Fundamental Concepts and Laws of Thermodynamics, Study notes of Physics

Download Fundamental Concepts and Laws of Thermodynamics and more Study notes Physics in PDF only on Docsity! 1 Phys 1111K – Lecture 15 C&J Ch15 – Thermodynamics What is Thermodynamics? Thermodynamics is the branch of physics that is built upon the fundamental laws that heat and work obey. The collection of objects on which attention is being focused is called the system, while everything else in the environment is called the surroundings. Walls that permit heat flow are called diathermal walls, while walls that do not permit heat flow are called adiabatic walls. To understand thermodynamics, it is necessary to describe the state of a system. Thermal Equilibrium Two systems are said to be in thermal equilibrium if there is no heat flow between then when they are brought into contact. Temperature is the indicator of thermal equilibrium in the sense that there is no net flow of heat between two systems in thermal contact that have the same temperature. Zeroth Law of Thermodynamics Two systems individually in thermal equilibrium with a third system are in thermal equilibrium with each other. 2 Heat and Work Suppose that a system gains heat Q and that is the only effect occurring. Consistent with the law of conservation of energy, the internal energy of the system changes: QUUU if =−=∆ Heat is positive when the system gains heat and negative when the system loses heat. If a system does work W on its surroundings and there is no heat flow, conservation of energy indicates that the internal energy of the system will decrease: WUUU if −=−=∆ Work is positive when it is done by the system and negative when it is done on the system. First Law of Thermodynamics The internal energy of a system changes due to heat and work: WQUUU if −=−=∆ Work is positive when it is done by the system and negative when it is done on the system. Heat is positive when the system gains heat and negative when the system loses heat. Positive & Negative Work Example In part a of figure, the system gains 1500J of heat and 2200J of work is done by the system on its surroundings. In part b, the system also gains 1500J of heat, but 2200J of work is done on the system. In each case, determine the change in internal energy of the system. (a) ( ) ( ) J 700J 2200J 1500 −=+−+= −=∆ WQU (b) ( ) ( ) J 3700J 2200J 1500 +=−−+= −=∆ WQU Ideal Gas Example The temperature of three moles of a monatomic ideal gas is reduced from 540K to 350K as 5500J of heat flows into the gas. Find (a) the change in internal energy and (b) the work done by the gas. nRTU 23=WQUUU if −=−=∆ ( ) ( )( )( ) J 7100K 540K 350KmolJ 31.8mol 0.323 2 3 2 3 −=−⋅= −=∆ if nRTnRTU(a) ( ) J 12600J 7100J 5500 =−−=∆−= UQW(b) 5 Isothermal Example Two moles of the monatomic gas argon expand isothermally at 298K from and initial volume of 0.025m3 to a final volume of 0.050m3. Assuming that argon is an ideal gas, find (a) the work done by the gas, (b) the change in internal energy of the gas, and (c) the heat supplied to the gas. (a) ( ) ( )( )( ) J 3400 m 250.0 m 050.0lnK 298KmolJ31.8mol 0.2 ln 3 3 +=      ⋅=       = i f V V nRTW 02323 =−=∆ if nRTnRTU(b) WQU −=∆(c) J 3400+==WQ Adiabatic Processes Adiabatic expansion or compression of a monatomic ideal gas ( )fi TTnRW −= 23 Adiabatic expansion or compression of a monatomic ideal gas γγ ffii VPVP = VP cc=γ adiabatic: no transfer of heat Specific Heat To relate heat and temperature change in solids and liquids, we used: TmcQ ∆= specific heat capacity The amount of a gas is conveniently expressed in moles, so we write the following analogous expression: TCnQ ∆= molar specific heat capacity Specific Heat(s) for Gases For gases it is necessary to distinguish between the molar specific heat capacities which apply to the conditions of constant pressure and constant volume: PV CC , ( ) ( ) TnRTTnRTTnRWUQ ifif ∆=−+−=+∆= 2523pressureconstant first law of thermodynamics nRTU 23= VPW ∆= constant pressure for a monatomic ideal gas RCP 25= 6 Specific Heat(s) cont. ( ) TnRTTnRWUQ if ∆=+−=+∆= 2323olumeconstant v 0 first law of thermodynamics nRTU 23= constant pressure for a monatomic ideal gas RCV 23= monatomic ideal gas 3 5 2 3 2 5 === R R C C V Pγ any ideal gas RCC VP =− Second Law of Thermodynamics THE SECOND LAW OF THERMODYNAMICS: THE HEAT FLOW STATEMENT Heat flows spontaneously from a substance at a higher temperature to a substance at a lower temperature and does not flow spontaneously in the reverse direction. The second law is a statement about the natural tendency of heat to flow from hot to cold, whereas the first law deals with energy conservation and focuses on both heat and work. Heat Engine A heat engine is any device that uses heat to perform work. It has three essential features. 1. Heat is supplied to the engine at a relatively high temperature from a place called the hot reservoir. 2. Part of the input heat is used to perform work by the working substance of the engine. 3. The remainder of the input heat is rejected to a place called the cold reservoir. heatinput of magnitude =HQ heat rejected of magnitude =CQ done work theof magnitude =W Efficiency The efficiency of a heat engine is defined as the ratio of the work done to the input heat: HQ W e = If there are no other losses, then CH QWQ += H C Q Q e −= 1 7 Automobile Efficiency Example An automobile engine has an efficiency of 22.0% and produces 2510 J of work. How much heat is rejected by the engine? HQ W e =CH QWQ += WQQ HC −= ( ) J 89001 220.0 1J 2510 11 =      −=       −=−= e WW e W QC e W QH = Carnot’s Principle A reversible process is one in which both the system and the environment can be returned to exactly the states they were in before the process occurred. CARNOT’S PRINCIPLE: AN ALTERNATIVE STATEMENT OF THE SECOND LAW OF THERMODYNAMICS No irreversible engine operating between two reservoirs at constant temperatures can have a greater efficiency than a reversible engine operating between the same temperatures. Furthermore, all reversible engines operating between the same temperatures have the same efficiency. Carnot Engine The Carnot engine is useful as an idealized model. All of the heat input originates from a single temperature, and all the rejected heat goes into a cold reservoir at a single temperature. Since the efficiency can only depend on the reservoir temperatures, the ratio of heats can only depend on those temperatures. H C H C T T Q Q e −=−= 11 H C H C T T Q Q = Tropical Sea Heat Engine Example Water near the surface of a tropical ocean has a temperature of 298.2 K, whereas the water 700 meters beneath the surface has a temperature of 280.2 K. It has been proposed that the warm water be used as the hot reservoir and the cool water as the cold reservoir of a heat engine. Find the maximum possible efficiency for such and engine. H C T Te −=1carnot 060.0 K 298.2 K 2.28011carnot =−=−= H C T Te 10 2nd Law of Thermo Using Entropy Entropy, like internal energy, is a function of the state of the system. Consider the entropy change of a Carnot engine. The entropy of the hot reservoir decreases and the entropy of the cold reservoir increases. R      =∆ T QS 0=−+=∆ H H C C T Q T Q S Reversible processes do not alter the entropy of the universe. Available Energy Example Example Suppose that 1200 J of heat is used as input for an engine under two different conditions (as shown on the right). Determine the maximum amount of work that can be obtained for each case. H C T Te −=1carnot HQ W e = Available Energy Solution The maximum amount of work will be achieved when the engine is a Carnot Engine, where (a) 77.0 K 650 K 15011carnot =−=−= H C T Te ( ) ( )( ) J 920J 120077.0carnot === HQeW (b) 57.0K 350 K 15011carnot =−=−= H C T Te ( ) ( )( ) J 680J 120057.0carnot === HQeW The irreversible process of heat through the copper rod causes some energy to become unavailable. More on Entropy universeeunavailabl STW o∆= 11 Third Law of Thermodynamics THE THIRD LAW OF THERMODYNAMICS It is not possible to lower the temperature of any system to absolute zero in a finite number of steps.