gains Q of heat and therefore increases its entropy by the amount
QSΔ =earthearth
ST
Δ =
The total entropy change is:S S SΔ Δ + Δ
1 1Q
= 1 11000 J 3 27J / K = = sun earthS S SΔ = Δ + Δ
earth sun
QT T
= − 1000 J 3.27J / K
290 K 5700 K= − =
Note that the entropy change is always positive when you transferpy g y p yQ from a hot object to a cold one. The total change in Entropy is nevernegative. Heat does not naturally flow from a cold object to a hot one.
Example
Wh t i th h i t f 1 00 k f li id t t 100°C
ST
Δ = =
• What is the change in entropy of 1.00 kg of liquid water at 100°C as it changes to steam at 100°C?
QST
Δ = vapL mT
=( ) ( )5
322.6x10 J / kg 1kg
6x10 J / K373K
= =
EntropyHigher entropy statesHigher entropy states are more likely. Systems naturally evolve to states ofevolve to states of higher entropy.
Slide 11-33
Second Law of Thermodynamics
Slide 11-34
Example• A power plant has been proposed that would make use of the
temperature gradient in the ocean. The system is to operate between 20.0°C (surface-water temperature) and 5.00°C (water temperature at a depth of about 1 km). (1) What is the maximum ffi i f h t ? (2) If th f l t t f thefficiency of such a system? (2) If the useful power output of the
plant is 75.0 MW, how much energy is absorbed per hour? (3) In view of your answer to (1), do you think such a system is worthwhile (considering that there is no charge for fuel)?worthwhile (considering that there is no charge for fuel)?
• Answers for 1 and 2:– a) .025, 3.2x106 J
max carnota) e e= cT1= −
2781 0.051293
= − =– b).051, 5.3x1012 J– c) . 25, 3.2x107 J– d) 051, 5.3x1013 J
max carnot)hT 293
h
Wb) eQ
= ( ) ( ) 12h
75MW 3600sWQ 5.3x10 Je 0.051
= = =) , hQ
c) What is the energy required to pump the water?
Story of Hawaiian deep water project
• Keahole sits at a point where underwater land slopes sharply down into the sea, it was a place where warm water can be
• KAILUA, HAWAI'I — Koyo USA Corp., a company selling deep-sea water from Keahole Hawai'i, iswas a place where warm water can be
piped from the surface of the sea and cold water can be piped from depths of about a half-mile.
water from Keahole Hawai i, is expanding its plant and has applied to sell the water in the United States
• The company is producing more than • A process called ocean thermal energy
conversion, or OTEC, used the temperature difference between hot and cold sea water to produce 50 KW of
200,000 bottles a day and says it can't keep up with demand in Japan, where it sells 1.5 liter bottles of its MaHaLo brand for $4 to $6 each.cold sea water to produce 50 KW of
electricity at Keahole in 1993. The process worked but it was uneconomical.
MaHaLo brand for $4 to $6 each.
Exam 1 problem 1• Concepts/equations:Co cep s/equa o s:
0
0
for constant acceleration: assuming t 0v v at
== +
( )o
20
1x v v t2
1x v t at
Δ = +
Δ = +
• Solution is posted.
0
2 20
x v t at2
v v 2a x
Δ +
− = ΔA rocket, starting from rest, undergoes a uniform acceleration of 52 m/s2 forp
• Examples of alternate formulations– Give acceleration ask for
uniform acceleration of 52 m/s for distance of 900 m. What is its final velocity?
2 20v v 2a x− = ΔGive acceleration ask for
the distance.– Ask for the final velocity.
• Key is that the physical
0
0v 0.
v 2a x
=
= Δ• Key is that the physical
situation is the same, but the question is different.
v 2 52 900m / sv 305m / s
==
Exam 1 problem 3• Concepts/equations:Co cep s/equa o s:
( )gtvv yy 0
2
1
downwards m/s 9.8 g−=
=
( )
gttvy
tvvyyy
y
yoy
20
0
21
21
−=Δ
+=−≡Δ
A pitcher throws a ball horizontally to a catcher 17 7 m away When the ball is
tvxvv
ygvv
xx
yy
0
20
2 22
Δ=
Δ−=−catcher 17.7 m away. When the ball is caught, its height has decreased by 0.3 m. What is the initial speed of the ball?
• Examples of alternate formulations
tvx x0=Δ 2
2
g 9.8 m/s downwards 1y gt 0.3m2
=
Δ = − = −formulations– Give drop distance and ask
for speed.Give drop distance and ask x0
2 0.3mt sqrt 0.25sg
x v t
= =
Δ =– Give drop distance and ask for horizontal displacement
x0
x0x 17.7mv 71.5m / st 0.25s
Δ= = =