Energy Transfer and Thermodynamics - Assignment Example

Energy Transfer and Thermodynamics Student’s Name Institutional Affiliation Course Date of Submission Energy Transfer and Thermodynamics 1) Discuss the four laws of thermodynamics using words, diagrams and equations where appropriate. Include a discussion on entropy and how this is related to the laws of thermodynamics. Zeroth law of thermodynamics states that whenever two thermodynamic bodies are both in thermal equilibrium with another one, then the three bodies are in thermal equilibrium with each other. This means that the three systems are at the same temperature. The temperature scale is based on the zeroth law of thermodynamics[Won11]. The zeroth law of thermodynamics also introduces the notion of heat capacity, which is expressed mathematically as follows: q = CvΔT where, q = heat transferred Cv = constant volume ΔT = change in temperature The first law of thermodynamics is concerned with energy conservation, and it says that energy cannot be created or destroyed. According to the law, may only transform its forms and this makes the universe’s total energy to remain the same. In a thermodynamic cycle, the total heat added to a body equals the total work done by the body[Won11]. The first law of thermodynamics is expressed mathematically as: ΔU = ΔQ + ΔW Where, U = internal energy Q = heat W = work The second law of thermodynamics is concerned with the creation of entropy, and it says that it is possible to create entropy, but impossible to destroy it. According to the law, the entropy of an isolated body not in equilibrium will have a tendency of increasing throughout a period of time, coming nearly to a maximum value at equilibrium. A change in entropy over a period of time in an isolated system is either zero or positive. Second law of thermodynamics can also be expressed in terms of heat flow or heat engines[Won11]. The third law of thermodynamics is the basis of absolute temperature scale, and it states that as temperature moves towards absolute zero, the entropy of a body moves towards a constant minimum. In other words, it is not imaginable to cool a system to absolute zero. (15 Marks) 2) Two cylindrical metal rods of 1 meter in length, one made from aluminium and one made from iron, are heated from 20°C to 90°C. Given the coefficient of thermal expansion for the aluminium and iron rods are: 23.1x10-6K-1 and 11.8x10-6K-1 respectively, what is the difference in length of the two rods after heating (answer in mm)? ΔL = αL0ΔT For aluminium ΔLa = αaLoaΔT ΔLa = (23.1x10-6) x (1 x 1000) x (90-20) = 1.617 mm For Iron ΔLi = αiLioΔT ΔLi = (11.8 x 10-6) x (1 x 1000) x (90-20) = 0.826 mm Therefore, the difference is 1.617 – 0.826 = 0.791mm (6 Marks) 3) Gold has a specific heat of 0.129J/(g°C). How many joules of heat energy are required to raise the temperature of 22.0g of gold from 27°C to 93°C? ΔQ = m c Δt = 22 x 0.129 x (93 - 27) = 187.308J (2 Marks) 4) 25.0g of mercury is heated from 25°C to 155°C, and absorbs 455J of heat in the process. Calculate the specific heat capacity of mercury. c = ΔQ / m Δt = (455) / (25 x (155 - 25) = 0.14J/(goC) (2 Marks) 5) Predict whether entropy increases or decrease for the following reaction and explain why. (Do not calculate entropy): a) CaCO3(s) → CaO(s) + CO2(g) Increases, the system has increased the number of particles. In other words, one particle has decomposed into two. Additionally, one of the resultant products is a gas, which is more disordered than the initial solid. b) N2(g) + 3H2(g) ↔2NH3(g) Decreases, this is because the system has decreased the number of particles. That is, the reactants gas had 4 moles but the resultant products have only 2 moles. c) NH4NO3(s) → NH4+(aq) + NO3¯(aq) Increases, this is because the system has increased its particles after the reaction. That is, one reactant has decomposed into 2 ion particles. d) H2O(g) ↔ H2O(l) Decreases, because the reactant has changed its form from a gas to a liquid. Gases are very disordered as compared to liquids. (4 Marks) 6) Explain how work and a change in energy are related. Explain how energy change and force are related. Work involves the transfer of energy. When force is employed through a distance it causes energy change between objects and between points and also do some work[Mor10]. (2 Marks) 7) Use the table of thermodynamic data to calculate ΔS° and ΔH° for the following reactions, then calculate ΔG° (at 25.0°C) using the Gibbs Equation, ΔG° = ΔH° -TΔS°. a. NaCl(s) → Na+(aq) + Cl¯(aq) 43.9J/K ΔSo = ∑ nSo products - ∑ mSo reactants ΔSo = (59 + 56.5) – 72.13 = 43.37J/K ΔHo = ∑ n products - ∑ m reactants ΔHo = ((-240.1) + (-167.2)) – (-411.2) = 3.9kJ ΔG° = ΔH° -TΔS° ΔG° = 3.9 – (298.15)(0.04337) = -9.03kJ b. 2NO(g)+O2(g) → N2O4(g), -322.4J/K ΔSo = ∑ nSo products - ∑ mSo reactants ΔSo = 304.4 – (2(210.8) +205.2) = -322.4J/K ΔHo = ∑ n products - ∑ m reactants ΔHo = 11.1 – (2(91.3) + 142.7) = -314.2kJ ΔG° = ΔH° -TΔS° ΔG° = -314.2 – (298.15)(-0.3224) = -218.08kJ c. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l). -243J/K ΔSo = ∑ nSo products - ∑ mSo reactants ΔSo = (213.8 + 2(70)) – (186.3 + 2(205.2)) = -242.9J/K ΔHo = ∑ n products - ∑ m reactants ΔHo = (-393.5 + 2(-285.8)) – (-74.6 + 2(142.7)) = -1175.9kJ ΔG° = ΔH° -TΔS° ΔG° = -1175.9 – (298.15)(-0.2429) = -1103.48kJ d. 2NO2(g) ↔ N2O4(g) -175.9J/K ΔSo = ∑ nSo products - ∑ mSo reactants ΔSo = 304.4 – 2(240.1) = -175.8J/K ΔHo = ∑ n products - ∑ m reactants ΔHo = 11.1 – 2(33.2) = -55.3kJ ΔG° = ΔH° -TΔS° ΔG° = -55.3 – (298.15)(-0.1758) = -2.89kJ (12 Marks) 8) 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? In K= 37 + 273.15 = 310.15K in F = 9/5(37) + 32 = 98.6oF in Ra = 37(1.8) + 32 + 459.67 = 558.27oRa b. What is absolute zero in Celsius, Fahrenheit and Rankine scales? In 0C = -273.15oC in F = 9/5(-273.15) + 32 = -459.670F in Ra = -273.15 (1.8) + 32 + 459.67 = 00Ra c. The temperature of a system rises by 45°C during a heating process. Express this rise in temperature in Kelvins. 45 + 273.15 = 318.15K (3 Marks) 9) What is Fourier’s Law? Mathematically express Fourier’s 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 Fourier's law have a minus sign? Fourier’s Law is an experimental law that was established through observation. It is stated as: the rate of heat transfer through a solid is directly proportional to the cross-section area through which the heat flows, and to the temperature difference along the heat transfer path[Pal14]. It serves to define the thermal conductivity of the material. Fourier’s Law is mathematically expressed as follows: Where, is the rate of heat transfer refers to the thermal conductivity of the material, which can be described as the measure of the material’s ability to transfer heat (thermal) energy by conduction. is the cross-sectional area that the heat is flowing along in the solid. is the temperature difference present in the solid Thermal conductivity refers to the measure of the material’s ability to transfer heat or thermal energy by conduction. There are differences of thermal conductivities of various materials. Pure metals give the highest value of thermal conductivity, while gases give the lowest value. Insulating materials have thermal conductivities that lie between metals and gases. Thermal conductivity varies with temperature; for metals, thermal conductivity decreases with temperature, while that of insulating materials and gases increases with temperature[Ozi02]. Thermal conductivities of metals also depend on their purity as any form of impurity in metals normally causes a huge reduction in thermal conductivity[Ozi02]. Fourier’s Law has a minus sign because of three major reasons: 1) to provide an explanation for the systems that have lost heat instead of gaining; 2) to provide an explanation for materials with a negative thermal conductivity; and 3) because the heat energy flows in the opposite direction of the temperature difference or gradient[Pal14]. (10 Marks) 10) Explain the Stefan-Boltzmann 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? Stefan-Boltzmann Law is a law that is applied in black bodies only. It is used to give a description of the power that a black body radiates in regards to its temperature[Hol11]. It expresses the total energy emitted across all wavelengths of a black body as being directly proportional to the 4th power of the thermodynamic temperature of the black body. Stefan-Boltzmann Law is expressed mathematically as: = σT4 j/m2s Where σ = 5.6703 x 10-8 watt/m2K4 and is referred to as the Stefan-Boltzmann constant For other hot objects except ideal radiators, the Stefan-Boltzmann Law is mathematically stated as: = eσT4 Where, e is referred to as the emissivity of the object in question. For ideal radiators, emissivity (e) is equal to 1[Hol11]. Emissivity is a measure used to determine an object’s ability to emit radiation at a specified temperature as compared to a blackbody at similar temperature. It is the main basis of errors in the measurement of infrared temperature. The emitted infrared energy determines the object’s temperature. The range of values for the emissivity of a surface is from 0 (shiny mirror) to 1 (blackbody)[Rat11]. A black surface is a diffuse emitter that absorbs all radiation that is incident on it. According to Planck’s Law, a black surface has the capability of emitting the maximum possible energy possible for a specific wavelength and temperature of radiation. In other words, a black surface is considered to be the perfect absorber and emitter of radiation. A grey surface refers to a surface which the emissivity and the spectral absorptivity are not dependent of wavelength over the spectral areas of surface emission and irradiation. A view factor plays an important role in in determining the rate of heat transfer as it helps in expressing the fraction radiation exiting a surface that hits another surface in regard to the orientation of the two surfaces proportional to one another. A black body is an object that has the capability of entirely absorbing the radiation dropping on it. That is, its absorptivity is equal to 1[Rat11]. (12 Marks) 11) A diatomic gas does 300J of work and also absorbs 2.50kJ of heat. What is the change in internal energy of the gas? ΔEint = Qin +Won ΔEint = 2.50kJ – 0.30kJ, because the work done by the gas is equal to the negative work done on the gas = 2.20kJ (2 Marks) 12) What is the pressure exerted by 3.6moles of an ideal gas when it occupies a volume of 12.0L at 373K? P = nRT/V R = 0.08206Latm/molK P = (3.6mol x 0.08206Latm/molK x 373K)/12.0L = 9.1825atm 1atm = 1.013 x 105 Pa = 9.1825atm x (1.013 x 105) = 9.30188 x 105 Pa (2 Marks) 13) A flashbulb of volume 2.6cm3 contains O2 gas at a pressure of 2.3atm and a temperature 26°C. How many moles of O2 does the flashbulb contain? n = PV/RT = (2.3 x 0.0026)/ (0.08206 x 299) = 0.000244 moles = 2.44 x 10-4 moles (2 Marks) 14) If 0.20moles of helium occupies a volume of 64.0L at a pressure of 0.15atm, what is the temperature of the gas? T = PV/nR = (0.15 x 64)/ (0.20 x 0.08206) = 584.94K = 311.79oC (2 Marks) 15) What is the volume of 0.35moles of gas at 1.7atm of pressure and a temperature of 100K? V = nRT/P = (0.35 x 0.08206 x 100)/1.7 = 1.69L (2 Marks) 16) What is the pressure of 1.5moles of an ideal gas at a temperature of 150K and occupies a volume of 20.0L? P = nRT/V = (1.5 x 0.08206 x 150)/20.0 = 0.923atm (2 Marks) 17) A 1600kg car travels at a speed of 12.5m/s. What is its kinetic energy? KE = 0.5 mv2 = 0.5 x 1600 x 12.52 = 125000J = 125kJ (2 Marks) 18) A bar of an unknown metal has a length of 0.975m at 45°C and a length of 0.972m at 23°C. What is the coefficient of linear expansion? ΔL = αL0ΔT α = ΔL/L0ΔT = 0.003/ (0.972 x 22) = 0.0001403 = 1.403 x 10-4m/mK (2 Marks) 19) The accepted transition Reynolds number for flow in a circular pipe is: For flow through a pipe, at what velocity will this occur at 20°C for: a. Oil flow (ρoil = 861kg/m3 , µoil = 0.01743Ns/m2) with a diameter of 19mm; V = Reµ/ρd = (2300 x 0.01743)/(861 x 0.019) = 2.45 m/s b. Water flow (ρwater = 998kg/m3 , µwater = 0.001003Ns/m2 ) with a diameter of 17mm? V = Reµ/ρd = (2300 x 0.001003)/(998 x 0.017) = 1.36 x 10-1 m/s (10 Marks) 20) What is viscosity? What is the relationship between viscosity and intermolecular forces? What is the relationship between viscosity and temperature? Viscosity is a term used to give a description of the internal friction of a fluid under motion. It defines how resistant a fluid is to flow. In other words, it represents opposition to flow. When the intermolecular forces are stronger, viscosity is higher. In other words, increasing strength of intermolecular forces increases viscosity. On the hand, viscosity may decrease or increase with increasing temperature in fluids. In gases, viscosity increases with increasing temperature because as the gases’ molecules move more rapidly, the number of collisions between the molecules increases, which minimizes the capability of flowing (reduces viscosity). In contrary, an increase in temperature in liquids decreases viscosity[Vis07]. (6 Marks) Reference List Won11: , (Wong, 2011, p. 15), Won11: , (Wong, 2011, p. 17), Mor10: , (Moran, et al., 2010), Pal14: , (Pal, 2014, p. 233), Ozi02: , (Ozisik, 2002, p. 97), Pal14: , (Pal, 2014, p. 234), Hol11: , (Holzner, 2011, p. 299), Rat11: , (Rathore & Kapuno, 2011, p. 350), Rat11: , (Rathore & Kapuno, 2011, p. 351), Vis07: , (Viswanath, et al., 2007, p. 3), Read More