Energy Transfer and Thermodynamics - Report Example

of Central Lancashire School of forensic and Investigative Sciences FV1201: Energy Transfer and Thermodynamics Assignment 1)Define the four laws of thermodynamics using words, diagrams and equations where appropriate. (6 Marks) Ans: Zeroth Law: Two systems in thermal equilibrium with a third are in thermal equilibrium with each other. First Law: Energy cannot be created nor destroyed. Further, E = mc2, such that energy (E) is equal to matter (m) times the square of a constant (c).  Second Law: Heat cannot be transfer from a colder to a hotter body. Third Law: If all the thermal motion of molecules (kinetic energy) could be removed, a state called absolute zero would occur. Absolute zero results in a temperature of 0 Kelvins or -273.15° Celsius. 2) What is entropy? Explain why liquids are more disordered than solids and why gases are more disordered than liquids? (2 Marks) Ans: Entropy is the measure of the disorder or randomness of energy and matter in a system. The disorderliness of liquids and gases are caused by their particles which are more separated from each other compared to the molecules or particles of gases. 3) Predict whether entropy increases or decrease for the following reaction and explain why. (Do not calculate entropy): a. CaCO3(s) → CaO(s) + CO2(g) Ans: Increase b. N2(g) + 3H2(g) ↔2NH3(g) Ans: Decrease c. NH4NO3(s) → NH4+(aq) + NO3¯(aq) Ans: Increase d. H2O(g) ↔ H2O(l) Ans: Decrease (4 Marks) 4) Calculate ΔS for the following reaction, using the thermodynamic data provided. Ans: ΔS = S(products) – S(reactants) a. NaCl(s) → Na+(aq) + Cl¯(aq) ΔS = (51.2 + 223.5 ) – (72.1) = 202.6 kJ/mol b. 2NO(g)+O2(g) → N2O4(g), ΔS = (219.9) – [2(211.2) + 205.1] = -407.6 kJ/mol c. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l). ΔS = [213.7 + 2(69.9)] – [186.3 + 2(205.1)] = -243 kJ/mol d. 2NO2(g) ↔ N2O4(g) ΔS = (304.3) – [2(240.1)] = -175.9 kJ/mol (8 Marks) 5) These questions test your understanding of temperature measurements and temperature scales. a. Body temperature is 37°C what is this in Kelvin, Fahrenheit and Rankine scales? Ans: 37°C = 98.6 °Fahrenheit = 310 °Kelvin = 558 °Rankine b. What is absolute zero in Celsius, Fahrenheit and Rankine scales? Ans: Absolute zero: 273°Celsius - 459.4°Fahrenheit 0°Rankine c. The temperature of a system rises by 45°C during a heating process. Express this rise in temperature in Kelvins. Ans: °Kelvin = °Celsius + 273 = 45°C + 273 = 318°C d. The temperature of a system rises by 180°F during a heating process. Express this rise in temperature in R, K and °C. Ans: 180°F = 82.22°C = 355.22°K = 639°R 6) The mass flow rate is 4kg/s, the heat of combustion for C3H8 is 46 450kJ/kg. Determine the heat release rate. (1 Mark) Ans: Heat release rate = heat of combustion * mass flow rate = 46 450 * 4 = 185800 kJ/s 7) What is Fourier’s Law? Mathematically express Fouriers Law defining all the terms used within it. What is thermal conductivity? Compare the values of thermal conductivity of metals, insulating materials and gases. Why does Fouriers law have a minus sign? (8 Marks) Ans: Heat flow, dQ/dt, through a homogeneous solid is directly proportional to the area, A, of the section at right angles to the direction of heat flow, and to the temperature difference along the path of heat flow. Mathematically, where dQ/dt = heat flow rate; A = area of the section; dT/dx = rate of change of temperature along the path of the heat flow; λ = proportionality constant 8) Explain the Stefen-Boltzman Law. What is emissivity? What are the range of values for the emissivity of a surface? Define the terms “black surface” and “grey surface”. What role does the view factor play in determining the rate of heat transfer? What is a blackbody? (8 Marks) Ans: The Stefen-Boltzman Law states that “[in a black body], the emissive radiations or radiating heat flux is proportional to the 4th power of temperature on the absolute scale (Kelvin).” This means that for every Kelvin degree increase in temperature of a black body, the radiation emitted is raised to the fourth power. The emissivity of a body is its relative power of emitting radiation in comparison with a blackbody. Emissivity values range from 0 to 1. A black surface is a surface that is a perfect emitter of heat energy and has an emissivity value of 1. On the other hand, a grey surface does not emit heat energy fully and has an emissivity value less than 1. The view factor affects the rate of heat transfer because the total energy intercepted by an area from a point on another area is dependent on the solid angle with which the point views the area. A black body is a material that is a perfect emitter of heat energy and has an emissivity value of 1. 9) Define heat of combustion, heat release rate and combustion reaction giving appropriate equations. Explain the different types of combustion and definitions of the following: Specific heat capacity, latent heat, calorimetry, combustion temperature and chemical equilibrium. (8 Marks) Ans: Definitions: Heat of combustion is the amount of heat released per unit mass or unit volume of a substance when the substance is completely burned. Heat release rate is the rate at which an object releases heat. Combustion reaction is when all substances in a compound combines with oxygen and burns. Specific heat capacity is the amount of heat energy required to raise the temperature of a body per unit of mass. Latent heat is the quantity of heat absorbed or released by a substance undergoing a change of state. Calorimetry is the measurement of the amount of heat evolved or absorbed in a chemical reaction, change of state, or formation of a solution. Combustion temperature is the lowest temperature at which a substance will burn. Chemical equilibrium is a condition in which a chemical reaction is occurring at equal rates in its forward and reverse directions, so that the concentrations of the reacting substances do not change with time. Different types of combustion: Spontaneous - happens at even room temperature when enough vapor of the material and enough oxygen have collected in a confined space to allow rapid oxidation (combustion) to start and continue. Flash - happens when enough vapor has collected to burn at a certain elevated temperature but not enough to continue. Fire - happens at a higher temperature when the combustion for the moment heats the material to continually give off vapor from the material. 10) Determine the rate of heat transfer per unit area for a blackbody at 20°C. Is a good absorber of radiation a good emitter or a poor emitter? (2 Marks) Ans: First, express the temperature in absolute scale: T = 20°C = 293 K σ = 5.67 x 10-8 W/m2 K4 rate of heat transfer per unit area: q = (σ*T4) = [5.67 x 10-8 * (293)4] = 417.88 W/m2 11) Explain Newton’s Law of cooling and give the mathematical equation defining all the terms used. How is natural convection different from forced convection? (4 Marks) Ans: Newtons Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the temperature of its surroundings. By mathematical equation, where dT/dt is the rate of change of temperature of an object, T is the object’s temperature and Tα is the temperature of its surroundings. Natural convection occurs when a warmer less dense fluid is pushed away by a cooler denser fluid. On the other hand, forced convection occurs when an outside force pushes a fluid, such as water of air, to make it move and transfer heat. 12) Aluminium has a specific heat of 0.902 J/goC.   How much heat is lost when a piece of aluminium with a mass of 28.984 g cools from a temperature of 615.0oC to a temperature of 122.0oC? (2 Marks) Ans: Heat lost = mass * specific heat * change in temperature = 28.984 * 0.902 * (615 – 122) = -12 888.78 J 13) A heat engine draws heat from a combustion chamber at 300°C and exhausts to atmosphere at 10°C. What is the maximum thermal efficiency that could be achieved? (2 Marks) Ans: First, change to absolute temperature: 300°C = 573 K 10°C = 283 K Maximum thermal efficiency = (283/573)*100% = 49.39% 14) The temperature of a sample of water increases by 39.5oC when 24 500 J are applied.  The specific heat of liquid water is 4.18 J/goC.  What is the mass of the sample of water? (2 Marks) Ans: Heat lost = mass * specific heat * change in temperature 24 500 = mass * 4.18 * 39.5 mass = 24 500/(4.18 * 39.5) = 148.39 15) How much energy does it take to raise the temperature of 80 g of copper by 30 °C? Specific heat of copper is 0.385 J/g ºC. (2 Marks) Ans: q = mass * specific heat * change in temperature = 80 * 0.385 * 30 = 924 J 16) Define the following terms: a. Heat capacity, b. Specific heat, c. Isothermal, isobaric, and isochoric processes. (4 Marks) Ans: Heat capacity is the change in temperature of a system as it is heated. Specific heat is the amount of heat required to raise the temperature of a substance by 1ºC. An isothermal process is a change of a system in which the temperature remains constant. An isobaric process is a change of a system in which the pressure remains constant. An isochoric process is a change of a system in which the volume remains constant. 17) Heat is added to a system, and the system does 56 J of work. If the internal energy increases by 17J, how much heat was added to the system? (2 Marks) Ans: q = 17 J 18) A 60kg block of iron is heated from 20°C to 125°C. How much heat had to be transferred to the iron? (2 Marks) Ans: q = mass * specific heat * change in temperature = 60 * 0.46 * (105) = 2898 J 19) 150J of heat are injected into a heat engine, causing it to do work. The engine then exhausts 45J of heat into a cool reservoir. What is the efficiency of the engine? (2 Marks) Ans: Efficiency = (Heat output / Heat input)*100% = (45/150)*100% = 30% 20) What is kinetic energy and how does it relate to the temperature of a system? (2 Marks) Ans: The object’s kinetic energy is the energy it possesses in motion. As kinetic energy increases, pressure exerted by the gas also increases. By Boyle’s Law, pressure is directly proportional to temperature. Thus, the higher kinetic energy a gas has, the higher is its temperature. 21) 61.6 ml of milk at 18.6 °C are added to 455.5 ml of coffee at 90.2 °C. What is the final temperature in degrees Celsius of this liquid mixture when thermal equilibrium is reached? Assume coffee has the same properties as pure water. The average density of milk is 1032 kg/m3. The specific heat of milk is 1.97 J/g °C. (6 Marks) Ans: Qmilk = - Qcoffee m*cp*(Tf –Ti) = - m*cp*( Tf –Ti) 61.6*1.97*(Tf – 18.6) = - 455*1*(Tf – 90.2) Tf = (61.6*1.97*18.6+455*90.2)/(61.6*1.97+455) Tf = 75.12 °C 22) Gold has a specific heat of 0.129 J/g °C. If 15.0 g of gold absorbs 1.33 J of heat, what is the change in temperature of the gold? (2 Marks) Ans: q = mass * specific heat * change in temperature 1.33 = 15 * 0.129 * change in temperature change in temperature = 1.33 / (15 * 0.129) = 0.69 °C 23) A gas absorbs 3.5J of heat and then performs 1.5J of work. What is the change in internal energy of the gas? (2 Marks) Ans: change in internal energy = 3.5 – 1.5 = 2 J 24) Explain the ideal gas law, give the mathematical equation and define all the terms used. (2 Marks) Ans: The state of an amount of gas is determined by its pressure, volume, and temperature. That is, at a fixed temperature and in the limit of low pressure, the product of P and V of a given amount of any gas approaches the same constant value k. This relationship can be expressed as: where P is the absolute pressure of the gas measured in atmospheres; V is thevolume (in this equation the volume is expressed in liters); N is the number of particles in the gas; k is Boltzmanns constant relating temperature and energy; and T is the absolute temperature. 25) Explain what intensive and extensive properties are, giving examples of each to support your answer. (1 Mark) Ans: An intensive property is independent of the amount of mass. On the other hand, the value of an extensive property varies directly with the mass. Examples of intensive properties are temperature, pressure, and density. Examples of extensive properties are mass and total volume. 26) Discus the different types of systems encountered in thermodynamics. What is the state postulate? (3 Marks) Ans: Types of thermodynamic systems: a. Open system – a system with one or more openings that allow the transfer of mass or energy. b. Closed system – a system with closed boundaries such that a substance can neither enter or leave the system. c. Isolated system – a system that is not influenced by its surroundings. Neither the mass nor the energy crosses the boundary of the system. *State postulate - defines the given number of properties to a thermodynamic system in a state of equilibrium. 27) A can of soft drink at room temperature is put into the refrigerator so that it will cool. Would you model the can of soft drink as a closed system or as an open system? Explain. (2 Marks) Ans: Since heat flows out of the can of softdrinks to the fridge and cool air from the refrigerator goes into the can, forming a loop, the system is closed. 28) For a system to be in thermodynamic equilibrium, do the temperature and the pressure have to be the same everywhere? (1 Mark) Ans: Since there are many states of thermodynamic equilibrium, it is possible that only the temperature or only the pressure remains the same. 29) What is a quasi-equilibrium process? What is its importance in engineering? (2 Marks) Ans: If, in the passing from one state to the next, the deviation from equilibrium is small, and thus negligible, a quasiequilibrium process occurs; in this case, each state in the process can be idealized as an equilibrium state. The quasi-equilibrium process is important in engineering because it is instrumental in deducing relationships that exist among the properties of systems at equilibrium. 30) Consider two closed systems A and B. System A contains 3000 kJ of thermal energy at 20°C, whereas system B contains 200 kJ of thermal energy at 50°C. Now the systems are brought into contact with each other. Determine the direction of any heat transfer between the two systems. (2 Marks) Ans: Heat transfers from hot to cold. System A = 3000 / 20 = 150 kJ / °C System B = 200 / 50 = 4 kJ / °C Therefore, heat transfers from A to B. Works Cited and other References: Books: Ball, D., 2003. Physical chemistry. Mason, OH: Cengage Learning. Bergman, T., Incropera, F., DeWitt, D., & Lavine, A., 2011. Introduction to heat transfer. Hoboken, NJ: John Wiley & Sons. Cengel, Y., & Boles, M., 2006. Thermodynamics: an engineering approach. New York, NY: McGraw-Hill Higher Education. Cheng, Y.C., 2006. Macroscopic and statistical thermodynamics. World Scientific. Cutler, C., 2006. Dictionary of energy. Elsevier. Holbrow, C., Lloyd, J., & Amato, J., 2006. Modern introductory physics. Springer. Moran, M., Shapiro, H., Boettner, D., & Bailey, M., 2010. Fundamentals of engineering thermodynamics. Hoboken, NJ: John Wiley & Sons. Potter, M., 2010. Thermodynamics demystified. New York, NY: McGraw-Hill Professional. Rao, D., 2010. Fundamentals of food engineering. New Delhi: PHI Private Ltd. Wasserman, A., 2011. Thermal physics: concepts and practice. Cambridge, MA: Cambridge University Press. Websites: Microwaves101. [online] (16 October 2009) Available at: http://www.microwaves101.com/encyclopedia/thermal_analysis.cfm [Accessed 8 November 2011] The physics classroom. [online] 2011 Available at: http://www.physicsclassroom.com/class/thermalP/u18l2b.cfm [Accessed 5 November 2011] Chemical thermodynamics. [online] (2008) Available at: http://www.shodor.org/unchem/advanced/thermo/ [Accessed 10 November 2011] Read More