2010
Thermodynamics Base competence : analize the change of ideal gas condition by applying the laws of thermodynamics
Aminatur Rahmawati 1/1/2010
Thermodynamics
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. 18. Social skill : cooperating, sharing opinion, being good listener. And commenting other’s opinion 1
Thermodynamics
Contents Standard Competence, Base Competence, Learning Objectives .................................... 1 Contents ........................................................................................................................... 2 How to Use This Learning Material .................................................................................. 3 Chapter Opening : Thermodynamics ................................................................................. 6 Concept Map ...................................................................................................................... 7 A. Thermodynamics System ..........................................................................................8 B. The First Law of Thermodynamiccs ...........................................................................8 C. Thermodynamics Processes for an Ideal Gas ......................... ...................................11 1. Isothermal ...........................................................................................................14 2. Isobaric ...............................................................................................................15 3. Isochoric .............................................................................................................16 4. Adiabatic ............................................................................................................17 D. The Second Law of Thermodynamics .......................................................................19 Entropy .......................................................................................................................21 Reversible and Irreversible Process ............................................................................21 E. Application of Second Law of Thermodynamics ......................................................21 1. Heat Engines ........................................................................................................21 2. Refrigerator and Air Conditioner ........................................................................22 3. Carnot Engine ......................................................................................................23 Thermodynamics and The Human Body ...........................................................................25 Summary ............................................................................................................................26 Evaluation ..........................................................................................................................27 Glossary .............................................................................................................................30 References ..........................................................................................................................31 Answer of Evaluation .........................................................................................................32
2
Thermodynamics
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 that we want.
is a place where we can find any information
Some matters will be given here to answer reader’s inquiring taste.
3
Thermodynamics
Balcony is an upper floor projecting from the wall . We can think clearly here, becuse we feel fresh air, and enjoying the view. So, 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.
Kitchen is place where we can do hand activity. So, there will be a simple experiment by reader.
4
Thermodynamics
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.
Bedroom before we go to bed, it’s good to reflect our activity. we memorize what we’ve done and learn. So, there will be eveluation.
5
Thermodynamics
THERMODYNAMICS
A
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 find out in this chapter.
6
Thermodynamics
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, because there are seven district or “kecamatan” covered by hot mud. 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 produce work in human body and in machines. So, do you still think “what for i learn Thermodynamics”? Word Thermodynamics comes from greek, that means Transfer of heat (therme : heat, dynamics : transfer). The development of thermodynamics started about 200 years ago. It is developed until now, so 7
Thermodynamics
you can enjoy the air conditioner, refrigerator, and another heat machines. Let’s go to library 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. Real power does not hit hard , but straight to the point
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)
8
Thermodynamics
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. 2.
If heat transferred to the system, value of Q is positive (+Q). If heat transferred from the system, value of Q is negative (-Q). 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 balcony and 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 = 9.3 x 103 kcal/kg = (9.3 x 103 kcal/kg)(4186 J/kcal)
9
Thermodynamics
= 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 People can be stronger by eating and wiser by reading
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 energy will cause mud’s smoke.
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 10
Thermodynamics
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 Thermodynamics Process. Now let’s help him to find what are the Thermodynamics Processes through exploring the given data below.
a)
Number
Pressure (N/m2)
Volume (m3)
1 2 3
7 x 105 4 x 105 3 x 105
5 8 12
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 Xaxis Based on the graphic you have made, what process happens to the system?
11
Thermodynamics
b)
Number Pressure (N/m2) 1 2 3 4 5
10 x 105 8 x 105 6 x 105 4 x 105 2 x 105
Volume (m3) 10 10 10 10 10
From the data above, you can find what process happens to the system. Just follow this way:
Takes 20 years to build a good reputation but needs 2 seconds to ruin it
Plot graphic from the given data, Pressure as Y-axis and Volume as Xaxis Based on the graphic you have made, what process happens to the system?
c)
Number Pressure (N/m2)
Volume (m3)
1
6 x 105
2
2
6 x 105
4
3
6 x 105
6
4
6 x 105
8
5
6 x 105
10
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 Xaxis Based on the graphic you have made, what process happens to the system? d)
Number Pressure (N/m2) 1 10 x 105 2 8 x 105 3 6 x 105 4 4 x 105 5 3 x 105
Volume (m3) 5 6 7 8 10 12
Thermodynamics
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 Xaxis Based on the graphic you have made, what process happens to the system?
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=∫ By substituting to equation P =
we get 13
Thermodynamics
W=
∫
P P2
W = nRT ∫
T1 = T2
P1 V
W = nRT ln (
)
V1
V1
P-V diagram of an isothermal process 2. Isobaric Processes You have to endure caterpillars if you want to see butterflies
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 = P 2
V1
V2
V 14
Thermodynamics
Diagram above is a P-V diagram for Isobaric process. Now you are invited to go to balcony and thinking about a match. Balcony
Isotherms
VS Isobars
Let’s consider a match between Isotherms and Isobars. Each of them want 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, and the isoterm’s is hyperbola. 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 15
Thermodynamics
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 To be wrong is nothing, unless you continue to remember it
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
16
Thermodynamics
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 factor is the ratio of the gas’s molar specifics 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 To answer the question you must try it and you will surprised to find that the answer is (b). 17
Thermodynamics
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 Now you know about thermodynamics processes. Let’s save it in the attic. Thermodinamics Processes 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
Adiabatic
Q=0
∆U = -W
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. 18
Thermodynamics
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 danger of small mistakes is that those mistakes are not always small
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 entropy. The change in system’s entropy is ∆S = Where ∆S is change of system’s entropy 19
Thermodynamics
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 10 5 J/Kg. Solution Given m = 0.4 Kg T = 34 + 273 = 307 K Lv = 24.2 x 105 J/Kg Find change in entropy (∆S ) Q = m Lv = 0.4 Kg 24.2 x 105 J/Kg = 9.68 x 105 J ∆S =
=
= 3.15 x 105 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? Hint : the answer is -1.22 x 103 J/K 20
Thermodynamics
Reading Room Reversible and Irreversible Process In order to learn about reversible and irreversible process, 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. And the example of reversible process is tawaf around ka’bah, or when you get out from home to go to school,then go to your friend’s home, then 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 High temperature reservoir Qin
Wout system
Q out
Low temperature reservoir
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 beneath the eartth’s surface. For example diesel using fuel as source. The mechanism of heat engine can be seen in this picture.
21
Thermodynamics
Pressure
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=
Ƞ = 1-
x 100% 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. 22
Thermodynamics
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=
Ƞ= ( Dig a well before you become thirsty
x 100% x 100%
The bottom of refrigerator is cooler than the upper one. That’s why 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.
23
Thermodynamics
P THot QHot Compression
Expansion
T Cold Q Cold
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
T Q = THot- T Cold
Because Q = T ∆S, then
THot
∆Sinitial = ∆Sfinal Q
T Cold Siniti
Sfin
al
al
The T-S diagram of carnot cycle
S
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. 24
Thermodynamics
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 from two contributions. One is from food, and the other from body’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. The human body is example of biological heat engine. The enrgy source of this engine is energy metabolized from food and fat. Some of this energy is converted to work, and the rest loss to environtment in form of heat. This situation is analogous of heat engine that taking heat from hot reservoir, doing work, and 25
Thermodynamics
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 In the gas system, firt law of thermodynamics is applied. It is a conservation of heat energy.
26
Thermodynamics
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.
2.
3.
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 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 Which of the following determines the thermal efficiency of a heat engine ? a. Qc x Qh b. Qc / Qh c. Qc - Qh 27
Thermodynamics
4.
5.
6.
d. Qc + Qh 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 Which of the following reservoir’s temperature relationship which 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 From the P-V diagram below, which process will produce work? a. P
P1 = P 2 V1
V2
V
b. P2 P1
V1 c
V2
P P2
P1
V1 = V2 d
V
P P2 P1 28
Thermodynamics
V V1 7.
V2
How much work done by system on this P-V diagram ? P 2
5
2 a. b. c. d.
5
V
6J 9J 10.5 J 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. In isochoric process, system doesn’t do work IV. In isobaric process, system do or accept work a. I,II b. I, IV c. III, IV d. I, II, III Conceptual Question 29
Thermodynamics
1. 2. 3.
4.
In adiabatic process, there is no heat transfer between system and environtment. But the temperature changes. How can it be? Is leaving refrigerator with door opened a practical way to cool the room? 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? 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.
2.
3.
4. 5.
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? 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? 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? If an engine does 200 J of work and produce 600 J of heat per cycle, what is its thernal efficiency? 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 After study this chapter, you have more vocabs and physical terms. Don’t forget them, just put them in the attic, so you can find them easily when you need them.
30
Thermodynamics
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. Humaidi, Abdul Haris & Maksum. 2009. Fisika SMA / MA kelas XI. Jakarta : Departemen Pendididkan Nasional.
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Thermodynamics
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. 2. 3. 4. 5.
D A B C A
6. A 7. D 8. D 9. C 10. C
Conceptual Question 1.
2.
3. 4.
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 or ∆U = -W. So the pressure, volume, and temperature all change in the process. 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 Between 50°C and 250°C. Because it is more efficient. a. increase b. decrease c. increase d. decrease
Essay 1. 2. 3. 4. 5.
5.5 x 105 J a. Remains the same b. ∆U = 0 (isothermal process) a. Positive, because heat is added to the ice. b. 1.2 x 103 j/K 25% a. 3000 J b. 83.33°C 32