ቴርሞዳይናሚክስ - ችግሮች እና መፍትሄዎች
1. Based on graph P-V below, what is the ratio of the ሥራ done by the gas in the process I, to the work done by the gas in the process II?
የሚታወቅ፦
Process 1 :
ግፊት (P) = 20 N/m2
Initial volume (V1) = 10 ሊትር = 10 ዲሜ3 = 10x10-3 m3
Final volume (V2) = 40 ሊትር = 40 ዲሜ3 = 40x10-3 m3
Process 2 :
Process (P) = 15 N/m2
Initial volume (V1) = 20 ሊትር = 20 ዲሜ3 = 20x10-3 m3
Final volume (V2) = 60 ሊትር = 60 ዲሜ3 = 60x10-3 m3
የሚፈለግ፦ The ratio of the work done by gas
መፍትሔው
The work done by gas in the process I :
W = P ΔV = P (V2–V1) = (20)(40-10)(10-3 m3) = (20)(30)(10-3 m3) = (600)(10)-3 m3) = 0.6 ሜትር3
The work done by gas in the process II :
W = P ΔV = P (V2–V1) = (15)(60-20)(10-3 m3) = (15)(40)(10-3 m3) = (600)(10)-3 m3) = 0.6 ሜትር3
The ratio of the work done by gas in the process I and the process II :
0.6 ሜትር3 : 0.6 ሜ3
1: 1
2.
Based on the graph below, what is the work done by helium gas in the process AB?
የሚታወቅ፦
Pressure (P) = 2 x 105 N / m2 = 2x105 ፓስካል
Initial volume (V1) = 5 ሴ.ሜ3 = 5x10-6 m3
Final volume (V2) = 15 ሴ.ሜ3 = 15x10-6 m3
የሚፈለግ፦ Work done by gas in process AB
መፍትሔው
W = ∆P ∆V
ወ = ፒ (ቪ)2 - ቪ1)
ወ = (2 x 105)(15 x 10-6 - 5 x 10-6)
ወ = (2 x 105)(10 x 10-6) = (2 x 105)(1 x 10-5)
ወ = 2 ጁል
3.
Based on the graph below, what is the work done in process a-b?
የሚታወቅ፦
Initial pressure (P1) = 4 Pa = 4 N/m2
Final pressure (P2) = 6 Pa = 6 N/m2
Initial volume (V1) = 2 ሜትር3
Final volume (V2) = 4 ሜትር3
የሚፈለግ፦ work done I process a-b
መፍትሔው
Work done by gas = area under curve a-b
W = area of triangle + area of rectangle
W = ½ (6-4)(4-2) + 4(4-2)
W = ½ (2)(2) + 4(2)
W = 2 + 8
ወ = 10 ጁል
4. Based on graph below, what is the work done in process A-B-C-A.
መፍትሔው
Work (W) = Area of the triangle A-B-C
W = ½ (20-10)(6 x 105 - 2 x 105)
W = ½ (10)(4 x 105)
W = (5)(4 x 105)
ወ = 20 x 105
ወ = 2 x 106 ዬሉል
5. An engine absorbs 2000 Joule of heat at a high temperature and exhausted 1200 Joule of heat at a low temperature. What is the efficiency of the engine?
የሚታወቅ፦
የሙቀት ግብዓት (Q)H) = 2000 ጁል
የሙቀት ውፅዓት (Q)L) = 1200 ጁል
Work done by engine (W) = 2000 – 1200 = 800 Joule
የሚፈለግ፦ efficiency (e)
መፍትሔው
e = W / QH
ኢ = 800/2000
e = 0.4 x 100%
e = 40%
6. An engine absorbs heat at 960 Kelvin and the engine discharges heat at 576 Kelvin. What is the efficiency of the engine.
የሚታወቅ፦
ከፍተኛ ሙቀት (ቲ)H) = 960 ኪ.ሜ
ዝቅተኛ የሙቀት መጠን (ቲ)L) = 576 ኪ.ሜ
የሚፈለግ efficiency (e)
መፍትሔው
Efficiency of Carnot engine = 0.4 x 100% = 40%
7. Based on the graph below, work done by the engine is 6000 Joule. What is the heat discharged by engine each circle?
የሚታወቅ፦
ስራ (ወ) = 6000 ጁል
ከፍተኛ ሙቀት (ቲ)H) = 800 Kelvin
ዝቅተኛ የሙቀት መጠን (ቲ)L) = 300 Kelvin
የሚፈለግ heat discharged by the engine
መፍትሔ :
Carnot (ideal) efficiency :
Heat absorbed by Carnot engine :
ወ = e Q1
6000 = (0.625) Q1
Q1 = 6000/0.625
Q1 = 9600
Heat discharged by Carnot engine :
Q2 = ጥ1 - ወ
Q2 = 9600 - 6000
Q2 = 3600 ጁል
8. The efficiency of a Carnot engine is 40%. If heat absorbed at 727°C then what is the low temperature.
የሚታወቅ፦
Efficiency (e) = 40% = 40/100 = 0.4
ከፍተኛ ሙቀት (ቲ)H) = 727oሲ + 273 = 1000 ኪ.ሜ
የሚፈለግ፦ ዝቅተኛ የሙቀት መጠን
መፍትሔው
TL = 600 Kelvin – 273 = 327oC
9. Based on graph below, if the engine absorbs 800 J of heat, what is the work done by the engine.
የሚታወቅ፦
ከፍተኛ ሙቀት (ቲ)H) = 600 Kelvin
ዝቅተኛ የሙቀት መጠን (ቲ)L) = 250 Kelvin
የሙቀት ግብዓት (Q)1) = 800 ጁል
የሚፈለግ Work (W)
መፍትሔው
The efficiency of Carnot engine :
Work was done by the engine :
ወ = e Q1
W = (7/12)(800 Joule)
ወ = 466.7 ጁል
10. The high temperature of a Carnot engine is 600 K. If the engine absorbs 600 J of heat and the low temperature is 400 K, what is the work done by the engine.
የሚታወቅ፦
ዝቅተኛ የሙቀት መጠን (ቲ)L) = 400 ኪ.ሜ
ከፍተኛ ሙቀት (ቲ)H) = 600 ኪ.ሜ
የሙቀት ግብዓት (Q)1) = 600 ጁል
የሚፈለግ ስራው የተከናወነው በካርኖት ሞተር (W) ነው
መፍትሔው
የካርኖት ሞተር ውጤታማነት፡
Work was done by Carnot engine :
ወ = e Q1
ወ = (1/3)(600) = 200 ጁሎች
- What is the primary focus of thermodynamics? መልስ: Thermodynamics focuses on the study of energy, its transformations, and its relationship with matter, especially in systems at equilibrium.
- How is the zeroth law of thermodynamics related to temperature? መልስ: The zeroth law states that if two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other. This implies the existence of a property called temperature, which is the same for all systems in thermal equilibrium.
- What does the first law of thermodynamics describe? መልስ: The first law, also known as the law of energy conservation, states that energy cannot be created or destroyed, only converted from one form to another. In a closed system, the change in internal energy is equal to the heat added to the system minus the work done by the system on its surroundings.
- Why is the second law of thermodynamics crucial for understanding the direction of natural processes? መልስ: The second law states that the entropy (or disorder) of an isolated system always increases or remains constant. This dictates that energy spontaneously disperses if not hindered from doing so, providing a direction to natural processes and essentially explaining why certain processes occur spontaneously while others do not.
- What is entropy, and how does it relate to disorder in a system? መልስ: Entropy is a measure of the amount of energy in a system that is unavailable to do work. It is also often described as a measure of the system’s disorder or randomness. In general, higher entropy corresponds to greater disorder or randomness.
- How does the third law of thermodynamics describe the entropy of a perfect crystal at absolute zero? መልስ: The third law states that the entropy of a perfect crystal is exactly zero at absolute zero temperature (0 Kelvin). This means that at this temperature, the system is perfectly ordered.
- Why can’t heat flow from a colder body to a hotter body on its own? መልስ: This behavior is a consequence of the second law of thermodynamics. If heat were to flow from a colder body to a hotter one spontaneously, it would lead to a decrease in the overall entropy of the system, which is not favored by natural processes.
- What is the difference between an isolated, closed, and open system in thermodynamics? መልስ: An isolated system does not exchange energy or matter with its surroundings. A closed system can exchange energy but not matter with its surroundings. An open system can exchange both energy and matter with its surroundings.
- How is the concept of “work” in thermodynamics different from the everyday use of the term? መልስ: In thermodynamics, “work” refers to the process of energy transfer where forces applied to an object move it in a direction parallel to the force. For example, when a gas expands against a piston, it does work on the piston. This is a more specific definition compared to the everyday use of “work,” which might simply mean any task or activity.
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What is a Carnot cycle, and why is it significant in thermodynamics? መልስ: The Carnot cycle is an idealized thermodynamic cycle that provides an upper limit on the efficiency that any classical thermodynamic engine can achieve during the conversion of heat into work (or vice versa). It’s significant because it sets a fundamental efficiency limit based on the temperatures of the heat reservoirs between which an engine operates.
