I need help understanding the easy to solve this problem. Explain everything, including the linear interpolation. ||!PDF *Elementary Principles of Chemic X PDF Elementary Principles of Chemica X +158°FSunnyFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDF Drive%20).pdfDrawD T Read aloudExample 8.3-3Solution←→When enthalpies must be calculated frequently for a species, it is convenient to prepare a tableof H(T) for the species (as was done for water in the steam tables) to avoid having to integrate theformula for C₂(T) again and again. Tables B.8 and B.9 in Appendix B list specific enthalpies ofspecies involved in combustion reactions-air, O₂, N2, H₂ (a fuel), CO, CO2, and H₂O(v). Thevalues in these tables were generated by integrating C₂(7) from the specified reference state (25°Cfor Table B.8, 77°F for Table B 9) to the listed temperatures. The next example illustrates the useof these tables.Evaluation of AH Using Heat Capacities and Tabulated EnthalpiesFifteen kmol/min of air is cooled from 430°C to 100°C. Calculate the required heat removal rate using (1) heatcapacity formulas from Table B.2, (2) specific enthalpies from Table B.8, and (3) the Enthalpy function ofAPEX.Q SearchΔΗ437With AEK, AĖp, and W, deleted, the energy balance iskJmolAssume ideal-gas behavior, so that pressure changes (if there are any) do not affect AĤ.1. The hard way. Integrate the heat capacity formula in Table B.2.Cp(T) dT[0.02894 +0.4147 × 10-³T +0.3191 × 10-87² — 1.965 × 10-¹²7³] AT0.4147 x 10-52==of 695 3= (0.497 mol)(-1.250 kJ/mol) = -0.621 KJQ = AH = nair Hair,out - nairĤair.in = nair AĤ100°C100°Cair(g, 430°C) → air(g, 100°C)J430°C= 0.02894(100 - 430) +-(100² - 430²)0.3191 x 10-81.965 × 10-12-(100³ - 430³) -34= (-9.5502-0.3627 -0.0835 +0.0167) kJ/mol = -9.98 kJ/molH(100¹ - 430¹) kJ/mol{JIDENGSign in10:22 AM5/6/2023•la+O ||!PDF *Elementary Principles of Chemic X PDF Elementary Principles of Chemica X +158°FSunnyFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDF Drive%20).pdfDrawD T Read aloud418 CHAPTER 8Balances on Nonreactive ProcessesCofaOD(100¹ - 430¹³) -438 of 695(2) TThe easy way. Use tabulated enthalpies from Table B.8.Ĥ for air at 100°C can be read directly from Table B.8 and Ĥ at 430°C can be estimated by linearinterpolation from the values at 400°C (11.24 kJ/mol) and 500°C (14.37 kJ/mol).Q SearchΔΗ = ἡΔῇ =34= (-9.5502-0.3627 -0.0835+ 0.0167) kJ/mol = -9.98 kJ/mol3. The easiest way. In a spreadsheet cell, insert =Enthalpy("air",430,100). The value -9.98 (in kJ/mol)will be returned. In using the spreadsheet, it is absolutely essential to keep track of units by addingexplicit notation where appropriate; otherwise the probability of getting units wrong increasessubstantially.Whichever way AĤ is determined15.0 kmolminĤ(100°C) = 2.19 kJ/molĤ(430°C) = [11.24 +0.30(14.37 - 11.24)] kJ/mol = 12.17 kJ/mol||AĤ = (2.19 - 12.17) kJ/mol = -9.98 kJ/mol(100* — 430*)] kJ/mol10³ mol1kmol- 9.98 kJmol1 min60 s1 kW1 kJ/s-2500 kWReminder: (The enthalpies listed in Tables B.8 and B.9(and for that matter, the heat capacityformulas of Table B.2 and the Enthalpy function of APEx apply strictly to heating and cooling ata constant pressure of 1 atm. The tabulated enthalpies and heat capacities may also be used fornonisobaric heating and cooling of ideal or nearly ideal gases; however, at pressures high enoughfor temperatures low enough) for the gases to be far from ideal, enthalpy tables on more accurateheat capacity formulas should be used.

Question

I need help understanding the easy to solve this problem.  Explain everything, including the linear interpolation. ||!PDF *Elementary Principles of Chemic X PDF Elementary Principles of Chemica X +158°FSunnyFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDF Drive%20).pdfDrawD T Read aloudExample 8.3-3Solution←→When enthalpies must be calculated frequently for a species, it is convenient to prepare a tableof H(T) for the species (as was done for water in the steam tables) to avoid having to integrate theformula for C₂(T) again and again. Tables B.8 and B.9 in Appendix B list specific enthalpies ofspecies involved in combustion reactions-air, O₂, N2, H₂ (a fuel), CO, CO2, and H₂O(v). Thevalues in these tables were generated by integrating C₂(7) from the specified reference state (25°Cfor Table B.8, 77°F for Table B 9) to the listed temperatures. The next example illustrates the useof these tables.Evaluation of AH Using Heat Capacities and Tabulated EnthalpiesFifteen kmol/min of air is cooled from 430°C to 100°C. Calculate the required heat removal rate using (1) heatcapacity formulas from Table B.2, (2) specific enthalpies from Table B.8, and (3) the Enthalpy function ofAPEX.Q SearchΔΗ437With AEK, AĖp, and W, deleted, the energy balance iskJmolAssume ideal-gas behavior, so that pressure changes (if there are any) do not affect AĤ.1. The hard way. Integrate the heat capacity formula in Table B.2.Cp(T) dT[0.02894 +0.4147 × 10-³T +0.3191 × 10-87² — 1.965 × 10-¹²7³] AT0.4147 x 10-52==of 695 3= (0.497 mol)(-1.250 kJ/mol) = -0.621 KJQ = AH = nair Hair,out - nairĤair.in = nair AĤ100°C100°Cair(g, 430°C) → air(g, 100°C)J430°C= 0.02894(100 - 430) +-(100² - 430²)0.3191 x 10-81.965 × 10-12-(100³ - 430³) -34= (-9.5502-0.3627 -0.0835 +0.0167) kJ/mol = -9.98 kJ/molH(100¹ - 430¹) kJ/mol{JIDENGSign in10:22 AM5/6/2023•la+O ||!PDF *Elementary Principles of Chemic X PDF Elementary Principles of Chemica X +158°FSunnyFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDF Drive%20).pdfDrawD T Read aloud418 CHAPTER 8Balances on Nonreactive ProcessesCofaOD(100¹ - 430¹³) -438 of 695(2) TThe easy way. Use tabulated enthalpies from Table B.8.Ĥ for air at 100°C can be read directly from Table B.8 and Ĥ at 430°C can be estimated by linearinterpolation from the values at 400°C (11.24 kJ/mol) and 500°C (14.37 kJ/mol).Q SearchΔΗ = ἡΔῇ =34= (-9.5502-0.3627 -0.0835+ 0.0167) kJ/mol = -9.98 kJ/mol3. The easiest way. In a spreadsheet cell, insert =Enthalpy(&#34;air&#34;,430,100). The value -9.98 (in kJ/mol)will be returned. In using the spreadsheet, it is absolutely essential to keep track of units by addingexplicit notation where appropriate; otherwise the probability of getting units wrong increasessubstantially.Whichever way AĤ is determined15.0 kmolminĤ(100°C) = 2.19 kJ/molĤ(430°C) = [11.24 +0.30(14.37 - 11.24)] kJ/mol = 12.17 kJ/mol||AĤ = (2.19 - 12.17) kJ/mol = -9.98 kJ/mol(100* — 430*)] kJ/mol10³ mol1kmol- 9.98 kJmol1 min60 s1 kW1 kJ/s-2500 kWReminder: (The enthalpies listed in Tables B.8 and B.9(and for that matter, the heat capacityformulas of Table B.2 and the Enthalpy function of APEx apply strictly to heating and cooling ata constant pressure of 1 atm. The tabulated enthalpies and heat capacities may also be used fornonisobaric heating and cooling of ideal or nearly ideal gases; however, at pressures high enoughfor temperatures low enough) for the gases to be far from ideal, enthalpy tables on more accurateheat capacity formulas should be used.<JDENGSign in10:22 AM5/6/2023•la+O

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Fifteen kmol/min of air is cooled from 430°C to 100°C. Calculate the required heat removal rate using (1) heat capacity formulas from Table B.2, (2) specific enthalpies from Table B.8, and (3) the Enthalpy function of APEX.