The Second Law of Thermodynamics

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Physics 102 Professor Lee Carkner Lecture 7. The Second Law of Thermodynamics. PAL #6 First Law. 1 mole of gas at 300 K and 2 m 3 compressed to 1 m 3 , constant pressure P = nRT/V = (1)(8.31)(300)/(2) = 1246.5 Pa W = P D V = (P) (V f -V i ) Sign of work? Volume decreased. Engines. - PowerPoint PPT Presentation

Transcript of The Second Law of Thermodynamics

The Second Law of Thermodynamics

Physics 102Professor Lee

CarknerLecture 7

PAL #6 First Law 1 mole of gas at 300 K and 2 m3 compressed to

1 m3, constant pressure P = nRT/V = (1)(8.31)(300)/(2) = 1246.5 Pa W = PV = (P) (Vf-Vi)

Sign of work?

Volume decreased

Engines

General engine properties: An input of heat An output of heat

Heat and Work Over the Cycle Four parts of the cycle:

compression output of heat QC

Over the course of one cycle positive work is done and heat is transferred

Since the engine is a cycle, the change in internal energy is zero

U=(QH-QC)-W =0W = QH - QC

Efficiency

In order for the engine to work we need a source of heat for QH

e = W/QH An efficient engine converts as much of the

input heat as possible into work

Today’s PAL If an automobile engine outputs

149200 joules per second to the drive train and burns fuel at a rate of 746000 joules per second, what is the efficiency?

If gas is $2.00 per gallon, how much money per gallon are you wasting?

Efficiency and Heat

e = 1 - (QC /QH) The efficiency depends on how

much of QH is transformed into W and how much is lost in QC:

Reducing the output heat means

improving the efficiency

The Second Law of Thermodynamics

This is one way of stating the second law:It is impossible to build an engine that

converts heat completely into work

Engines get hot, they produce waste heat (QC)

Carnot Engine In 1824 Sadi Carnot related the maximum efficiency to the

temperature of the reservoirs:eC = 1 - (TC / TH)

A hot input reservoir and a cold output reservoir make it “easier” to

move heat in and out

e < eC Another way to state the second law is:

There is a limit as to how efficient you can make your engine

The First and Second Laws The first law of thermodynamics says:

The second law of thermodynamics

says:

The two laws imply: W < QH W QH

Dealing With Engines Most engine problems can be solved by knowing how to

express the efficiency and relate the work and heats:W = QH - QC

e = W/QH = (QH - QC)/QH = 1 - (QC/QH)

eC = 1 - (TC/TH)

For individual parts of the cycle you can often use the ideal gas law:

PV = nRT

P-V Diagram for Engine

The total work output per cycle

Positive work is clockwise

Refrigerators

A refrigerator is a device that uses work to move heat from low to high temperature

The refrigerator is the device on the back of the box Your kitchen is the hot reservoir

Heat QC is input from the cold reservoir, W is input power, QH is output to the hot reservoir

How a Refrigerator Works

The fluid is pumped into the hot chamber (coils on the back) and compressed, adding work W

Need special fluid that can evaporate and condense in the right place

Refrigerator Cycle

Liquid

Gas

Compressor (work =W)

Expansion Valve

Heatremovedfrom fridgeby evaporation

Heat added to room bycondensation

HighPressure

Low Pressure

QC QH

Refrigerator Performance Input equals output:

The equivalent of efficiency for a refrigerator

is the coefficient of performance COP:COP = QC / W

Unlike efficiency, COP can be greater than 1

Today’s PAL Lets say you wanted to cool your

house on a hot day so you buy a refrigerator, plug it in and open the door.

Does the temperature of the house, increase, decrease or stay the same? Why? (assume insulated house)

Heat Pumps

It removes QC from your house and exhausts QH to the outside

It removes QC from the outside and adds QH to your house

Heat pump COP = QH / W Want the most heat output for the work

Refrigerators and Temperature

We can relate the coefficient of performance to the temperature:

COP = TC /(TH-TC)

This is the maximum COP for a fridge operating between these two temperatures

Refrigerators and the Second Law

You cannot move heat from low to high temperature without the addition of work COP cannot be infinite

Heat doesn’t flow “uphill” by itself, although this would not violate the first law

Statements of the Second Law

It is impossible for any device which

operates in a cycle to convert heat completely to work

For refrigerators:

Next Time Read: 15.7-15.11 Homework: Ch 15, P 26, 31, 35, 37 I will also post some practice

problems Won’t count for grade