July 17, Week 7 Today: Heat Pumps and Engines, Chapter 11...

July 17, Week 7

Thermodynamics 21st July 2014

Today: Heat Pumps and Engines, Chapter 11

Final Homework #7 now available. Due Monday at 5:00PM.

First Law of Thermodynamics


First Law of Thermodynamics: W +Q = ∆Eth

First Law of Thermodynamics


First Law of Thermodynamics: W +Q = ∆Eth

There are two ways to change the thermal energy of on object -Work being done to the object (W ) and heat (Q)

First Law Signs


In applying the first law of thermodynamics, we have to think aboutthe “system” = the object that is of interest. Everything else iscalled the environment

First-Law Followup


Process W Q ∆Eth ∆T

First-Law Followup



Steam is used to spin a turbine.(Assume the turbine’s temperatureremains constant)

First-Law Followup




− 0 − −

First-Law Followup




− 0 − −

An expanding gas inflates a balloon

First-Law Followup




− 0 − −

An expanding gas inflates a balloon − ? ? ?Heat depends on how the gas expands

First-Law Followup




− 0 − −


After 30 minutes of baking, a pan isremoved from the oven and sits on acounter

First-Law Followup




− 0 − −


After 30 minutes of baking, a pan isremoved from the oven and sits on acounter

0 − − −

Heat Engine


Heat Engine - Device that uses the transfer of heat from a highertemperature to lower temperature to extract work

Heat Engine



W +Q = ∆Eth

Heat Engine



W +Q = ∆Eth

The system is the engine

Heat Engine



W +Q = ∆Eth


W = −Wout

Heat Engine



W +Q = ∆Eth


W = −Wout

Q = QH −QC

Heat Engine



W +Q = ∆Eth


W = −Wout

Q = QH −QC

∆Eth = 0 (In a perfect situation,the engine simply uses heat todo work. It doesn’t absorb theheat itself.)

Heat Engine II



W +Q = ∆Eth

Heat Engine II



W +Q = ∆Eth

−Wout +QH −QC = 0

Heat Engine II



W +Q = ∆Eth


Wout = QH −QC

Heat Engine II



W +Q = ∆Eth


Wout = QH −QC

Better Form: QH = Wout +QC

Heat Engine II



W +Q = ∆Eth


Wout = QH −QC


Efficiency: e =WoutQH

Heat Engine II



W +Q = ∆Eth


Wout = QH −QC



e =QH −QC

QH= 1−

QCQH

Heat Engine Exercise


What is the efficiency (Wout/QH) of the following heat engine?

HOT

COLD

100 J

75 J




HOT

COLD

100 J

75 J

(a) 0%




HOT

COLD

100 J

75 J

(a) 0%

(b) 25%




HOT

COLD

100 J

75 J

(a) 0%

(b) 25%

(c) 50%




HOT

COLD

100 J

75 J

(a) 0%

(b) 25%

(c) 50%

(d) 75%




HOT

COLD

100 J

75 J

(a) 0%

(b) 25%

(c) 50%

(d) 75%

(e) 100%




HOT

COLD

QH = 100 J

QC = 75 J

Wout

(a) 0%

(b) 25%

(c) 50%

(d) 75%

(e) 100%




HOT

COLD

QH = 100 J

QC = 75 J

Wout = 25 J

(a) 0%

(b) 25%

(c) 50%

(d) 75%

(e) 100%

Heat Pump


Heat Pump - Device that does work in order to move heat from coldto hot

Heat Pump



W +Q = ∆Eth = 0

Heat Pump



W +Q = ∆Eth = 0

W = Win

Heat Pump



W +Q = ∆Eth = 0

W = Win

Q = QC −QH

Heat Pump



W +Q = ∆Eth = 0

W = Win

Q = QC −QH

Win +QC −QH = 0 ⇒

Win +QC = QH

Heat Pump



W +Q = ∆Eth = 0

W = Win

Q = QC −QH


Win +QC = QH

Coefficient of Performance:

COP =QCWin

=QC

QH −QC

Heat Pump Exercise


If the COP (QC/Win) of the following heat pump is 2, how muchheat was removed from the cold reservoir and how much heat wasreleased to the hot reservoir?

HOT

COLD

20 J

Heat Pump Exercise



HOT

COLD

20 J

(a) QC = 10 J , QH = 30 J

Heat Pump Exercise



HOT

COLD

20 J

(a) QC = 10 J , QH = 30 J

(b) QC = 10 J , QH = 10 J

Heat Pump Exercise



HOT

COLD

20 J

(a) QC = 10 J , QH = 30 J

(b) QC = 10 J , QH = 10 J

(c) QC = 20 J , QH = 20 J

Heat Pump Exercise



HOT

COLD

20 J

(a) QC = 10 J , QH = 30 J

(b) QC = 10 J , QH = 10 J

(c) QC = 20 J , QH = 20 J

(d) QC = 40 J , QH = 60 J

Heat Pump Exercise



HOT

COLD

20 J

(a) QC = 10 J , QH = 30 J

(b) QC = 10 J , QH = 10 J

(c) QC = 20 J , QH = 20 J

(d) QC = 40 J , QH = 60 J

(e) QC = 40 J , QH = 20 J

Heat Pump Exercise



HOT

COLD

Win = 20 J

(d) QC = 40 J , QH = 60 J

Heat Pump Exercise



HOT

COLD

Win = 20 J

COP = QC/Win ⇒ QC = COP ×Win ⇒ QC = 2× 20 J

(d) QC = 40 J , QH = 60 J

Heat Pump Exercise



HOT

COLD

Win = 20 J

COP = QC/Win ⇒ QC = COP ×Win ⇒ QC = 2× 20 J

40 J

QH

(d) QC = 40 J , QH = 60 J

Entropy


Heat engine have a maximum efficiency (that is MUCH less than100%) because of the second law of thermodynamics

Entropy



The second law is based on the concept of entropy

Entropy



The second law is based on the concept of entropy

Entropy - Measures the amount of disorder in a system.

Entropy Exercise


Which of the following processes would involve a decrease inentropy? In each case the system has been underlined.

Entropy Exercise



(a) A gas is allowed to escape from its container.

Entropy Exercise




(b) A college student throws his clothes on the room’s floorlooking for the perfect outfit.

Entropy Exercise





(c) Your physics instructor spills coffee on himself.

Entropy Exercise






(d) A child neatly stacks her blocks to make a tower.

Entropy Exercise







(e) A piece of food warms while sitting under a heat lamp.

Entropy Exercise




A more ordered system, like stacked blocks, has less entropy.

Entropy Exercise




A more ordered system, like stacked blocks, has less entropy.

Disordered systems are more likely to occur, so entropy is also ameasure of the probability that a state of a system will occur.

The Second Law of Thermodynamics


The Second Law of Thermodynamics - The entropy of an isolatedsystem never decreases. It increases until the system reachesequilibrium and then stays constant.




The second law does NOT imply that the entropy never decreases.The system must be isolated. If there is an external force doingwork on the system, its entropy can decrease.




The second law does NOT imply that the entropy never decreases.The system must be isolated. If there is an external force doingwork on the system, its entropy can decrease.

On the macroscopic scale, any process that obeys the second lawwill be irreversible. (Irreversible processes increase entropy.)

Second Law Exercise


Which of the following processes violates the second law ofthermodynamics?

Second Law Exercise



(a) A freshly-baked pie sits on the counter and cools.

Second Law Exercise




(b) The wind gathers up the leaves in your front yard and blowsthem into a neat pile.

Second Law Exercise





(c) Liquid water freezes to form ice.

Second Law Exercise






(d) A man piles logs into a neat pyramid.

Second Law Exercise






(d) A man piles logs into a neat pyramid.

(e) An acorn grows into a very tall tree.

Second Law Exercise




This is the “reverse” of an everyday event. The wind can scatterthe leaves from a neat pile, but it cannot gather them.

Heat Engine II



W +Q = ∆Eth


Wout = QH −QC



e =QH −QC

QH= 1−

QCQH

Heat Pump



W +Q = ∆Eth = 0

W = Win

Q = QC −QH


Win +QC = QH

Coefficient of Performance:

COP =QCWin

=QC

QH −QC

Maximum Efficiencies


Using the mathematical version of the second law, we can findequation for the maximum efficiency of heat engine and COP of aheat pump.




emax = 1−TCTH




emax = 1−TCTH

COPmax =TC

TH − TC

July 17, Week 7First Law of ThermodynamicsFirst Law SignsFirst-Law FollowupHeat EngineHeat Engine IIHeat Engine ExerciseHeat PumpHeat Pump ExerciseEntropyEntropy ExerciseThe Second Law of ThermodynamicsSecond Law ExerciseHeat Engine IIHeat PumpMaximum Efficiencies

July 17, Week 7 Today: Heat Pumps and Engines, Chapter 11...

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