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Understanding Heat, Work, and Energy in Thermodynamic Systems

Learn about the fundamental concepts of heat, work, and internal energy in thermodynamic processes. Discover the relationships between heat transfer, work done, and energy conservation in different systems.

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Understanding Heat, Work, and Energy in Thermodynamic Systems

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  1. Section 1 Relationships Between Heat andWork Chapter 10 Preview • Objectives • Heat, Work, and Internal Energy • Thermodynamic Processes

  2. Section 1 Relationships Between Heat andWork Chapter 10 Objectives • Recognizethat a system can absorb or release energy as heat in order for work to be done on or by the system and that work done on or by a system can result in the transfer of energy as heat. • Compute the amount of work done during a thermodynamic process. • Distinguishbetween isovolumetric, isothermal, and adiabatic thermodynamic processes.

  3. Section 1 Relationships Between Heat andWork Chapter 10 Heat, Work, and Internal Energy • Heatandworkareenergy transferredto or from a system. An object never has “heat” or “work” in it; it has only internal energy. • A systemis a set of particles or interacting components considered to be a distinct physical entity for the purpose of study. • Theenvironmentthe combination of conditions and influences outside a system that affect the behavior of the system.

  4. Section 1 Relationships Between Heat andWork Chapter 10 Heat, Work, and Internal Energy, continued • In thermodynamic systems, work is defined in terms of pressure and volume change. • This definition assumes that P is constant.

  5. Section 1 Relationships Between Heat andWork Chapter 10 Heat, Work, and Internal Energy, continued • If the gasexpands,as shown in the figure, DV is positive, andthe work done by the gas on the piston is positive. • If the gas iscompressed,DV is negative, andthe work done by the gas on the piston is negative.(In other words, the piston does work on the gas.)

  6. Section 1 Relationships Between Heat andWork Chapter 10 Heat, Work, and Internal Energy, continued • When the gas volume remainsconstant,there is no displacement andno workis done on or by the system. • Although the pressure can change during a process,workis done only if thevolumechanges. • A situation in which pressure increases and volume remains constant is comparable to one in which a force does not displace a mass even as the force is increased. Work is not done in either situation.

  7. Section 1 Relationships Between Heat andWork Chapter 10 Thermodynamic Processes • Anisovolumetric processis a thermodynamic process that takes place at constant volume so that no work is done on or by the system. • Anisothermal processis a thermodynamic process that takes place at constant temperature. • Anadiabatic processis a thermodynamic process during which no energy is transferred to or from the system as heat.

  8. Section 2 The First Law of Thermodynamics Chapter 10 Energy Conservation • Iffrictionis taken into account,mechanical energy is not conserved. • Consider the example of a roller coaster: • A steady decrease in the car’s total mechanical energy occurs because of work being done against the friction between the car’s axles and its bearings and between the car’s wheels and the coaster track. • If the internal energy for the roller coaster (the system) and the energy dissipated to the surrounding air (the environment) are taken into account, thenthe total energy will be constant.

  9. Section 2 The First Law of Thermodynamics Chapter 10 Energy Conservation

  10. Section 2 The First Law of Thermodynamics Chapter 10 Energy Conservation, continued • The principle ofenergy conservationthat takes into account a system’s internal energy as well as work and heat is called thefirst law of thermodynamics. • The first law of thermodynamics can be expressed mathematically as follows: DU = Q – W Change in system’s internal energy = energy transferred to or from system as heat – energy transferred to or from system as work

  11. Section 2 The First Law of Thermodynamics Chapter 10 Signs of Q and W for a system

  12. Section 2 The First Law of Thermodynamics Chapter 10 Sample Problem The First Law of Thermodynamics A total of 135 J of work is done on a gaseous refrigerant as it undergoes compression. If the internal energy of the gas increases by 114 J during the process, what is the total amount of energy transferred as heat? Has energy been added to or removed from the refrigerant as heat?

  13. Section 2 The First Law of Thermodynamics Chapter 10 Sample Problem, continued 1. Define Given: W = –135 J DU = 114 J Diagram: Tip: Work is done on the gas, so work (W) has a negative value. The internal energy increases during the process, so the change in internal energy (DU) has a positive value. Unknown: Q = ?

  14. Section 2 The First Law of Thermodynamics Chapter 10 Sample Problem, continued 2. Plan Choose an equation or situation: Apply the first law of thermodynamics using the values for DU and W in order to find the value for Q. DU = Q – W Rearrange the equation to isolate the unknown: Q = DU + W

  15. Section 2 The First Law of Thermodynamics Chapter 10 Sample Problem, continued 3. Calculate Substitute the values into the equation and solve: Q = 114 J + (–135 J) Q = –21 J Tip: The sign for the value of Q is negative. This indicates that energy is transferred as heat from the refrigerant.

  16. Section 2 The First Law of Thermodynamics Chapter 10 Sample Problem, continued 4. Evaluate Although the internal energy of the refrigerant increases under compression, more energy is added as work than can be accounted for by the increase in the internal energy. This energy is removed from the gas as heat, as indicated by the minus sign preceding the value for Q.

  17. Section 2 The First Law of Thermodynamics Chapter 10 Cyclic Processes • Acyclic processis a thermodynamic process in which a system returns to the same conditions under which it started. • Examples includeheat enginesandrefrigerators. • In a cyclic process, the final and initial values of internal energy are the same, and the change in internal energy is zero. DUnet = 0 and Qnet = Wnet

  18. Section 2 The First Law of Thermodynamics Chapter 10 Cyclic Processes, continued • Aheat engineuses heat to do mechanical work. • A heat engine is able to dowork(b)by transferring energy from ahigh-temperature substance (the boiler) atTh(a)to a substance at alower temperature (the air around the engine) atTc(c). • The internal-combustion enginefound in most vehicles is an example of a heat engine.

  19. Section 2 The First Law of Thermodynamics Chapter 10 The Steps of a Gasoline Engine Cycle

  20. Section 2 The First Law of Thermodynamics Chapter 10 The Steps of a Refrigeration Cycle

  21. Section 2 The First Law of Thermodynamics Chapter 10 Thermodynamics of a Refrigerator

  22. Section 3 The Second Law of Thermodynamics Chapter 10 Objectives • Recognizewhy the second law of thermodynamics requires two bodies at different temperatures for work to be done. • Calculatethe efficiency of a heat engine. • Relatethe disorder of a system to its ability to do work or transfer energy as heat.

  23. Section 3 The Second Law of Thermodynamics Chapter 10 Efficiency of Heat Engines • Thesecond law of thermodynamicscan be stated as follows: No cyclic process that converts heat entirely into work is possible. • As seen in the last section,Wnet = Qnet = Qh – Qc. • According to the second law of thermodynamics, W can never be equal to Qh in a cyclic process. • In other words, some energy must always be transferred as heat to the system’s surroundings (Qc > 0).

  24. Section 3 The Second Law of Thermodynamics Chapter 10 Efficiency of Heat Engines, continued • A measure of how well an engine operates is given by the engine’sefficiency (eff ). • In general,efficiencyis a measure of theuseful energytaken out of a process relative to thetotal energythat is put into the process. • Note that efficiency is a unitless quantity. • Because of the second law of thermodynamics, the efficiency of a real engine is always less than 1.

  25. Section 3 The Second Law of Thermodynamics Chapter 10 Sample Problem Heat-Engine Efficiency Find the efficiency of a gasoline engine that, during one cycle, receives 204 J of energy from combustion and loses 153 J as heat to the exhaust. 1. Define Given: Diagram: Qh = 204 J Qc = 153 J Unknown eff = ?

  26. Section 3 The Second Law of Thermodynamics Chapter 10 Sample Problem, continued 2. Plan Choose an equation or situation: The efficiency of a heat engine is the ratio of the work done by the engine to the energy transferred to it as heat.

  27. Section 3 The Second Law of Thermodynamics Chapter 10 Sample Problem, continued 3. Calculate Substitute the values into the equation and solve: 4. Evaluate Only 25 percent of the energy added as heat is used by the engine to do work. As expected, the efficiency is less than 1.0.

  28. Section 3 The Second Law of Thermodynamics Chapter 10 Entropy • In thermodynamics, a system left to itself tends to go from a state with a very ordered set of energies to one in which there is less order. • The measure of a system’s disorder or randomness is called theentropyof the system. The greater the entropy of a system is, the greater the system’s disorder. • The greater probability of a disordered arrangement indicates that an ordered system is likely to become disordered.Put another way, theentropyof a system tends to increase.

  29. Section 3 The Second Law of Thermodynamics Chapter 10 Entropy, continued • If all gas particles moved toward the piston, all of the internal energy could be used to do work. This extremely well ordered system is highly improbable. • Greater disorder means there is less energy to do work.

  30. Section 3 The Second Law of Thermodynamics Chapter 10 Entropy, continued • Because of the connection between a system’s entropy, its ability to do work, and the direction of energy transfer, thesecond law of thermodynamicscan also be expressed in terms of entropy change: The entropy of the universe increases in all natural processes. • Entropy can decrease for parts of systems, provided this decrease is offset by a greater increase in entropy elsewhere in the universe.

  31. Section 3 The Second Law of Thermodynamics Chapter 10 Energy Changes Produced by a Refrigerator Freezing Water Because of the refrigerator’s less-than-perfect efficiency, the entropy of the outside air molecules increases more than the entropy of the freezing water decreases.

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