Bahan Ajar Termodinamika

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Thermodynamics

Base competence : analize the change of ideal gas

condition by applying the laws of thermodynamics

Aminatur Rahmawati

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1

Standard Competence :

apply the laws of thermodynamics in heat engines

Base Competence :

analize the change of ideal gas condition by applying the laws of

thermodynamics

Learning Objectives :

Products

1. To define thermodynamics systems

2. To explain the concept of internal energy, heat, and work as expressed by the first law of thermodynaics.

3. To calculate work done by gases

4. To explain the condition of hot water poured to the cold water 5. To explain the temperature change of mixed water

6. To state and explain second law of thermodynamics. 7. To explain the concept of entropy

8. To define reversible and irreversible process

9. To explain the principle of carnot engine and its ideal efficiency 10.To analize the ideal gas processes based on a P-V diagram

11.To explain gas condition because of the change in temperature, pressure, and volume 12.To mention two examples for each process (isothemal, isobaric, isochoric, adiabatic) 13.To apply second law of thermodynamics in everyday life.

Psychomotor

14.To discuss the thermodynamics processes

15.To do simple experiment to show second law of thermodynamics

Affective

16.Character : being creative, critical, logic, working accurately, honestly, and getting polite behaviour

17.Keterampilan sosial : bekerjasama, menyampaikan pendapat, menjadi pendengar yang baik, dan menanggapi pendapat orang lain.

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How to Use This Learning material

This learning material consists of two aspect, they are conceptual basic and

application. In order to interest the reader, this book is designed like a house.

The readers are invited to learn physics while exploring the house.

Veranda

is a front side of a house. It containts everyday phenomena

related to the chapter. Reader will starting to have inquiring taste here.

Reading room

is a place where we can find any information

that we want.

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Balcony

is an upper floor projecting from the wall . We can think

lea l he e, e use e feel f esh ai , a d e jo i g the ie . “o, it’

s perfect

place to do mind activity. Problems will be given, sometimes conceptual, but at

he other time will be exercise problems.

Attic

is a space directly below

the roof of a house. It is usually used

for storage. Some important information will be saved in the attic.

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Living Room

is a room fore casual activity, like spending time with family

or just relax. So there will be some application of the matter in our everyday

life.

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THERMODYNAMICS

Refrigerator can cooling your food, drinks, vegetables, also fruits. It even can make you a cube of ice. Do you know how it works? Why your mom puts fruits and vegetables in the bottom and ice in the top of refrigerator? Let’s fi d out i this chapter.

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Veranda

You must be know about this tragedy in our nation. Suddenly hot mud rises up to the earth surface in Sidoarjo, East java. It was a big tragedy,

e ause the e a e se e dist i t o ke a ata o e ed hot ud. Many people lose their home, land, and farm. Some people think that it is a sign that the nature is angry o us. The others said that it is a recurring legend. But do you know that it is an application of fist law of

thermodynamics?

At certain location on the Earth, heat water in the interior rises up to the surface as hot springs. In Yellowstone National Park, USA it produces geysers like the picture above. In Iceland, the hot water warms the ocean and create warm lagoon which is surrrounded by glaciers.

The uses of hot springs is not only for recreation. It also used for a renewable source to generate electric energy. In this chapter you will learn about in what codition, and with what efficiency heat can be exploited to p odu e o k i hu a od a d i a hi es. “o, do ou still thi k hat fo i lea The od a i s ?

Word Thermodynamics comes from greek, that means Transfer of heat (therme : heat, dynamics : transfer). The development of

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you can enjoy the air conditioner, refrigerator, and another heat machines. Let’s go to li rary to explore more.

Reading Room

A.

Thermodynamics System

In thermodynamics, an amount of matters that we observe is called

system. And everything arround it is called environtment. If there is no heat transfer into or out of the system, the system said to be thermally isolated system.

When you learn about heat and temperature, you find that heat is generated by a change of temperature. Heat transfers from higher temperature to the lower temperature. So, heat is a form of energy transfer.

B.

The First Law of Thermodynamics

If a gas with constant volume is heated, the temperature will increase. So the molecules of gas move faster than before. It cause more collution between molecules and wall. The collutions cause the pressure of gas increase, also the kinetic energy. It means the internal energy also incrrease.

In order to raise gas temperature, an amount of heat (Q) is needed. If an amount of heat is added to the system, the heat will be used to do work. But there is some of heat is used to raise internal energy of the system. If an amount of heat is given to the system, the internal energy of the system will increase.

The heat given to the system is expressed by equation

Q = ∆U + W

Q : heat transferred to or from the system (J)

∆U : change of internal energy (J)

W : work done by system (J)

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The equation above is the formulation of first law of thermodynamics. We have to follow some rules to apply first law of thermodynaics. The rules are

1. If heat transferredto the system, value of Q is positive (+Q). If heat

transferred from the system, value of Q is negative (-Q).

2. If the system do work, value of W is positive (+W). If the system accept work, value of W is negative (-W).

You can also find work done by system using curve. Work done is equal to under the process curve on a P-V diagram.

P

V

V1 V2

Now let’s go to al o a d take a look at application of first law of thermodynamics.

Balcony

A worker with weight 65 kg shovels coal for 3 hours. During the shoveling, the worker did work at average rate of 20 W and loss heat to the

environtment at average rate of 480 W. How much fat will the worker lose? Energy value of fat Ef is 9.3 kcal/g.

Solution

Listing the given value, then converting power to owrk and heat.

Given: W = Pt = (20 W)(3h)(3600s) = 2.16 x 105 J

Q = - (480 W)(3 h)(3600 s) = -5.18 x 106 J (Q is negative because heat is lost)

Ef = 9.3 kcal/g

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= 3.89 x 107 J/kg

Find: mass of fat burned

Answer from the first law of thermodynamics, Q = ∆U + W, we have

∆U = Q – W = - 5.18 x 106 J – 2.16 x 106 J = - 5.4 x 106 J

The mass of fat loss is

m =

=

= 0.14 kg

follow up exercise

How much fat will be lost if the worker were playing basketball, doing work at rate of 120 W and generating heat at rate of 600 W?

Hint : the answer is 0.2 Kg

Living Room

Lapindo and First Law of Thermodynamics

The example of The first Law of Thermodynamics is mud spray in sidoarjo. The mud inside the earth blows out because it was at a higher pressure than the atmosphere. So it did positive work on its surroundings. So its internal energy decreased (negative). ∆U = Q – W. Because the work is

positive, and ∆U is negative, so the Q is zero. The reduction in internal e e g ill ause ud’s s oke.

Reading Room

C.

Thermodynamics Processes for an Ideal Gas

The first law of thermodynamics can be applied to several processes for an ideal gas system. In three of the processes, one thermodynamics

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processes is kept konstant. He three processes have names that begin with

iso- (from the greek, isos means equal). To learn the characteristic of the processes, just do the steps below.

Balcony

Work sheet

“

Thermodynamics processes

”

Name : Class_:

Imagine that you are asked by a detective to help him to analize The od a i s P o ess. No let’s help hi to fi d what are the Thermodynamics Processes through exploring the given data below.

a)

From the data above, you can find what process happens to the system. Just follow this way:

 Plot graphic from the given data, Pressure as Y-axis and Volume as X-axis

 Based on the graphic you have made, what process happens to the system?

Number Pressure (N/m2) Volume (m3)

1 2 3

7 x 105 4 x 105 3 x 105

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b)

c)

d)

1 2 3 4 5

10 x 105 8 x 105 6 x 105 4 x 105 2 x 105

10 10 10 10 10

1

2

3

4

5

6 x 105

2

4

6

8

10

Number Pressure (N/m2) Volume (m3) 1

2 3 4 5

10 x 105 8 x 105 6 x 105 4 x 105 3 x 105

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Based on your simple investigation, now you consider what are the Thermodynamics Process happens to the system. Just write them here, also the characteristics.

CASE SOLVED! Thanks for your help..

Reading Room

1. Isothermal Process

Isothermal process is a constant-temperature process. It means ∆T =

0. From the equation ∆U = n R ∆T, we know that ∆U = 0

So the first law of thermodynamics become

Q = ∆U + W Q = W

And the work done by Isothermal system is stated in this equation.

W =

∫

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W

=

∫

P

P2 T1 = T2

W = nRT

∫

P1

V

W = nRT ln

(

)

V1 V1

P-V diagram of an isothermal process

2. Isobaric Processes

A constant-pressure process is called isobaric process. On a P-V diagram, an isobaric process is represented by a horizontal line called isobar. When heat is added to system, the volume will increase. It means system doing work. So

W = P ∆V

W = P (V2 – V1)

We can write the internal energy as

Q = ∆U + W

Q = ∆U + P (V2 – V1)

Because of P is constant,

P

P1 = P2

V1 V2 V

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Diagram above is a P-V diagram for Isobaric process. Now you are invited to go to balcony and thinking about a match.

Balcony

VS

Let’s o side a at h et ee Isothe s a d Iso a s. Ea h of the a t to be the greatest process by doing the largest work. So they use two moles of an ideal gas, initially at 0°C and 1 atm. The gases are expanded to twice their original volume using two different processes. First they are expanded isothermally, and then starting Isobarically from the same initial state.

During which process does the gas do more work and be the winner? isothermal process, isobaric process, or both process do the same work? Explain.

Conceptual Reasoning

Based on a P-V diagram, the curve of isobaric process is horizontal line, a d the isote ’s is h pe ola. So, the gas doing more work during isobaric process because there is more area under the curve.

Follow-Up Exercise

Now try to determine the work done by the gas in each process.

Hint : for isothermal process W = 3.14 x 103 J

for isobaric process W = 4.53 x 103 J

Library

3. Isochoric Process

An isochoric process is a volume-constant process. As illustrated in figure(bla) the process path on a P-V diagram is vertical line.

P

Isotherms

[image:16.595.157.505.115.231.2]

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P2

P1

V1 = V2 V

No work is done, because the area under curve is zero. Because the gas can not do work, if heat is added, it is completely used to increase internal energy of the system.

W = 0

In term of first law of thermodynamics,

Q = ∆U + W

Q = ∆U + 0

Q = ∆U

4. Adiabatic Process

In adiabatic process, there is no heat transferred to or from system. That is,

Q = 0

This condition ocuurs on a thermally isolated system. For real life condition, we only approximate adiabatic process if the changes occur rapidly enough. Because if you wait long enough, there will be heat transferred to or out of system.

The curve of this process is called adiabat. During an adiabatic

process, al thermodynamics coordinates (P, V, T ) change. For example, if a pressure reduced, the gas expands. However there is no heat flows to the gas. So, the change of internal energy is equal to work done by system, but change of internal energy is negative (internal energy decrease).

Q = ∆U + W

0 = ∆U + W

-∆U = W

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Because the internal energy and temperature decrease, the process is cooling process. Similarly, an adiabatic compression process is warming process.

We need to state other relationship in adiabatic process. An important fa to is the atio of the gas’s ola spe ifi s heat, defined by quantity

ϒ = Cp / Cv

Where Cp is specific heats at constant pressure, and Cv is specific heats at constant volume. For monoatomic and diatomic gas, the value of ϒ is about 1.67 and 1.40. The volume and pressure at any two points on an adiabatic are related by

P1 V1ϒ = P2 V2ϒ

The work done by an ideal gas during an adiabatic process can be shown to be

W =

(

P1 V1– P2 V2)

The curve of adiabatic is similar with the curve of isothermal process, but for adiabatic is steeper.

Balcony

Conceptual Example : Exhaling : Blowing Hot or Cold?

The air in your lungs is warm. This can be demonstrated by putting your forearm near your mouth and blow air with your mouth opened wide. If you repeat this with your lips puckered, the air will feel : a) warmer b) cooler c) the same

Reasoning and Anwer

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When you blow with mouth opened, you gush a warm air. However, when you blow with lips puckered, the stream of air is compressed. Then the air expand, doing work againts the atmosphere. The process is approximately adiabatic, because it takes place quickly. Because Q = 0, ∆U = -W.

Therefore ∆U is negative, and the temperature decrease.

Follow-Up Exercise

Even during winter days with snow on slopes, It is common in the Rocky Mountain to experiance blast of warm air coming down the slopes. Explain how this winds could experience a significant rise in temperature while there is still snow and ice on the ground.

Hint : the process occurs is adiabatic.

The Attic

No ou k o a out the od a i s p o esses. Let’s sa e it i the atti .

Thermodinamics Processes

The Kitchen

In order to understand about the second law of thermodynamics, you can do a simple experiment below.

 place cold water in a glass

 pour hot water to the same glass

 observe what happen, if necessary check the mixed water with your finger.

Process Characteristics Result First Law of

Thermodynamics

Isothermal T = Constant ∆U = 0 Q = W

Isobaric V = Constant W = P ∆V Q = ∆U + P ∆V

Isochoric P = Constant W = 0 Q = ∆U

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What happen to the mixed water? Is it still cold or hot? Why it can be happen?

Reading Room

D.

The Second Law of Thermodynamics

Suppose that a piece of hot metal is placed in an insulated container of cool water. Heat will be transferred from the metal to the water, and they will come to thermal equilibrium at certain temperature.

The second law of thermodynamics says that certain processes do not happen, or never been observed to happen, even though they may be consistent with the first law. There are several equivalent statements of the second law, they are:

Heat will not flow spontaneously from a colder body to a warmer body.

Clausius state that it is impossible to transfer haet from cooler body to warmer body without any work. An equivalent statements of the second law involves thermal cycle. A thermal cycle consist of several separate thermal processes which the system ends back at its starting condition. It means the finall coordinate (P, V, T ) is the same with inial coordinate.

In a thermal cycle, heat energy can not be completely transformed into mechanical work

Mechanical work can be done by a machine that transform heat com pletely into work and motion, with no energy loss. However, real machines are always have less than 100 % efficiency. So it is impossible to transform heat completely into mechanical work.

Entropy

A property that indicates the natural directions of a process was first described by Rudolph Clausius, a German physicist. This property is called e t op . The ha ge i s ste ’s e t op is

∆S =

Whe e ∆“ is ha ge of s ste ’s e t op

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Q is heat added to or removed from system T is temperature.

The statements of second law of thermodynamics in terms of entropy are



∆



Balcony

Change in entropy : an isothermal process

While doing physical exercise at 34°C, an Cristiano Ronaldo loses 0.4 Kg water per hour by the evaporation from his skin. Count change of entropy of water as it vaporizes. The latent heat of vaporization is about 24.2 x 105 J/Kg.

Solution

Given m = 0.4 Kg

T = 34 + 273 = 307 K Lv = 24.2 x 105 J/Kg

Find ha ge i e t op ∆“

Q = m Lv = 0.4 Kg 24.2 x 105 J/Kg = 9.68 x 105 J ∆S = =

= 3.15 x 10

5 J/K

Q is positive, bacause heat is added to the water. The change of entropy also positive, and the entropy of water increase.

Follow-Up Exercise

What is change in entropy of a 1 Kg water when it freezes to ice at 0°C?

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Reading Room

Reversible and Irreversible Process

I o de to lea a out e e si le a d i e e si le p o ess, let’s go to the balcony and think it through.

Balcony

The example of irreversible process is explotion of a bomb, or a glass break or a paper burned.

A d the e a ple of e e si le p o ess is ta af a ou d ka’ ah, o he you get out from home to go to school,the go to ou f ie d’s ho e, the go home again.

Now, can you explain what is reversible and irreversible process? Can yo give an example of reversible and irreversible process in thermodynamics?

Can you draw the path?

Reading Room

E.

Application of Second Law of Thermodynamics

1. Heat Engines

Heat engine is any device that convert heat energy into work. It takes heat from a high temperature source (a hot reservoir), confert some of it into work, and transfer the rest into its surroundings (a low temperature reservoir). This machine is used to produce work with operate in thermal cycle continously. Most turbines that generate electricity are heat engines. Using input heat fro various resources. These source including fossil, fuels (oil, gas, coal), nuclear reaction, and ttthermal energy

e eath the ea tth’s su fa e. Fo e a ple diesel using fuel as source. The mechanism of heat engine can be seen in this picture.

Qin

Q out Wout

High temperature reservoir

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Volume

if Q1 is heat transfered to the system, Q2 is heat removed from the system, and W is work done by the system, so the efficiency of the machine is

since the process is cyclic process, so ∆U = 0. ∆U = ∆Q – W, with ∆Q = Qin - Qout

W = ∆Q = Q1 – Q2 So the efficiency become

Ƞ =

x

100%

Ƞ = 1- x 100%

2. Refrigerator and Air Conditioner

One of apllication of second law of termodynamics is refrigerator and air conditioner. The

machanism of refrigerator and air conditioner explained by

statement of Clausius that to transfer haet from lower body to warmer body, work is needed.

Pressure

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You may be wondering why is intside refrigerator is cold but it is hot inside? Refrigerator and air conditioner takes heat from the cooler body (inside refrigerator) and transfer it to the warmer body (outside refrigerator). To transfer heat from cooler body to the warmer one work is needed. Bacause heat inside refrigerator decrease, the temperature also decrease. In the other hand, outside refrigerator become warmer because heat increase. For refrigerator and another cooling machines, the efficiency is

Where W = Q2 – Q1

Ƞ =

x

100%

Ƞ = ( x 100%

The otto of ef ige ato is oole tha the uppe o e. That’s h we usually put ice in the top side of refrigerator, and put the fruits and vegetable in the bottom. It will freeze if you put in the top.

3. Carnot engine

In order to optimize the work of heat engines, we have to pay attention to the desain. So we can minimize the heat loss. But how less is heat must loss? In the other words, what is possible

maximum efficiency of a heat engine?

Sadi Carnot, a french engineer solved the problem. First he

observed that thermodynamics cycle of heat engine will make the most efficient cycle. A haet engine will remove heat from high temperature reservoir then put the rest of the heat into loe temperature reservoir. This isothermal process is reversible. But it is not enough. We need to find another reversible processes to complete the cycle so it produces maximum efficiency. Carnot found that the two process is adiabatic.

So, Carnot cycle consist of two isoterms and two adiabats.

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P

V

If it changed to the T-S diagram, it will be like the picture in the left side. Work done by system is equal to square area inside the closed curve.

W = Q = Qhot - Qcold = ( Thot– Tcold ) ∆S

Because Q = T ∆S, then ∆Sinitial = ∆Sfinal

The equation can be used to find efficiency of carnot engine.

Ƞ = (

x 100%

Ƞ = (

x 100%

You have learned about laws of thermodynamics and the

applications. Do you know, our bodies are the applications too. Lets go to living room to find out more.

THot

T Cold

Compression

Expansion Q Cold

QHot

Q = THot- T Cold

THot

T Cold

Siniti Sfin

T

S

Q

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Living Room

Thermodynamics and The Human Body

Like the bodies of all another organism, the human body is not a closed system. We must consume food and oxygen to survive. First and second laws of thermodynamics have interesting implications for our bodies.

The human body metabolizes the chemical energy stored in food. This process is quite efficient, because 95% of energy content in food is metabolized. Some of this energy is converted to work (W) to do daily activities, circulate blood, and so on. The rest is loss to environtment in form of heat (Q).

The first law of thermodynamics, the law of conservation energy, can be written as ∆U = Q – W.

∆U is the change in internal energy of the body, which come

f o t o o t i utio s. O e is f o food, a d the othe f o od ’s

fat. So we can write ∆U = ∆Ufood + ∆Ufat. Hence ∆U is negative

quantity, because when the energy converted to work and heat, our bodies have less energy stored. Because Q is heat loss to

environtment, it is also a negative quantity.

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exhausting heat into the environtment. So the efficiency of human body is

Ɛ =

=

Efficiency often determined by using work per unit time (Power), and the energy consumed per unit time (metabolic rate).

The power used during p articular activity, such as running or

cycling can be measured by a device called dynamometer. The

metabolic rate is proportional to the arte of oxygen consumption, so this rate can be measured by using breathing devices. So, the efficiency of body that perform different activities can be measured by measuring the rate of oxygen consumption.

The efficiency of human body depends on muscle activity and which muscles are used. The larger muscles in the body are leg muscles, so if an activity uses these muscles, the efficiency is relatively high. For example some professional bicycle racers can achieve an aefficiency 20%, generating more than 2hp of power. Arm muscles are relatively small, so activities such as pressing bench have efficiencies of less than 5%. Like any other heat engine, the human body can never achieve 100% efficiency.

The Attic

Now let us summary this chapter and save it in the attic.

SUMMARY

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27  Processes of thermodynamics are Isothermal ( Temperature

constant ), Isobaric ( Pressure constant ), Isochoric (volume constant ), and Adiabatic ( Q = 0 ).

 The second law of thermodynamics states whether a process can occurs naturally or alternatively, and specifies the direction of the process.

 Entropy is a property that indicates the natural directions of a process.

The total entropy of universe increase in every natural process.

 Application of second law of thermodynamics are heat engines, refrigerator, and carnot machine.

 Heat engine is a machine that convert heat into work. Efficiency of heat engine is ratio between heat added to a system and work done by the system.

Bedroom

It is time to go to backyard and testing about what you have learned so far.

Evaluation

Multiple Choice

1. When heat is added to asystem of ideal gas during an isothermal process,

a. Work is done on the system b. The internal energy decreases

c. The effect is the same as for isochoric process d. None of the above

2. In any natural process, the overall change in the entropy of the universe could not be

a. Negative b. Zero c. Positive

d. None of the above

3. Which of the following determines the thermal efficiency of a heat engine ?

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d. Qc + Qh

4. The carnot cycle consist of

a. Two isobaric and two isothermal process b. Two isokhoric and two adiabatic process c. Two adiabatic and two isothermal process

d. Four different process that return the system into initial state 5. Whi h of the follo i g ese oi ’s te pe atu e elatio ship hi h

produce highest efficiency for carnot engine? a. Tc = 0.15 Th

b. Tc = 0.25 Th c. Tc = 0.50 Th d. Tc = 0.90 Th

6. From the P-V diagram below, which process will produce work? a. P

P1 = P2

V1 V2 V

b.

P2

P1

V1 V2

c P P2

P1

V1 = V2 V

d P

P2

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V

V1 V2

7. How much work done by system on this P-V diagram ? P

2

5

2 5 V

a. 6 J b. 9 J c. 10.5 J d. 15 J

8. If heat added to an ideal gas on isobaric process, which statements are true ?

I. Temperature of the gas changes II. The number of paricle constant III. Pressure of gas constant

a. I, II b. I, III c. II, III d. I, II, III

9. Change of state consist of . . .

a. Adiabatic and isothermal processes b. Adiabatic and isobaric processes c. Isobaric and isothermal processes d. Isochoric and isothermal processes 10. Which of the following statements are true?

I. In adiabatic process, system always do work

II. In isothermal process, the internal energy of the system changes

III. I iso ho i p o ess, s ste does ’t do o k IV. In isobaric process, system do or accept work a. I,II

b. I, IV c. III, IV d. I, II, III

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1. In adiabatic process, there is no heat transfer between system and environtment. But the temperature changes. How can it be? 2. Is leaving refrigerator with door opened a practical way to cool the

room?

3. If you have the choice to running your heat engine between following temperature, which one will you choose, between 100° C and 300°C or between 50°C or 250°C? Why?

4. Does the entropy of each this object increase or decrease? a) Ice as it melts

b) Water vapor as it condenses c) Water as it is heated

d) Food as it is cooled in refrigerator

Essay

1. While you playing football, you lost 6.5 x 105 J of heat, and your internal energy also decreased 1.2 x 106 J. How much work did you do in the match?

2. An ideal gas expands from 1.0 m3 to 3.0 m3 at atmospehere pressure. It absorbs 5.0 x 105 J of heat in the process.

a) Is temperature of the gas increase, stay the same, or decrease? Why?

b) What is the change in internal nergy of the system? 3. 1.0 Kg of ice melts completely into water at 0° C.

a) Is the entropy of the process positive, zero, or negative? Explain. b) What is the change of entropy of the ice?

4. If an engine does 200 J of work and produce 600 J of heat per cycle, what is its thernal efficiency?

5. A carnot engine with efficiency of 40% operates with low temperature reservoir at 50°C and produce 1200 J of heat each cycle. What are a) The heat input per cycle

b) The temperature of high-temperature reservoir.

Attic

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GLOSSARY

Cycle : some processes that bring system to initial state.

Efiiciency : ratio between heat added to the system and work produced by system.

Heat : a form of energy that transferred by a difference in temperature

Internal energy : the sum of kinetic and potential energy in an object

Reservoir : a place for saving objects

Specific heat : the amount of heat taken by one garam substance to raise 1°C

Thermodynamics : knowledge about transfer of heat

References

Wilson, Jerry D; Buffa, Anthony J & Lou Bo. 2007. College Physics volume1. New Jersey : Pearson Prentice Hall.

Foster, Bob. 2004. Fisika SMA 2B. Jakarta : Erlangga.

Handayani, sri & Damari, Ari. 2009. Fisika untuk SMA dan MA kela Xi. Jakarta : Departemen Pendidikan Nasional.

Cari. 2009. Aktif Belajar Fisika. Jakarta : Departemen Pendidikan Nasional.

Sarwono; Sunarroso & Suyatman. 2009. Fisika 2. Jakarta : Departemen Pendidikan Nasional.

Siswanto & Sukaryadi. 2009. Kompetensi Fisika. Jakarta : Departemen Pendidikan Nasional.

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32 Peraturan Menteri Nomor 23Tahun 2006 tentang Standar Kompetensi

Lulusan. 2006.

General Certificate of Education (International) Syllabus Advanced Level and Advanced Subsidiary Level. 2010.

Answer of Evaluation

Multiple Choice

1. D 6. A

2. A 7. D

3. B 8. D

4. C 9. C

5. A 10. C

Conceptual Question

1. In adiabatic process, there is no heat transfer Q = 0. But during expansion or compression, work is done by or on the system. So W = -∆U o -∆U = -W. So the pressure, volume, and temperature all change in the process.

2. No, it is not. Refrigerator can only cooling narrow area near it. Because the room temperature is much bigger than the low temperature produced by refrigerator

3. Between 50°C and 250°C. Because it is more efficient. 4. a. increase

b. decrease c. increase d. decrease

Essay

1. 5.5 x 105 J

2. a. Remains the same

. ∆U = 0 isothe al p o ess

3. a. Positive, because heat is added to the ice. b. 1.2 x 103 j/K

4. 25% 5. a. 3000 J

Bahan Ajar Termodinamika

Thermodynamics

thermodynamics

Psychomotor

Contents

How to Use This Learning material

is a front side of a house. It containts everyday phenomena

s perfect

Living Room

Thermodynamics System

The First Law of Thermodynamics

Solution

follow up exercise

Thermodynamics Processes for an Ideal Gas

Work sheet

CASE SOLVED! Thanks for your help..

1. Isothermal Process

Conceptual Reasoning

3. Isochoric Process

4. Adiabatic Process

Conceptual Example : Exhaling : Blowing Hot or Cold?

Follow-Up Exercise

Thermodinamics Processes

The Second Law of Thermodynamics

Entropy

Change in entropy : an isothermal process

Follow-Up Exercise

Application of Second Law of Thermodynamics

Thermodynamics and The Human Body

SUMMARY

Evaluation

Multiple Choice

Essay

GLOSSARY

Answer of Evaluation

Essay