1 Chemistry for Engineers (CHEM 114) PRELIM: Weeks 1 to 3 MODULE 1 Heat and Energy: Thermochemistry Course Learning Outcomes • CLO 1. Discuss the application of chemistry in relation to the generation of heat and energy. •CLO 3. Identify key chemistry concepts related to the specific field of engineering. •CLO 4. Learn and apply concepts in engineering materials. Module Learning Outcomes •Recognize a system as separate from the universe and its surroundings •Explain the energy flow between a system and its surroundings, and associate it by solving problems involving the flow of energy. •Apply previous knowledge on the flow of energy in solving calorimetric problems as an exchange of heat energy. CONTENTS of this MODULE Topics Page Energy Transformation and Conservation 2 Flow of Energy: Heat and Work 3 Heat Quantity Units 5 Exercise Number 1 6 Calorimetry: Sensible Heat and Specific Heat 7 Sensible Heat 7 Specific heat 8 Method of Mixtures 12 Exercise Number 2 14 Exercise Number 3 15 Calorimetry: Phase Changes 17 Latent Heat 17 Exercise Number 4 21 Topic Learning Outcomes: •TLO 1. Describe the energy flow between a system and its surroundings. •TLO 2. Solve problems involving the flow of energy. •TLO 3. Define calorimetry and solve problems on exchange of heat. Time Frame (as applicable) 1 hour 15 minutes 30 minutes 45 minutes Please put your signature over printed name in the Student’s Honesty Clause at the bottom of each exercise. Keywords: energy, system, heat, work, calorimetry, sensible heat, specific heat, latent heat Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition. 2 Energy Transformation and Conservation of Energy Refer to the figure below: Universe Surroundings System boundary Figure 1.1 A system and the universe In studying energy, we must separate a specific area to be studied. A system is that part of the universe that is being considered for studying energy flows. Outside the system, is its surroundings. Beyond the surroundings, is the universe. The system and the surroundings is separated by a boundary. The reason why we should consider a system, is because we have to maintain obedience to the Law of Conservation of Energy: Energy can neither be created nor destroyed, only transformed, i.e., the energy of the universe is constant. This is expressed by the equation: ∆𝑬 𝐮𝐧𝐢𝐯𝐞𝐫𝐬𝐞 = ∆𝑬𝒔𝒖𝒓𝒓𝒐𝒖𝒏𝒅𝒊𝒏𝒈𝒔 + ∆𝑬 𝒔𝒚𝒔𝒕𝒆𝒎 = 𝟎 The equation above expresses the First Law of Thermodynamics. Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition. 3 Some forms of Energy Energy is the ability to do work. It takes many forms among which are: o o o o Potential Energy is energy at rest. Kinetic Energy is energy in motion; Electrical energy is the energy from the movement of charges; Heat energy or thermal energy is the energy that is transferred through differences in temperature (sensible heat) or is released or absorbed due to a change in phase (latent heat). (Your lecturer will provide the equations and units for each of these forms of energy). Flow of Energy: Heat and Work Based from Figure 1.1, energy flowing into a system should be equal to energy that flows out of a system, to obey the law of conservation of energy. Surroundings Energy or Heat in System Energy out or Work Done boundary In equation form, this applies that energy is expressed as: ∆𝑬 = 𝒒 + 𝒘 Where: ΔE q w = Energy transferred in a system Units: = heat added to (or removed from) a system Joule, J or kiloJoule, KJ = work done by (or done to) a system BTU (British Thermal Units Calorie (cal) Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition. 4 Take note of the sign conventions: q is POSITIVE (+) when ADDED TO a system; NEGATIVE (-) when REMOVED FROM or RELEASED BY a system. w is POSITIVE (+) when DONE TO a system; NEGATIVE (-) when DONE BY a system. Heat Quantity Units Calorie, cal - refers to the amount of heat required to raise the temperature of 1 gram of water by 1 C0. British Thermal Unit, BTU - is defined as the amount of heat required to raise the temperature of 1 pound of water by 1 F0 Conversion Units for Heat: 1 calorie, cal 1 kilocalorie (kcal) = 4.186 Joules, J = 1000 cal = 4,186 J = 4.186 kiloJoules, KJ 1 British Thermal Unit, BTU = 778 ft.lb = 252 cal = 1055 J *The calorie is not a fundamental SI unit. The International Committee on Weights and Measures recommends using Joule as the basic unit of energy in all forms. Examples: 1. Convert 2000 calories to Joules 2000 calories 4.186 J 1 cal = 8, 372 J 2. Convert 300 BTU to Joules 300 BTU 1055 J 1 BTU = 316, 500 J Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition. 5 Examples: (to be solved by your lecturer) 1. If 500 J of heat is added to a gas system and it did 215 J of work, how much is its change in energy? Convert your answer to BTU, then to calories. 2. If a machine does 3.8 x 103 KJ of work after an input of 4.6 x 104 KJ of heat from fuel, how much is its change in internal energy? Convert your answer to BTU, then to calories. 3. Calculate the heat in a system if it does 27 J of work, and its energy decreases by 40 J. Answer Exercise No.1 to be uploaded by your faculty-in-charge later. Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition. 6 PART 2 CALORIMETRY: Heat Transfer, Quantity of Heat (Sensible Heat), Specific Heat Thermal Equilibrium - The state where two bodies have the same temperature. An example of this state is when a hot metal shot is placed in a glass containing water at room temperature. The heat of the metal shot will be transferred to the water until such time that there will be a constant temperature reading. The temperature achieved is the equilibrium temperature, Te, also called mixing temperature, Tm. Temperature - A physical property that depends on the physical state of a material and is a quantitative description of its hotness or coldness. Heat - refers to energy in transit (energy transferred) from one body to another because of a temperature gradient (temperature difference). It is not the amount of energy contained within a system. Sir James Prescott Joule (1818 – 1889) He studied how water is warmed by stirring the water with a paddle wheel (refer to figure 1.2). The paddle wheel adds energy to the water by doing work on it. He found out that the temperature rise is directly proportional to the amount of work done on the water. ∆T α W It is for this reason that the energy transferred as heat is also the amount of energy consumed as work. Take note that the units of heat, energy and work are the same. The basic unit of energy, the Joule, is named after him. Figure 1.2: Joule’s paddle wheel Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition. 7 Quantity of Heat as Sensible Heat, Q Heat transfer or heat flow takes place solely because of a temperature difference. Heat flows occurs between bodies when their temperatures are different, from a body with the higher temperature going towards the body with a lower temperature. The quantity of heat, Q required to increase the temperature of a mass m of a certain material is proportional to its temperature change, ∆T. It is also dependent on the nature of the material. Hence; Q = mc∆T Where: Q m c ∆T = quantity of heat transferred or absorbed = mass of substance = specific heat = temperature difference or temperature gradient Sign Conventions: + Q = add heat - Q = release or remove heat Temperature Change, ∆T (also called temperature difference, or temperature gradient). This refers to the difference between two temperatures, usually final temperature, (Tf) minus the initial temperature (Ti) or higher temperature (TH) minus the lower temperature (TL). A temperature change can be calculated using the following formulas: ∆T = Tf - Ti ∆T = TH – TL Note: change in temperature is denoted by the degree symbol AFTER the unit symbol. Problem-Solving Tips: ✓ For a decrease in temperature, ∆T is negative. ✓ For an increase in temperature, ∆T is positive. ✓ 1 C0 change in temperature is equal to 1 K change. ✓ 1 C0 change in temperature is equal to 1.8 F0 change. ✓ Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition. 8 Specific Heat, c Specific heat is the amount of heat required to raise a unit mass of a substance by one degree of temperature. It is dependent on the nature of material. The specific heat, c, of water is: 4190 J/kg K = 4.19 KJ/kg K = 1 cal/g C0 = 1 BTU/lb F0 TABLE 1.1 Approximate Specific and Molar Heat Capacities at Constant Pressure Substance Specific Heat Molecular Molar Heat Capacity, c Mass, M Capacity, C (J/ kg K) (kg/mol) (J/mol K) Aluminum 910 0.0270 24.6 Beryllium 1970 0.00901 17.7 Copper 390 0.0635 24.8 Ethanol 2428 0.0460 112.0 Ethylene glycol 2386 0.0620 148.0 Ice 2000 0.0180 36.5 Iron 470 0.0559 26.3 Lead 130 0.207 26.9 Marble (CaCO3) 879 0.100 87.9 Mercury 138 0.201 27.7 Salt (NaCl) 879 0.0585 51.4 Silver 234 0.108 25.3 Water (liquid) 4190 0.0180 75.4 Source: University Physics 9th edition (p. 473) 1996 by H.D. Young and R.A. Freeman. Addison-Wesley Publishing Company, Inc. Massachusetts, USA. Problems: Example 1. An ice cube tray of negligible mass contains 0.450 kg of water at 220C. How much heat must be removed to cool the water to 00C without freezing? Solution: Q = mc ΔT = (0.450 kg)(4190 J/ kg C0)( 00C - 220C) = - 41,481 J Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition. 9 Example 2. How much heat is required to raise the temperature of 30 kg of silver from 250C to 900C? From Table 1.1 : Csilver = 234 J/kg C0 Solution: Q = mcΔT = (30 kg)( 234 J/kg C0)( 900C - 250C) = 456,300 J Example 3. How much heat must be removed to cool 8.5 kg of mercury from 1000C to 450C? From Table 1.1: c mercury = 138 J/ kg C0 Solution: Q = mc ΔT = (8.5 kg)(138 J/ kg C0)(450C - 1000C) = - 64,515 J Conceptual question: A cold block of metal feels colder than a cold block of wood at the same temperature. Likewise, a hot block of metal feels hotter than a hot block of wood at the same temperature. Why? Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition. 10 Method of Mixtures In heat transfer, the hotter object loses heat while the cooler object absorbs the heat released to reach thermal equilibrium (thermodynamic equilibrium). The final temperature of the resulting mixture of these two objects is the equilibrium temperature, Tm (or Te). If the masses of the objects, the initial and equilibrium temperatures are measured, the specific heat of the materials can be calculated using the Method of Mixtures. Q lost =Q gained mcΔTH = mcΔTL Tip: Te or Tm should be between TH and TL Figure 1.3 Principle of Heat Transfer to achieve thermal equilibrium (thermodynamic equilibrium) Problems: Example 1. Steam Burns vs. Water Burns. What is the amount of heat input to your skin when it receives the heat released by an engine boiler? Which of them will cause a more severe burn? a) 20 g of steam initially at 1000C when it is cooled to 340C. (c of steam= 2,260,440 J/kg C0) Q = mc ΔT = (0.02 kg)( 2,260,440 J/ kg C0)(340C - 1000C) = - 2, 983,780.8 J The answer is negative because released by the engine boiler. Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition. 11 b) 20 g of water initially at 1000C when it is cooled to 340C. Q = mc ΔT = (0.02 kg)( 4190 J/ kg C0)(340C - 1000C) = - 5,530.8 J A more severe burn will be caused by steam. Example 2. An engineer is working on a new engine design. One of the moving parts contains 1.40 kg of aluminium and 0.50 kg of iron, and is designed to operate at 1500C. How much heat is required to raise its temperature from 200C to 1500C? Q total = Q aluminium + Q iron = mcΔT aluminium + mcΔT iron From Table 1.1 : caluminum = 910 J/kg C0 ciron = 470 J/kg C0 Q total = (1.40 kg)(910 J/kg C0)( 1500C - 200C) + (0.50 kg)( 470 J/kg C0)( 1500C - 200C) Q total = 196,170 J Example 3. A technician measures the specific heat capacity of an unidentified liquid by immersing an electrical resistor in it. Electrical energy is converted to heat, and is transferred to the liquid for 120 seconds at a constant rate of 65 W. The mass of the liquid is 0.780 kg, and its temperature rises from 18.550C to 21.320C. Find the average specific heat of the liquid at this temperature range. Solution: Step 1: Power, P = 65 W = 65 J/s time, t = 120 seconds Recall in Physics 1: P = work, W = energy, E t t E = Pt = (65 J/s)( 120 seconds) = 7,800 J = Q (amount of heat) Step 2: Q = mcΔT c = __Q__ mΔT = ________7,800 J ________ (0.780 kg)( 21.320C - 18.550C) = 3,610.11 J/kg C0 or K Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition. 12 Example 4. A 0.0356 kg aluminum calorimeter contains 0.200 kg of water at a temperature of 24 C. Lead shots with a mass of 0.241 kg at 1290C were dropped into the water. Find the equilibrium temperature of the mixture. 0 caluminum = 910 J/kg C0 cwater = 4190 J/kg C0 clead = 130 J/kg C0 Q lost =Q gained mcΔTH = mcΔTL mc ΔT lead = mc ΔTaluminum + mc ΔTwater (0. 241 kg)( 130 J/kg C0)( 1290C – Te) = (0.0356 kg)( 910 J/kg C0)(Te – 240C) + (0. 200kg)( 4190 J/kg C0)(Te – 240C) Te = 27.650C Answer Exercises Nos. 2 & 3 to be uploaded by your faculty-in-charge later. Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition. 13 PART 3: CALORIMETRY: PHASE CHANGES (Latent Heat) Refer to the figure below: Freezing Evaporation Condensation Melting Sublimation Deposition Figure 1.4: Phase Changes (Changes of State) For any given pressure, a phase change takes place at a definite temperature, usually accompanied by absorption or emission of heat, and a change in volume and density. PHASE CHANGES Latent Heat of Fusion, Lf The heat required per unit mass to melt (or freeze) a given material. Q = mLf To melt 1 kg of ice at 00 C to 1 kg of liquid water at 00C and normal atmospheric pressure, requires an absorption of a latent heat of fusion of 3.34 x 105 Joules, or 334,000 J. The process is reversible, so to freeze water, heat is removed, but the magnitude is the same. For any given material at any given pressure, the freezing temperature is the same as the melting temperature. The latent heat of fusion, Lf of water at normal atmospheric pressure is: 3.34 x 105 J/kg = 334,000 J/kg = 80 cal/g = 143 BTU/lb Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition. 14 Latent Heat of Vaporization, Lv The heat required per unit mass to evaporate (or condense) a given material. Q = mLv For any given material at any given pressure, the boiling temperature is the same as the condensation temperature. The latent heat of vaporization, Lv of water at normal atmospheric pressure is: 2.256 x 106 J/kg = 2,256,000 J/kg = 540 cal/g = 970 BTU/lb To vaporize 1 kg of water at 1000 C to 1 kg of water vapour or steam at 1000C and normal atmospheric pressure, requires an absorption of a latent heat of vaporization of 2.256 x 106 Joules. The process is reversible, so to condense the steam or water vapor, heat is removed, but the magnitude is the same. Sign Conventions: Positive (+) = absorption of heat Negative (-) = release of heat Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition. 15 Quantity of Heat with Phase Changes and Temperature Changes, Q Phase change caused by the release or absorption of heat involves the masses of the materials, and the latent heat of fusion, Lf and latent heat of vaporization, Lv. The process also involves the transfer of heat across a temperature gradient (temperature difference) between the hotter and colder objects (refer to figure 1.5). Q gained + Q change of phase = Q lost + Q change of phase mcΔTL + mLf (or Lv) = mcΔTH+ mLv(or Lf) Figure 1.5 Concept diagram of Specific Heat and Latent Heat for water During change of state in a substance, for example, water becomes ice, its internal energy changes, so the kinetic energy of its particles changes. As it is changing from one state to another, the change in energy is reflected in the bonds between the particles, and therefore the temperature of the object doesn’t change. These bonds either break or bind to melt (or vaporize) or solidify (or condense) the substance. Once the change of state is complete, however, changes in energy are again observed in the form of changes in temperature. Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition. 16 Substance Helium Hydrogen Nitrogen Oxygen Ethanol Mercury Water Sulfur Lead Antimony Silver Gold Copper TABLE 1.2 Heats of Fusion and Vaporization Normal Melting Heat of Fusion, Normal Boiling Point Lf (in J/ kg) Point K * 13.84 0 C * 259.31 63.18 209.97 54.36 218.79 159 -114 234 -39 273.15 0 392 119 600.5 327.3 903.65 630.50 1233.95 960.80 1336.15 1063 1356 1083 Heat of Vaporization, Lv (in J/ kg) 0 * 58.6 x 103 K 4.216 20.26 C -268.93 -252.89 20.9 x 103 452 x 103 25.5 x 103 77.34 -195.8 201 x 103 13.8 x 103 90.18 -183.0 213 x 103 x 103 x 103 x 103 x 103 x 103 x 103 x 103 x 103 x 103 351 630 373.15 717.75 2023 1713 2466 2933 1460 78 357 100 444.60 1750 1440 2193 2660 1187 854 272 2256 326 871 561 2336 1578 5069 104.2 11.8 334 38.1 24.5 165 88.3 64.5 134 x 103 x 103 x 103 x 103 x 103 x 103 x 103 x 103 x 103 Source: University Physics 9th edition (p. 475)1996 by H.D. Young and R.A. Freeman. Addison-Wesley Publishing Company, Inc. Massachusetts, USA. • A pressure in excess of 25 atmospheres (25 atm) is required to make helium solidify. At 1 atmosphere pressure, helium remains a liquid down to absolute zero (-273 K) Problems: Example 1. How much heat is required to convert 8 grams of ice at -150C to steam at 1000C? Given: cice below 00C = 2000 J/kg K Q total = Q gained + Q change of phase = mcΔT + mLf + mcΔT Ice at -150C to ice at 00C Melting of ice at 00C Melted ice heated to water at 1000C + mLv water at 1000C to steam at 1000C Q total = (0. 008 kg)( 2000 J/kg K)(00C - - 150C) + (0. 008 kg )(3.34 x 105 J/kg) + (0. 008 kg)( 4190 J/kg K)(1000C – 00C) + (0. 008 kg )( 2.256 x 106 J/kg) Q total = 24,312 J Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition. 17 Example 2. A calorimeter (vessel whose walls is thermally insulated) contains 2.10 kg of water and 0.250 kg of ice, all at a temperature of 00C. The outlet of a tube coming from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel to raise the temperature of the system to 340C? Neglect the heat transferred to the container. Figure 1.6 Concept of a calorimeter containing water and ice. TH = 1000C Tube from a boiler m steam = ? TL = 00C Solution: Q gained + Q change of phase Te = 340C 2.10 kg water + 0.250 kg ice = Q lost + Q change of phase Water and ice mcΔT water + mLf Melting of ice at 00C steam + mcΔT Melted ice to water at Te = 340C = mLv steam at 1000C + mcΔT steam to water at Te = 340C (2.10 kg)( 4190 J/kg K)(340C - 00C) + (0.250 kg )( 3.34 x 105 J/kg) + (0.250 kg)( 4190 J/kg K)(340C - 00C) = (m steam)(2.256 x 106 J/kg)+ (m steam) ( 4190 J/kg K)(1000C – 340C) By factoring: (2.10 kg)( 4190 J/kg K)(340C - 00C) + (0.250 kg )( 3.34 x 105 J/kg) = (m steam) [(2.256 x 106 J/kg)+ ( 4190 J/kg K)(1000C – 340C) ] Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition. 18 m steam = (2.10 kg)( 4190 J/kg K)(340C - 00C) + (0.250 kg )( 3.34 x 105 J/kg)+ (0.250 kg)( 4190 J/kg K)(340C - 00C) [(2.256 x 106 J/kg)+ ( 4190 J/kg K)(1000C – 340C) ] m steam = 0.165163 kg ≈ 165.16 grams of steam Answer Exercises No. 4 to be uploaded by your faculty-in-charge later. SUMMARY: In studying the flow of energy, the system must be identified first as separate from its surroundings and the universe. Energy is the ability to do work, in the same way that work is the use of energy. Hence, the change of energy is the sum (or difference) of these two. Calorimetry demonstrates the exchange of energy in two ways: as sensible heat and as latent heat. As sensible heat, the temperature difference (temperature gradient) between hot and cold substances, and their specific heats are considered. In latent heat, the amount of heat absorbed (or released) by substances to change their state or phase, are considered. In most instances, sensible heat and latent heat occur together in calorimetry. References: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition. Potter, Merle and Somerson, Craig (2006). Schaum’s Outline of Thermodynamics for Engineers. 2nd Edition. McGraw-Hill Companies Inc. USA https://phet.colorado.edu and https://www.merlot.org for laboratory simulations Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition. 19 Enrichment Activity/Web Based Task: Laboratory Simulation on “Calorimetry” In real engineering practice, a bomb calorimeter is used to measure the heat generated by fuels. It is a thick-walled vessel filled with water, with an inner vessel wherein the fuel sample is placed with an ignition wire. The heat generated by the ignited fuel sample is measured as the change in the temperature of the surrounding water. This laboratory simulation demonstrates the said measurement. 1. Connect to this link: http://employees.oneonta.edu/viningwj/sims/calorimetry_s.html 2. Using this laboratory simulation, provide data for the following table: (remember to “reset” every time you change the substance used, before clicking “ignite”) Final Water Change in Substance Mass Mass of water Initial Water Temperature Temperature Water in the in Temperature in degree Celsius in degree Celsius milligrams calorimeter 0 0 ( C) ( C) in Celsius in grams degree (C0) Benzoic acid 500 1000 20 Trinitrotoluene 500 1000 20 Nitroglycerin 500 1000 20 3. Based on your gathered data, write a conclusion for this experiment. Conclusion:_______________________________________________________________ ________________________________________________________________________. Engr. M. C. Javier-Tala, MSEE.CLAUSE: Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition. Student’s HONESTY I hereby declare that all solutions and answers in this activity were done by me. I pledge to practice the highest degree of Academic Honesty at all times, as expected from all students, indicated in the Course Policies of the syllabus for the course, Chemistry for Engineers (CHEM114). 20 Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.