11/29/2017 IRREVERSIBLE PROCESSES. UNIT 2 Thermodynamics: Laws of thermodynamics, ideal gases, and kinetic theory

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1 11/9/017 AP PHYSICS UNIT Thermodynamics: Laws of thermodynamics, ideal gases, and kinetic theory CHAPTER 13 SECOND LAW OF THERMODYNAMICS IRREVERSIBLE PROCESSES The U G of the water-earth system at the top of a waterfall is converted into KE as the water falls. Then KE is converted to U THERMAL when the water hits the rocks at the bottom. But the warmer water at the bottom never cools while it flows back up in the reverse direction. 1

2 11/9/017 This is the outcome that always occurs. This outcome never occurs. = This is the outcome that always occurs. This outcome never occurs.

3 11/9/017 ice cube This is the outcome that always occurs. U This outcome never occurs. OBSERVATIONS In the outcomes that are observed (consistent with the first law of thermodynamics), organized energy converts into less organized energy. In the experiments in which the outcomes do not occur (even though they are also consistent with the first law of thermodynamics), disorganized energy would convert into more organized energy. Processes that occur in one direction but never occur in the opposite direction are said to be irreversible. It seems that an isolated system s thermal energy (random kinetic energy) cannot be converted into organized kinetic or gravitational energy. 3

4 11/9/017 Irreversible processes The Law of Conservation of Energy The total energy of a system and its environment is constant. some processes, although allowed by the first law of thermodynamics, do not occur in nature. If a part of the universe chosen as the system gains energy because of work done it by external forces or because of energy transfer through heating, then the environment surrounding the system loses an equal amount of energy, or vice versa. The Law of Conservation of Energy The total energy of a system and its environment is constant. (K i +U gi +U si +...) + (W env on sys +Q env to sys ) (K f +U gf +U sf +...+DU internal ) REVERSIBLE AND IRREVERSIBLE PROCESSES Processes that occur in one direction but never occur in the opposite direction are said to be irreversible. In the processes that never occur, energy would have to be converted from disorganized thermal energy into an equal amount of organized kinetic or potential energy. The unallowed processes do not contradict conservation of energy. 4

5 11/9/017 Use the ideas of organized and random energy to decide which of the following processes are reversible. Then compare your results with your everyday experience. (a)a car skids to a stop. (b)a rock is thrown upward and reaches a height h above the ground. (c)two fast-moving cars collide and stick together while stopping. m cart = 10 kg v cart = 5 m/s The organized motion of the cart can do work on the spring, compressing it. U th J o o mc T ( 1kg)(4,180 )(100 C 0 C) o kg C Each of the systems shown below has the same amount of energy, and can interact with another system by doing work on it. Arrange these systems in approximate order of how much work they can do on another system, listing the best system for doing work first. The much greater thermal energy of the hot water cannot convert into the elastic potential energy of the spring. 1,000 J of chemical potential energy stored in the molecules of gasoline 1,000 J of thermal energy in the air particles of the classroom 1,000 J of gravitational potential energy between the Earth and a 50-kg barbell held.0 m above the ground. 5

6 11/9/017 USEFULNESS OF ENERGY As the energy of a system becomes less organized, the amount of work it can do on other systems decreases Observation You place a room temperature spoon in a cup of hot tea. The spoon gets warmer and the tea gets cooler until they reach the same temperature. Outcome Energy is transferred from the hot tea to the cold spoon until they are at the same temperature. Observation The air conditioner in a building fails on a hot day. The temperature in the building increases until it is a similar temperature to outside temperature. Outcome Energy is transferred through the walls until the inside and outside are at the same temperature. The second law of thermodynamics SECOND LAW OF THERMODYNAMICS sort of The second law of thermodynamics is a statement about which processes occur and which do not. There are many ways to state the second law; here is one: Heat can flow spontaneously from a hot object to a cold object; it will not flow spontaneously from a cold object to a hot object. 6

7 11/9/017 The second law of thermodynamics NICOLAS LÉONARD SADI CARNOT The second law of thermodynamics describes what processes can and cannot occur in an isolated system. As the energy of a system becomes less organized, the amount of work it can do on other systems decreases. Carnot's principle tells us it is impossible to build an engine that uses all of its thermal energy to do mechanical work. France, (Father of Thermodynamics) Carnot s Principle It is impossible to build an engine that uses all of its thermal energy to do mechanical work on its environment THERMODYNAMIC ENGINES (HEAT ENGINES) HEAT ENGINE A heat engine is a machine that takes thermal energy from an external source to a substance (Q in ), the substance does useful work (W net ), and the substance releases some excess thermal energy (Q out ). Q in = Q hot = Q H = Q H to sub W net = W = W out Q out = Q cold = Q L = Q sub to c Thermodynamic engines HEAT ENGINE A heat engine has three main parts: 1. An input reservoir with hot material at a higher temperature. Some kind of working substance that can expand and contract and do useful mechanical work. 3. An output reservoir with cool material at a lower temperature. 7

8 11/9/017 HEAT ENGINES HEAT ENGINE This is a representation of a heat engine; mechanical energy can be obtained from thermal energy only when heat can flow from a higher temperature to a lower temperature. CONSERVATION OF ENERGY Q in (+Q) heat engine Q out (-Q) W net (-W) Thermodynamic engine The net positive work that the substance does on the environment is: W = Q H Q C All of the quantities in this equation have positive values. HEAT ENGINES Why does a heat engine need a temperature difference? Second Law of Thermodynamics Spontaneous flow of energy, Otherwise the work done on the system in one part of the cycle will be equal to the work done by the system in another part, and the net work will be zero. 8

9 11/9/017 Increasing the maximum efficiency of a thermodynamic engine Increase the temperature of the hot reservoir. Diesel fuel burns hotter than gasoline, making diesel car engines more efficient than gasoline engines. Reduce the temperature of the cold reservoir. Doing so requires a more advanced cooling system. For example, turbocharged and supercharged car engines have an intercooler. HEAT ENGINES: THE STEAM ENGINE HEAT ENGINES: INTERNAL COMBUSTION ENGINE The most common example is the modern combustion engine. Let s use our knowledge of thermodynamics to analyze what is going on in one of the engine s cylinders! Start by setting up your whiteboards with the following PV diagram and a chart to keep track of each step of the cycle. We ll start with the gas (a gasoline/air mixture) at state A, with the gas at its lowest volume, ready to be ignited. n = 0.8 moles A B C P V T U A B B C C A Cycle W Q ΔU 9

10 11/9/017 The spark plug ignites the gasoline/air mixture, and the gas starts to burn! n = 0.8 moles B A B A Assume that the temperature of the gas triples, and that the piston doesn t have enough time to move significantly during this part of the process. Draw the process on your graph, and fill in the first row of the chart. A B C P V T U A B B C C A Cycle W Q ΔU As the gasoline/air mixture continues to burn, it expands isobarically, pushing the piston and turning the crank shaft! n = 0.8 moles B C B -W (Work done by gas) C A Useful! Assume that the volume of the gas multiplies by.5 during this expansion. Draw the process on your graph, and fill in the second row of the chart. A B C P V T U A B B C C A Cycle W Q ΔU So, what has happened so far? A B C W Q ΔU A B B C C A Cycle Thermal energy has been transferred to the gas (+Q) during both processes. This thermal energy actually came from chemical bonds being broken in the gasoline molecules when the gas ignites! Chemical potential energy transformed into thermal energy As a result, the gas expands and we get some useful work (-W) done by the gas! So far, we have put in 8.5 kj of thermal energy (Q in ), and gotten out 9 kj of work done by the gas This seems very wasteful, but having some excess thermal energy is inevitable in every heat engine! 10

11 11/9/017 Finally, as the gas cools down by dissipating thermal energy into the surroundings, it is compressed back to its initial state. n = 0.8 moles C A +W (Work done to bring gas back to its initial state) Not useful! Assume that this process is a straight line on the PV diagram (for the sake of simplicity). Draw the process on your graph, and fill in the last row of the chart. P V T U A B C A B W Q ΔU B C C A Cycle B C n = 0.8 moles During the last part of the process, 6 kj of work is done on the gas to bring it back to its initial state (this is not useful work!) A +W (Work done to bring gas back to its initial state) Not useful! At the same time, 5.5 kj of thermal energy is dissipated by the gas into the environment (-Q) as the gas cools off. A B C P V T U A B B C C A Cycle W Q ΔU This is why your engine feels so hot! It dissipates tons of thermal energy into the surroundings! This is also wasted energy, but it is impossible to avoid at least some of it being dissipated! P V T U A B C Q in = 8.5 kj heat engine W Q ΔU A B B C C A Cycle Q out = 5.5 kj W net by gas = 3 kj CONSERVATION OF ENERGY Q Hot to Sub W sub on Env Q sub to cold Change int U sub

12 11/9/017 EFFICIENCY The efficiency of a heat engine is defined as W e Q net in Efficiency is how much net work you get out divided by how much energy you put in! The more net work that you get out of the process, the closer the efficiency will be to 1. 0 < e < 1 What is the efficiency of the combustion engine that we have been analyzing? e = W NET Q in W Q ΔU A B B C C A Cycle e = 3000J 8500 J e = 10.53% Although this sounds low, the average efficiency of a modern car engine is only about 15-0%! Efficiency Whiteboard! 1 mole of an ideal gas in a heat engine undergoes the process shown below. B C Calculate the efficiency of this engine! A B P V T U A B C C A W Q ΔU A B B C C A Cycle W Q ΔU A B B C C A Cycle e = W NET Q in e = e = 1.5% J J WHITEBOARD In a car engine, gasoline is ignited by a spark plug to generate an explosion in a cylinder. Each explosion transfers 700 J of energy from the burning fuel to the gas in the cylinder. During the expansion (and later contraction) of the gas, the gas does a net 00 J of mechanical work on the environment (on a piston that pushes outward and helps propel the car). How much energy is transferred through heating to the car's cooling system during this cycle? What is the efficiency of the engine? 1

13 11/9/017 WHITEBOARD - solution HEAT ENGINES: CARNOT ENGINE Q in = 700 J e = W NET Q H e = heat engine Q out = 500 J W net by gas = 00 J 00 J 700 J e = 8.6% The Carnot engine was created to examine the efficiency of a heat engine. It is idealized, as it has no friction. Each leg of its cycle is reversible. The Carnot cycle consists of: Isothermal expansion Adiabatic expansion Isothermal compression Adiabatic compression Carnot's principle 1. It is impossible to use an engine's fuel directly to do work. The fuel needs to first warm a gas. This gas then expands and does work on a piston.. If the fuel transfers a certain amount of energy to the gas, only a fraction of this energy can do work on the piston. There is a theoretical limit on the work that can be done. All real-world engines fail to reach the theoretical limit. e carnot EFFICIENCY OF CARNOT ENGINE T hot T T hot cold T cold is the temperature of the cold reservoir of the engine. EFFICIENCY OF CARNOT ENGINE From this we see that 100% efficiency can be achieved only if the cold reservoir is at absolute zero, which is impossible. e carnot 1 T T cold hot T hot is the temperature of the hot reservoir of the engine. Real engines have some frictional losses; the best achieve 60-80% of the Carnot value of efficiency. 13

14 11/9/017 WHITEBOARD Determine the maximum efficiencies of the thermodynamic engines described below: A) burning coal heats the gas in the turbine of an electric power plant to 700 K. After the blades of the generator, the gas is cooled in cooling towers to 350 K. 700 K 350 K e carnot = 700 K WHITEBOARD - solution e carnot = T hot T cold T hot e carnot = 50% B) an inventor claims to have a thermodynamic engine that attaches to a car s exhaust system. The temperature of the exhaust gas is 90 C and the temperature of the output of this proposed engine is 0 C K K e carnot = K e carnot = 19.8% STATISTICAL APPROACH TO IRREVERSIBLE PROCESSES When describing a system that is made up of a large number of particles, it becomes helpful to analyze the possible microstates and macrostates of the system. Imagine a box consisting of four gas particles bouncing around at random. At any moment in time, you can check the box to see how many particles are on the left side, and how many are on the right side. These are all possible macrostates of the system. These are all possible macrostates of the system. i = macrostates of the system. i = 0 i = 1 i = i = 3 i = 4 macrostate of a system refers to its macroscopic properties, such as its temperature, pressure, volume and density. In each possible macrostate, a particular number of atoms are on the left side of the box, and a particular number are on the right side. We don t care which atoms are on the left or the right sides, only how many in total on each side. 14

15 11/9/017 Each of the unique possible ways for a macrostate to exist is called a microstate. All microstates are equally probable. i = macrostate Microstates (ways in which this macrostate can occur) However, each macrostate is comprised of a different number of microstates. The most probable macrostate is the one with the highest number of microstates. In general, an isolated system with a large number of particles will evolve toward states that have a higher probability of occurring. The highest probability macrostate is the one in which the particles are evenly distributed, because that macrostate has the greatest number of microstates (since microstates are equally probable). This state of maximum probability is often called the equilibrium state of the system. ENTROPY ENTROPY ( S ) OF THE ith MACROSTATE Entropy is a measurement of the probability of a particular macrostate. S i = k B Ln W i High-entropy states are more disorganized than lower-entropy states and are less able to do work on their environment. Entropy is a measurement of the randomness or disorder in the system. In which Ln(W i ) is the natural logarithm of the count W i and k B is a proportionality constant (Boltzmann s constant) with value k B = 1.38x10-3 J/K 15

16 NUMBER OF PARTICLES 11/9/017 Microstates and macrostates n! W i = n left! n right! WHITEBOARD The possible states of 8 atoms found on each side of a divider after its insertion in the box. Determine (draw) all possible macrostates. Determine all possible microstates of a macrostate (W i ) i = Macrostate Determine the entropy of each macrostate (S i in terms of k B ). W = number of microstates of one macrostate. Determine the probability of each macrostate (W i / total) WHITEBOARD WHITEBOARD (SOLUTION) MACROSTATE i n left n right MICROSTATE W i ENTROPY S i PROBABILITY MACROSTATE i n left n right MICROSTATE W i ENTROPY S i PROBABILITY k 1/ k 1/ k 7/ k 7/ k 35/ k 7/ k 7/ k 1/3 TOTAL k 1/56 TOTAL Spontaneous processes in an isolated system tend to proceed in the direction of increasing entropy (equilibrium) TEMPERATURE 16

17 11/9/017 SECOND LAW OF THERMODYNAMICS Spontaneous processes in an isolated system tend to proceed in the direction of increasing entropy Connecting atoms in a box to physics An isolated system with a large number of particles will evolve toward states that have a higher probability of occurring. The highest-probability state is the one in which the particles are evenly distributed that is, the state with the highest level of randomness. This state of maximum probability is often called the equilibrium state of the system. Macroscopic definition of entropy change When the gas in this system expands, it pushes outward on the environment. The entropy of the system increases even though the internal energy does not change if the process is isothermal. DIRECTION OF THERMAL ENERGY TRANSFER In an isolated system, energy always transfers from a warmer region to a cooler region. Most probability of a state occurring Most probability of a state occurring Macroscopic definition of entropy change The entropy change S SYS of a system during a process equals the ratio of the energy transferred to the system from the environment through heating Q divided by the system s temperature T SYS while the process is occurring: S sys = Q env to sys T sys DETERMINING S A 0.5 kg ice cube at -5 C melts on a plate. The ice (water) reaches room temperature (0 C ) : ICE CUBE (Q) ENVIRONMENT Identify two different systems. Identify a transfer of energy (Q) between both systems 17

18 11/9/017 1.Find Q (if it is not already given to you) Heat to increase the temperature of ice to 0 C Q = m c T Q = [0 C (-5 C)] Q = 5,50 J Heat to melt the ice Q = +m LF Q = ,000 Q = 167,500 J Heat to increase the temperature of water to 0 C Q = m c T Q = [0 C 0 C] Q = 41,800 J Total Heat Q total = Q ice + Q F + Q water Q total = 5, , ,800 Q total = 14,550 J ENVIRONMENT (HOT OBJECT) Q = - 14,550 J ICE CUBE (COLD OBJECT) Q = + 14,550 J. Find the average temperature for each system Temperature always in Kelvins. Temperatures before and after reaching thermal equilibrium. ENVIRONMENT T env_avg = T env_f + T env_i 93.15K K T env_avg = ICE CUBE T ice_avg = T ice_f + T ice_i 93.15K K T ice_avg = 3. Find the total change in entropy S total S total = S ice + S env S total = Q cold to hot T cold_avg + Q hot to cold T hot_avg S total = 14, ( 14,550) T env_avg = K T ice_avg = K S total = 3.6 J K NOTES ABOUT ENTROPY: S for an individual system may be positive or negative. If S total > 0, the process is allowed by the second law of thermodynamics (spontaneous process moving to a state of higher entropy). If S total < 0, the process is NOT allowed by the second law of thermodynamics (NOT a spontaneous process). SECOND LAW OF THERMODYNAMICS During any process that involves the transfer of energy through heating, the net change in entropy of the system and its environment is always greater than zero. 18

19 11/9/017 WHITEBOARD 0.50 kg of water at 70 C is added to 0.50 kg of water at 30 C. Follow the three steps and determine if the process is allowed by the second Law of Thermodynamics. WHITEBOARD - SOLUTION 1.Find Q (if it is not already given to you) HOT WATER (HOT OBJECT) COLD WATER (COLD OBJECT) Q = m c T Q = (50 70) Q = 41,800 J Q = m c T Q = (50 30) Q = 41,800 J. Find the average temperature for each system 3. Find the total change in entropy S total HOT WATER (HOT OBJECT) T H_avg = T H_F + T H_i 33.15K K T H_avg = T H_avg = K COLD WATER (COLD OBJECT) T c_avg = T C_F + T C_i 33.15K K T C_avg = T C_avg = K S total = S H + S C S total = Q cold to hot T cold_avg + Q hot to cold T hot_avg S total = 41, ( 41,800) S total = 8.01 J K S total > 0 An 0.8 kg aluminum placed in.16 kg of water WHITEBOARD 900 J c = kg c = spoon at 90 C is 4180 J kg final temperature of the mixture is 8.50 C. at C. The WHITEBOARD - SOLUTION 1.Find Q (if it is not already given to you) SPOON (HOT OBJECT) COLD WATER (COLD OBJECT) Determine if the process is allowed by the second Law of Thermodynamics. Q = m c T Q = (8.5 90) Q = 58,680 J Q = m c T Q = (8.5 ) Q = 58,680 J 19

20 11/9/017. Find the average temperature for each system 3. Find the total change in entropy S total SPOON (HOT OBJECT) T H_avg = T H_F + T H_i K K T H_avg = T H_avg = 3.4 K COLD WATER (COLD OBJECT) T c_avg = T C_F + T C_i 75.15K K T C_avg = T C_avg = 78.4 K S total = S H + S C S total = Q cold to hot T cold_avg + Q hot to cold T hot_avg S total = 58, ( 58,860) 3.4 S total = 8.8 J K S total > 0 WHITEBOARD A house at C transfers 115,000 J of thermal energy to the outside air, which has a temperature of -18 C. WHITEBOARD - SOLUTION OUTSIDE Determine the entropy change of the houseoutside air system and decide if the process is allowed by the second Law of Thermodynamics. HOUSE 1.Find Q (if it is not already given to you) HOUSE OUTSIDE AIR Q = 115,000 J Q = 115,000 J. Find the average temperature for each system 3. Find the total change in entropy S total HOUSE (HOT OBJECT) OUTSIDE AIR (COLD OBJECT) S total = S H + S C S total = Q cold to hot T cold_avg + Q hot to cold T hot_avg T H_avg = T H_F + T H_i 95.15K K T H_avg = T H_avg = K T c_avg = T C_F + T C_i 55.15K K T C_avg = T C_avg = K S total = 115, ( 115,000) S total = 61.1 J K S total > 0 0

21 11/9/017 WHITEBOARD WHITEBOARD - SOLUTION A barrel containing 190 kg of water sits in a cellar during the winter. On a cold day, the water freezes (change of state), releasing thermal energy to the room. This energy passes from the cellar to the outside air, which has a temperature of -5 C. CELLAR BARREL Determine the entropy change of this process if the cellar remains at 0 C and decide if the process is allowed by the second Law of Thermodynamics. 1.Find Q (if it is not already given to you) BARREL Q = -m LF Q = ,000 Q = -63,650,000 J CELLAR Q = 63,650,000 J. Find the average temperature for each system 3. Find the total change in entropy S total BARREL (HOT OBJECT) CELLAR (COLD OBJECT) S total = S H + S C S total = Q cold to hot T cold_avg + Q hot to cold T hot_avg T H_avg = T H_F + T H_i 73.15K K T H_avg = T H_avg = K T c_avg = T C_F + T C_i 48.15K K T C_avg = T C_avg = K S total = 63,650,000 + ( 63,650,000) S total = 3,476 J K S total > 0 WHITEBOARD WHITEBOARD - SOLUTION A 8.0 kg block slides on a level surface and stops because of friction. Its initial speed is 10 m/s and the temperature of the surface is 0 C. Determine the entropy change of the block (Hint: Use conservation of energy). m v KE = KE = 8 10 KE = 400 J S block = Q T S = KE = DU Conservation of energy S total > 0 S = 1.36 J K 1

22 11/9/017 WHITEBOARD WHITEBOARD - SOLUTION A 8 kg block slides from an initial speed of 10 m/s to a final speed of zero. It travels 15 m down a plane inclined at 18 with the horizontal. sinθ = opp hyp sin18 = h 15 h = 4.64 m Determine the entropy change of the block-inclined plane-earth system for this process if originally the block and the inclined plane were at 7 C. m v KE = KE = 8 10 KE = 400 J UG = m g h UG = UG = J WHITEBOARD - SOLUTION S block = Q T S = THERMODYNAMIC PUMP KE + UG = DU Conservation of energy S =.54 J K S total > 0 (COOLING SYSTEM) Thermodynamic pumps: Refrigerators and air conditioners By reversing the thermodynamic engine process, we have invented a machine called a thermodynamic pump. A refrigerator is a type of thermodynamic pump. An air conditioner is also a thermodynamic pump. Thermal energy is taken from the food by heating up a colder gas (refrigerant) Refrigerant is compressed (increasing its temperature) Refrigerant cools with the outside temperature Q C to sub W Comp on sub Q Sub to H

23 11/9/017 THERMODYNAMIC PUMP REFRIGERATOR The heat energy transferred from the food to the refrigerant (the system): Q C to sub + W comp on sub + Q H to Sub = 0 Q C to sub + W comp on sub Q Sub to H = 0 Q C to sub = Q Sub to H W comp on sub All of the quantities in this equation have positive values. REFRIGERATION COEFFICIENT OF PERFORMANCE CoP or K Ratio of heat energy transferred to the system and the amount of work done on the system (how good the system is cooling) K = Q C to substance W comp on sub K = Q Sub to hot W comp on Sub W comp on sub MAXIMUM REFRIGERATION COEFFICIENT OF PERFORMANCE CoP MAX or K MAX The Second Law of Thermodynamics limits this performance coefficient (how good the system is cooling) K MAX T C T H T C Using a thermodynamic pump to warm a house THERMODYNAMIC PUMP (HEATING SYSTEM) Outside air heats the working substance (refrigerant) The compressor does work to make the process possible The refrigeratns heats up the house Q C to sub W Comp on sub Q Sub to H 3

24 11/9/017 AIR CONDITIONING The heat energy transferred from the refrigerant (the system) to the house (environment): Q C to sub + W comp on sub + Q H to Sub = 0 COEFFICIENT OF PERFORMANCE CoP or K Ratio of heat energy transferred from the system to the environment (how good the system is heating) Q C to sub + W comp on sub Q Sub to H = 0 Q Sub to H = Q C to sub + W comp on sub K = Q Sub to H W Comp on sub All of the quantities in this equation have positive values. K = Q C to sub + W comp on sub W comp on sub MAXIMUM REFRIGERATION COEFFICIENT OF PERFORMANCE CoP MAX or K MAX The Second Law of Thermodynamics limits this performance coefficient (how good the system is heating) K MAX T H T H T H WHITEBOARD An ice making machine needs to convert 0.0 kg of water at 0 C to 0.0 kg of ice at 0 C. The room temperature surrounding the ice machine is 0 C. A) How much thermal energy must removed from the water? B) Determine the coefficient of performance C) Determine the minimum work needed to extract this energy by the ice-making machine. D) How much energy is deposited in the room? WHITEBOARD - SOLUTION WHITEBOARD - SOLUTION Q = m L F Q = 0.kg 335,000 J kg Q = 67,000 J kg K K T c T H T C (73.15) [ ] K 13.7 K = Q C to sub W comp on sub W comp on sub = Q C to sub K W comp on sub = 67,000 J 13.7 W comp on sub = J 4

25 11/9/017 WHITEBOARD - SOLUTION WHITEBOARD Q C to sub + W comp on sub Q Sub to H = 0 Q Sub to H = Q C to sub + W comp on sub Q Sub to H = 67,000 J + 4,905.7 J Q Sub to H = 71,905.7 J A thermodynamic pump transfers 14,000 J of energy through heating into a house. The air temperature outside is 10 C and the temperature inside the house is 0 C. Determine the maximum performance coefficient of the thermodynamic pump. Determine the minimum work that the pump's compressor must do on the working substance to cause this energy transfer. WHITEBOARD - SOLUTION WHITEBOARD - SOLUTION K max T H T H T C K = Q Sub to H W Comp on sub K max ( ) [ ] K max 9.77 No units W Comp on sub = Q Sub to H K W Comp on sub = J 9.77 W Comp on sub = J THERMODYNAMIC PUMP PICTURES 014 Pearson Education, Inc. 5

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