Thermodynamic System - Assignment Example

University of Central Lancashire School of Engineering FV1201 Assignment Brief Question 1: Answers A thermodynamic system is a three (3)-dimensional visible space in the universe in which one or more thermodynamic process of energy conversion is taking place (i.e. from one form to another form) (Petrucci, et al, 2007). Surrounding is a 3 dimensional visible space external (outside) to the thermodynamic system (Petrucci, et al, 2007). Universe= system + surroundings. Since energy is constant in the universe. There are three types of thermodynamic system i.e. 1) open system; 2) closed system; and 3) isolated system (Petrucci, et al, 2007). In open system, the mass is transferred across the boundary i.e. the energy is transferred across the boundary (Petrucci, et al, 2007). For example, cooling tower, engine of any car, water heated in an open container, and desert cooler. In closed system, only the energy is transferred but not the mass (Petrucci, et al, 2007). For example, a mixer, an air conditioner, human body, a refrigerator and electric motor. In an isolated system, both the energy and mass is not transferred (Petrucci, et al, 2007). For example, a cylinder and a piston arrangement in which the fluid like gas or air is being expanded or compressed is insulated it becomes isolated system. Question 2: Answers The Zeroth Law of Thermodynamics states that if two systems are in thermal equilibrium separately with a third systems then they are also in equilibrium with each other (Petrucci, et al, 2007). For example, if system A is in thermal equilibrium with system B and system C is also in thermal equilibrium with system C (Petrucci, et al, 2007). Then, according to 0th law of thermodynamics system C and system A will be thermal equilibrium with each other. In zeroth 0th law of thermodynamics law, we are able to use a third system (system C) to compare the temperature of two systems (system A and system B) without making thermal contact (Petrucci, et al, 2007). For example, if we take a cold bottle of juice from the fridge and a cup of hot water and leave them in the open space for hours, both the juice and water will have the same temperature at room temperature (Petrucci, et al, 2007). That is they both reach thermal equilibrium with the room, we can conclude "Two systems which are equal in temperature to a third system are equal in temperature to each other" Question 3: Answers Where: A is surface area, d is thickness, k is the thermal conductivity, and T1 and T2 is the temperature of two systems. Rate=Q/t = [0.162 (5x3)(20+273.1)]/0.015 Rate=[0.162 x 15 x 293.1]/0.015 Rate=712.233/0.015 Rate= 47482.2 W Question 4: Answers Where: A = heat transfer surface area, m2 [ft2] Q = heat transfer rate, W=J/s [btu/hr] ΔTLM = logarithmic mean temperature difference, °C [°F] U = overall heat transfer coefficient, W/(m2°C) Q=125x 293.1 x 3 Q= 109912.5 W Question 5: Answers How many joules of energy did the water absorb? Answer H2O Metal m= 50g m=12.48g T1=25.0°C T1= 99.0°C T2=28.1°C T2=28.1°C Cp water=4.184J Cp metal =? Qgained water= (mass) (Δt) (Cp, water) Qgained water= 50 x (28.1-25) x 4.181 Qgained water=648.055J How many joules of energy did the metal lose? Answer Qlost metal = Qgained water Qlost metal= 648.055J What is the heat capacity of the metal? Answer Heat Capacity of metal = 648.055/[99-28.1] =577.155 j/0C What is the specific heat of metal? Answer Qlost metal = Qgained water (mass) (Δt) (Cp, metal) = (mass) (Δt) (Cp, water) 12.8 x (99-28.1) x Cp metal = 50 x (28.1-25) x 4.181 Cp metal =648.055/ [12.8 x (99-28.1)] Cp metal=648.055/ 1213.68 Cp metal=0.534 J/g 0C Question 6: Answers Temperature Coefficients - α of steel = steel: 0.000012 (m/moC) Change in Temperature= 50 °C dl = L0 α (t1 - t0)         where: dl = change in object length (m, inches) L0 = initial length of object (m, inches) α =  linear expansion coefficient (m/moC, in/inoF)  t0 = initial temperature (oC, oF) t1 = final temperature (oC, oF) dl=200 x 0.000012 x 50 change in height of the tower dl= 0.12 m Question 7: Answers dl = 10 x 0.000023 x [450-10] dl =10 x0.000023 x 440 dl =0.212 m Final length = 10+ 0.212 =10.212 m Question 8: Answers Figure 1 Copper pipe Answer Q= hc A (Ts - Ta) Where A= surface area in hc = convective heat transfer coefficient Ta = Temperature air Ts = Temperature surface Surface area, A= 2 x pi x radius x length A= [2 x 3.14 x 30 x 1]/1000 A=188.4/1000=0.188 m2 Q= 6 W/m2K x 0.188 m2 x (80-20) Q= 6 x 0.188 x 60 Q=67.68 K Question 9: Answers Endothermic reactions are those reactions that take in energy from the surroundings. The energy is usually transferred as heat energy, causing the reaction mixture and its surrounding to get colder (Sanders, 2010), for example, electrolysis; reaction between sodium carbonate and ethanoic acid. Exothermic reactions are those reactions that transfer energy to the surroundings (Sanders, 2010). The energy is transferred as heat energy causing the surroundings and reaction mixture to become hotter (Sanders, 2010), for example, burning, reaction between calcium oxide and water and neutralisation reaction between alkalis and acids (Raizen, 2001). Question 10: Answers Answer ΔS°reaction = ΣnpS°products - ΣnrS°reactants ΔS°reaction = (4 S°NO + 6 S°H2O) - (4 S°NH3 + 5 S°O2) ΔS°reaction =[(4 x 211) + (6 x 189)] – [(4 x193) + (5 x 205)] ΔS°reaction = 1978 J/K·mol - 1797 J/K·mol) ΔS°reaction = 181 J/K·mol Question 11: Answers 2) a) CaCO3(s) → CaO(s) + CO2(g) Products CO2=213.6 J/mol.K CaO=39.75 J/mol•K Reactants CaCO3=88.70 J/mol•K ΔSreaction = [(213.6+ 39.75) J/mol•K - 88.70 J/mol•K] ΔSreaction = [253.35-88.70] J/mol•K ΔSreaction = 164.65 J/mol•K Products CO2= -393.5 kJ/mol CaO=-635.1 kJ/mol Reactants CaCO3=-1207 kJ/mol ΔH=[(-393.5-635.1) kJ/mol -(-1207) kJ/mol] ΔH=[-1028.6 + 1207] kJ/mol ΔH=178.4 kJ/mol ΔG° = [(-604+-394.4)-(-1128)] kJ mol–1 ΔG° =-998.4 +1128 kJ mol–1 ΔG° = 129.6 kJ mol–1 ---------------------------------------------------------------------------------- b) N2(g) + 3H2(g) ↔2NH3(g) ΔSreaction = [(2 x 192.3) J/mol•K – (191.5+130.6) J/mol•K] ΔSreaction = 384.6 J/mol•K-322.1 J/mol•K ΔSreaction = 62.5 J/mol•K ΔH=[(2 x-80.29)-0) J/mJ/mol•K ΔH= -160.58 J/mJ/mol•K ΔG° = [2 x -16.48] kJ mol–1 ΔG° = -32.96 kJ mol–1 ----------------------------------------------------------------------------------- c) NH4NO3(s) → NH4+(aq) + NO3¯(aq) ΔSreaction = [(113.4+146.4) J/mol•K – (151.1) J/mol•K ΔSreaction = [259.8-151.1] J/mol•K ΔSreaction =108.7 J/mol•K ΔH= [(-132.5-205.0) kJ/mol -(-339.9) kJ/mol]] ΔH=[-337.5+ 339.9] kJ/mol ΔH= 2.4 kJ/mol ΔG° = [(-79-108.7)-(-184)] kJ mol–1 ΔG° =-187.7 +184 kJ mol–1 ΔG° = -3.7 kJ mol–1 -------------------------------------------------------------------------------- d) H2O(g) ↔ H2O(l) (24 Marks) ΔSreaction = 188.7 J/mol•K – 69.91 J/mol•K ΔSreaction = 118.79 J/mol•K ΔH= [(-285.8) kJ/mol - (-241.8) kJ/mol]] ΔH= 44 kJ/mol ΔG° = (-237.1 - 203.26) kJ mol–1 ΔG° = -440.36 kJ mol–1 Question 12: Answers ΔSsurr = -ΔH/T Reaction a ΔSsurr = -ΔH/T ΔSsurr = - (-2045 kJ)/ (25 + 273) ΔSsurr = 2045 kJ/298 K ΔSsurr = 6.86 kJ/K Reaction b ΔSsurr = -ΔH/T ΔSsurr = -(+44 kJ)/298 K ΔSsurr = -0.15 kJ/K The change in entropy of the surroundings of reaction ‘a’ and ‘b’ was 6.86 kJ/K and -0.15 kJ/K respectively. Reference List Raizen, M.G. (2001). Demons, Entropy and the Quest for Absolute Zero. Scientific American, March 2001, pp 54-59. Sanders, L. (April 10, 2010). Molecules Get Superchilly Reaction. Science News, p 11. Petrucci, et al. (2007). General Chemistry: Principles & Modern Applications. 9th ed. Upper Saddle River, New Jersey. Read More