Energy Transfer and Thermodynamics - Assignment Example

Energy Transfer and Thermodynamics Name: Course: Instructor: Institution: Date of Submission: 1. Thermodynamics systems refers to the radiative content or materials perceived in a macroscopic volume in a certain space. The thermodynamics systems is enclosed by the materials space which separate it from the surroundings (Goss & Ralph, 2007). Thus, the surroundings the changes occurring in the space of the thermodynamic systems as changes from the original to the ultimate state occur. The interaction of the material through to the final stage leads to the identification of the system (Gonick & Criddle, 2005). The universe is the combination of the thermodynamic system and the surroundings. The universe is attained when the energy in the system and surroundings is constant. The thermodynamic spaces differ from open, closed to isolated systems (Moran, et al., 2010). An Open system refers to the system where energy and matter interact freely (Halliday, et al., 2010). The best example of this is in kitchens, where when boiling something in through the stove, both matter and energy are transferred through steam. Also, through work energy is transferred such as when making coffee. The equipment for stirring coffee allows energy to be transferred freely when pouring it to the cup. A closed system is different from the open in that the energy is exchanged only through the surroundings of the space. For instance, when boiling meat in the kitchen and one places a lid on the pan, the energy is transferred only to the surroundings, depicting the closed system. Isolated systems are those systems where energy transfer does not occur, even among the surroundings of the space. For instance, coffee in a thermos or hot pot used for serving food. 2. The Zeroth law of thermodynamics demonstrates that when two thermodynamic spaces are at equilibrium with a third system, then the process of thermal equilibrium is attained (Kondepudi, 2008). The importance of the law in measuring temperature is that the systems at equilibrium can be used as thermometers, which can give the absolute thermodynamic temperature scale. 3. Conductivity rate Q = k A ∆T ∆x = Thermal conductivity = 0.162m/ wK Temperature = 20 oC Assuming that the heat travels in one direction and that the thermodynamic system is steady, the conduction rate is attained (5m) (3m) (0.162w/ mK) / 1.5m (20 oC) (15m) (0.162m/ wK) / 1.5 (20 oC) (10m) (0.162m/ wK) (20 oC) 1.62 * 20 32.4W 4. 3m2 window Temperature of the surface = 22oC Outside wind is blown at abut 2 oC Convection coefficient = 125W/ m2K Total heat transfer = q = h A ΔT 125 W/ m2K *3m2 * (22 -2) Q = 7500 W Which presents heat transfer out 5. a) q = m * c *Δt q = 4.18J/ g oC * C * 50.0ml (28.1 oC – 25.0 oC) q = 4.18 * C * 50ml * 3.1 c = 4.18 * 50 * 3.1 q = 647.9J b) According to the statistics, the energy lost is always equivalent, though the lost energy s given in the opposite direction. Therefore, the energy lost = -647.0J c) Heat capacity of the metal is given through: cp = q / m * Δt cp = 647.9 / 12.48 * (28.1 oC – 90.0 oC) cp= 647.9 / 12.48 * - 61.9 cp = 647.9 / -772.5 cp = 0.84J/ g oC d) Specific heat of the metal 12.48 * 0.84 10.48J/ oC 6. The vertical antenna of the tower is 200m high. Temparature change occurs from -20 oC to 30 oC. Length = 200m Temperature 2 = 30 oC Temperature 1 = - 20 oC (t2 – t1) = 30 oC – (-20 oC) = 50 o C Steel alpha = 0.000011/ C Thus, the change in the antenna length based on the above figures is based as: (0.000011/C (200m / 50 C) 0.000044m L2 – L1 The change = 0.000044m 7. Aluminium girder is 10 long; it is heated in a fire of about 450 oC from the original temperature of 15 oC. Specific heat of aluminium = 900J/ kg Co The final length of the aluminium = dl = L0 a (t1 – t0) dl = length change lo = initial length a= linear expansion coefficient aluminium linear co-efficient = 0.000023 (m/ m Oc) to = initial temperature t1 =final temparature dl = (10m) 0.000023 (450 – 15) dl = (10m) 0.000023 (435) dl = 10 * 0.0100 dl = 0.1001 Final Length = 10.1001m 8. 30mm copper pipe Heat transfer co-efficient through convective process = h= 6 W/m2K Temparature = 50 oC Surroundings temperature = 20oC Part 1 (Heat Loss through convention) Q = h (tsurface – tsurrounding) Q = 6 (80oC – 20 oC) Q Through convention = 360W/ m2 Q conv = A = Q conve * 2 πr = 360 * 2 * π * 0.01 Heat loss through the convention process is 22.6W/m Part 2 (Heat loss through radiation for a metre length of the pipe q radiation = σ (t40 – t41) 5.67 * 10-8 (3534 – 2934) Heat loss through radiation = 462 W/ m2 For a metre of the pipe, heat loss = 462 * 2 * π * 0.01 Heat loss through radiation per metre of the pipe is 29.1 W/ m2 9. Endothermic reactions refer to the process where energy is conveyed from the thermodynamic system surroundings of the space/system (Cengel, 2007). For instance, during a blast it can be described through the calcium carbonate thermal decomposition process. The surroundings have a lower heat compared to the heat they had initially (Bergman & Frank, 2011). On the other hand, exothermic reactions refers to the heat transfer to the surroundings of the thermodynamic system (Vidal, 2003). The surroundings have a higher energy transfer rate compared to what they started with such as the process of burning wood. 10. Calculating the entropy standard molar changes of the reactions given below? 4NH3(g) + 502(g) 4NOg + 6H2Og S°NH3 = 193J/Kmol S°O2 = 205J/Kmol S°NO = 211J/Kmol S°H2O = 189J/Kmol The standard molar entropy reaction is attained through the molar entropies differences of the sum of the products of the reactants of the molar entropies (Modest, 2013). ΔS°reaction = ΣnpS°products - ΣnrS°reactants ΔS°reaction = (4 S°NO + 6 S°H2O) - (4 S°NH3 + 5 S°O2) ΔS°reaction = (4(211 J/K·K) + 6(189 J/K·mol)) - (4(193 J/K·mol) + 5(205 J/K·mol)) ΔS°reaction = (844 J/K·K + 1134 J/K·mol) - (772 J/K·mol + 1025 J/K·mol) ΔS°reaction = 1978 J/K·mol - 1797 J/K·mol) ΔS°reaction = 181 J/K·mol 11. Calculate entropy ΔS, enthalpy ΔH and Gibbs Free Energy ΔG for the following equations: a) CaCO3(s) → CaO(s) + CO2(g) ΔG° = Σ ΔGf°(products)– Σ ΔGf°(reactants) The ΔGf° presents the reactions of the equation constituents. CaCO3(s): –1128 kJ mol–1, CaO(s): –603.5 kJ mol–1, CO2 (g): –137.2 kJ mol–1. ΔG° = (–603.5 –137.2) – (–1128) kJ mol–1 = +387.3 kJ mol–1 It indicates that under the standard conditions the course is not spontaneous. b) N2(g) + 3H2(g) ↔2NH3(g) ΔS°= 2S°(NH3) -[S°(N2) + 3S°(H2)] ΔS°= (2 mol) (192.5 J/mol-K) -[(1 mol) (191.5 J/mol-K) + (3 mol) (130.6 J/mol-K)] = -198.3 J/K Answer 198.3 J/K ΔS° is negative, in pact with our qualitative forecast constructed on the decrease in the sum of molecules of gas throughout the reaction. c) NH4NO3(s) → NH4+(aq) + NO3¯(aq) 1ΔHf (NH4+1 (aq)) + 1ΔHf (NO3-1 (aq)) - 1ΔHf(NH4NO3 (s))] [1(-132.51) + 1(-207.36)] - [1(-365.56)] = 25.69 kJ 25.69 kJ      [1ΔSf(NH4+1 (aq)) + 1ΔSf(NO3-1 (aq))] - [1ΔSf(NH4NO3 (s))] [1(113.39) + 1(146.44)] - [1(151.08)] = 108.75 J/K 108.75 J/K      From ΔGf° values: 1ΔGf (NH4 +1 (aq)) + 1ΔGf (NO3-1 (aq)) - [1ΔGf(NH4NO3 (s)) [1(-79.37) + 1(-111.34)] - [1(-184.01)] = -6.70000000000002 kJ -6.70 kJ     (spontaneous) From ΔG = ΔH - TΔS: -6.73 kJ     (The reaction from this equation is spontaneous) d) H2O(g) ↔ H2O(l) ∆ H = +6.02 kJ/mol 283 K ? (+10°C) sol → liq 263 K? (−10°C) sol ← liq The flow of the heat energy influences the process of spontaneous change. 12. Calculate the thermodynamic system surroundings entropy for the following two reactions, assume the temperature is 25°C a) C2H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) ΔH = -2045kJ ΔH°rxn= [ (2) (ΔH°CO2) + (4) (ΔH° H2O)] – [ (1) (ΔH°C3H8) + (5) (ΔH°O2) ΔH°rxn= [ (3 mol) (-393.5 kJ/mol) + (4 mol)  (-285.5 kJ/mol) ] – [ (1 mol) (-103.9 kJ/mol) + (5 mol) (0 kJ/mol) ΔH°rxn= [-1180.5 kJ + (-1142 kJ)] – [-103.9 kJ] ΔH°rxn= (-2322.5 kJ) + 103.9 kJ ΔH°rxn= -2218.6 kJ b) H2O(l) → H2O(g) ΔH = +44kJ ∆ S = 188.7 – 69.9 = 118.8 J/K/mol ∆ H = (-241.8)-(-285.8) = +44.0 kJ/mol ∆ G = ∆ H – T ∆ S = 44,000 – 298 x 118.8 J/mol = +8600 J/mol (8.60 kJ/mol) ∆ G > 0, therefore this is not a spontaneous reaction References Bergman, T. L. & Frank, P. I., 2011. Introduction to heat transfer. New York: John Wiley & Sons. Cengel, Y. A., 2007. Introduction to thermodynamics and heat transfer+ EES software. New York: McGraw Hill Higher Education Press. Gonick, L. & Criddle, C., 2005. The Cartoon Guide to Chemistry. 1 st ed. New York, NY: Harper Collins Publishers Inc.. Goss, D. J. & Ralph, H. P., 2007. General Chemistry Principles & Modern Applications, Petrucci, Harwood, Herring, Madura: Study Guide.. New York: Pearson/Prentice Hall,. Halliday, D., Jearl, W. & Robert, R., 2010. Fundamentals of Physics, Chapters 33-37.. New York: John Wiley & Sons. Kondepudi, D. K., 2008. Introduction to modern thermodynamics. NY: Chichester: Wiley. Modest, M. F., 2013. Radiative heat transfer.. New York: Academic press. Moran, M. J., Howard, N. S., Daisie, D. B. & Margaret, B. B., 2010. Fundamentals of engineering thermodynamics. New York: John Wiley & Sons. Vidal, J., 2003. Thermodynamics: Applications in chemical engineering and the petroleum industry. USA: Editions TechniP. Read More