Energy Transfer and Thermodynamics - Assignment Example

Thermodynamics INSTRUCTIONS: Please answer all 30 questions, showing workings out where appropriate. 1) Define the four laws of thermodynamics using words, diagrams and equations where appropriate. (10 Marks) Zeroth Law If two systems are in thermal equilibrium with a third, then they are in thermal equilibrium with each other. Thermometers are a working example of this law (The Zeroth Law of Thermodynamics, 1998). The First Law The first law of thermodynamics states that “The total energy of an isolated thermodynamic system is constant”. The law is also known as the conservation of energy. In other words, energy maybe lost from a system in only two ways, either as work or as heat, U cannot change in any other way. Thus, for a finite change: ΔE=Q-W or dE=dQ-dW (First Law of Thermodynamics, 2008) Second law: The second law of thermodynamics, states that the total entropy of any system remains unchanged, except for what flows outward across the boundary of the system. Consequently, the entropy of an isolated system cannot reduce (Second Law of Thermodynamics). Third Law The absolute zero cannot be realized, by any finite series of processes (Jones). 2) What is entropy? Explain what happens to the motion of water molecules when ice melts into water? What happens to the entropy in this situation? (3 Marks) With the passage of time, energy and matter progress towards disorder. This disorder is Entropy, S, is defined as a measure of disorder. The increasing disorder in a system, accounts for spontaneous change. Entropy as energy, Q, in relation to absolute temperature, T, is expressed as S =  . Hence, its units are JK-1. The random motion of water molecules is greater than that of the ice molecules, hence entropy increases and the change from ice to water is favored (Entropy as Time's Arrow). 3) Predict whether entropy will be 0 for the following processes: a. Dry ice melts, b. Water freezes, c. Gasoline evaporates. (3 Marks) a. Entropy increases when dry ice melts. b. Entropy decreases when water freezes. c. Entropy increases when gasoline evaporates. 4) Calculate ΔS for the following reaction, using the thermodynamic data provided. a. CO(g)+2H2(g) → CH3OH(l), b. 3H2(g) + N2(g) → 2NH3(g), c. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l). (6 Marks) a. CO(g)+2H2(g) → CH3OH(l) The standard molar entropies of reactants and products are So (CO) = 197.6 JK−1mol−1 So (H2) = 130.7Jmol−1K−1 So (CH3OH(l)) = 126.8 J/K mol ∆ So = Σ So(Products) - Σ So(Reactants) ∆ So = So (CH3OH(l)) – 2 × So (H2) - So (CO) ∆ So = 126.8JK−1mol−1 - 261.4 JK−1mol−1 – 197.6 JK−1mol−1 = -332.2 JK−1mol−1 b. 3H2(g) + N2(g) → 2NH3(g) Standard entropies: N2 (g) =192 Jmol-1K-1; H2 (g) =131 Jmol-1K-1 ; NH3 (g) =192 Jmol-1K-1 ΔS° (NH3) = [(2 x 192)] – [ (3 x 131) + 192 ] = - 201 JK-1mol-1 ÷ 2 = -101 JK-1mol c. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) The standard molar entropies of reactants and products are So (CH4) = 197.6 JK−1mol−1 So (O2) = 205.138 Jmol−1K−1 So (CO2) = 213.74 Jmol−1K−1 So (H2O (l)) = 69.91 J/K mol ∆ So = Σ So(Products) - Σ So(Reactants) ∆ So = So (CO2) +2 So (H2O (l)) – So (CH4) – 2So (O2) ∆ So =213.74JK−1mol−1 +139.82JK−1mol−1 - 197.6 JK−1mol−1 – 410.276 JK−1mol−1 = -254.316JK−1mol−1 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? The temperature of the human body in the Kelvin scale is 37 OC = 37 + 273.15 = 310.15K . 37°C = {[37x(9/5)] + 32} 0 F = 98.60 F Rankine = 98.60 F + 459.67 = 558.270 R b. The freezing point of water if 0°C what is this in Kelvin’s? K = C + 273.15 Hence, 00 C = 273.15K c. The temperature of a system rises by 20°C during a heating process. Express this rise in temperature in Kelvin’s. Each degree on the centigrade and Kelvin scales are equally spaced, hence a 200C rise is a 20K rise. d. The temperature of a system rises by 40°F during a heating process. Express this rise in temperature in R, K and °C. (8 Marks) The Fahrenheit and the Rankine scales are equally spaced; hence, 400F corresponds to 40R. 40F = [40] x5/9C = 22.22C As the Kelvin and Celsius scales are equally spaced, a 22.22C increase corresponds to a 22.22K increase. 6) The mass flow rate is 3kg/s, the heat of combustion for C3H8 is 46450kJ/kg. Determine the heat release rate. (1 Mark) Where  is the measured rate of heat release, is the mass loss rate, and HC is the known lower heat of combustion, in kJ/kg. Here  = 3kg/s,  is 46450 kJ/kg, and  is to be determined. kJ/s. Hence the rate of release of heat is 139350 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. What does Fourier"s law have a minus sign? (10 Marks) Fourier’s law states that the rate of conduction of heat, per unit cross – sectional area of a body is proportional to the negative temperature gradient of the body. Heat flow is in the direction of the negative temperature gradient; as a result, it flows from a hotter body to a colder body. Thus we have a negative sign in Fourier’s equation: QCond =  The constant of proportionality k is called the thermal conductivity of the material. It is an estimation of the heat conducting capacity of a substance, and it is measured in w/m-K. The range of values of k for metals, insulators and gases is given below: Type Of Material Thermal Conductivity Value Metals 52 ~ 415 w/m-K Insulating materials 0.035 ~ 0.173 w/m-K Gases 0.0069 ~ 0.173 w/m-K (Fourier's Law of Conduction, 1998). 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? (10 Marks) Heat radiation is an electromagnetic wave, whose propagation is as a transverse wave. The Stefan – Boltzmann law provides the relation between the maximum amount of radiation that can be emitted per second and the absolute temperature of the emitting surface (TS), as: Qe(max) = ATS4 Where Qe(max) is the amount of radiation emitted per second, A is the area of the radiating surface, TS is the absolute temperature of the surface in Kelvin and  is Stefan’s constant. = 5.67 × 10-8 W/m2-k4 The above relation applies to a black body or ideal radiator. As there are no ideal radiators, emissivity (ε), which reflects the true position to a greater degree was defined. The latter is the ratio between the energy radiated by some material to the energy radiated by a black body, at the same temperature. This can be expressed as: ε = ε is a dimensionless quantity that depends on temperature at which it is measured and the material. For an ideal radiator or blackbody, ε = 1; for all other surfaces 0< ε Read More

The latter is the ratio between the energy radiated by some material to the energy radiated by a black body, at the same temperature. This can be expressed as: ε = ε is a dimensionless quantity that depends on temperature at which it is measured and the material. For an ideal radiator or blackbody, ε = 1; for all other surfaces 0< ε

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