[Win,odd Dhamnekar]

During a recent winter month in Sheboygan, Wisconsin, it was necessary to obtain 3500 kWh of heat provided by a natural gas furnace with 89% efficiency to keep a small house warm (the efficiency of a gas furnace is the percent of the heat produced by combustion that is transferred into the house).

a)Assume that natural gas is pure methane and determine the volume of natural gas in cubic feet that was required to heat the house. The average temperature of the natural gas was 56 °F; at this temperature and a pressure of 1 atm, natural gas has a density of 0.681 g/L.

b)How many gallons of LPG (liquefied petroleum gas) would be required to replace the natural gas used? Assume the LPG is liquid propane [C₃H₈: density, 0.5318 g/mL; enthalpy of combustion, 2219 kJ/mol for the formation of CO₂(g) and H₂O(l)] and the furnace used to burn the LPG has the same efficiency as the gas furnace.

c)What mass of carbon dioxide is produced by combustion of the methane used to heat the house?

d)What mass of water is produced by combustion of the methane used to heat the house?



e)What volume of air is required to provide the oxygen for the combustion of the methane used to heat the house? Air contains 23% oxygen by mass. The average density of air during the month was 1.22 g/L.

f)How many kilowatt–hours (1 kWh = 3.6 × 10⁶ J) of electricity would be required to provide the heat necessary to heat the house? Note electricity is 100% efficient in producing heat inside a house.

g)Although electricity is 100% efficient in producing heat inside a house, production and distribution of electricity is not 100% efficient. The efficiency of production and distribution of electricity produced in a coal-fired power plant is about 40%. A certain type of coal provides 2.26 kWh per pound upon combustion. What mass of this coal in kilograms will be required to produce the electrical energy necessary to heat the house if the efficiency of generation and distribution is 40%?

My attempt to answer question

a) It was necessary to obtain 12.6 million kJ of heat provided by the combustion of CH₄ with 89% efficiency to keep small house warm. Combustion of  CH₄ at 25°C and 1 atm pressure releases -890.4 kJ/mol(1). But question states average temperature of CH₄ was 13.3333°C with density 0.681 g/L at 1 atm pressure.
  Assuming (1) and density of CH₄ 0.681 g/L   the volume of natural gas that was required to heat the house is 13227.84 ft³. Is ths answer correct?

Anawer to b) 139.756 U.S. gallons of LPG would be required to replace the natural gas used. Is this answer correct?

Answer to c) 699.74 Kg. of CO₂ is produced by a combustion of CH₄ used to heat the house. Is this answer correct?

Answer to d) 572.874 Kg. of water is produced by a combustion of CH₄ used to heat the house. Is this answer correct?

Answer to e) 3626273.45 liters of air is required to provide the oxygen for the combustion of methane used to heat the house. Is this answer correct?


Answer to f) 3500 kwh of electricity would be required to provide the heat necessary to heat the house assuming 100 % efficiency in producing the heat inside the house. Is this answer correct?

 I am working on all other questions. In a short span of time, I shall post my answers  to other questions if they are easy to answer.

[Win,odd Dhamnekar]

Answer to g) 1756.165 kg. of coal will be required to produce the electrical energy  necessary to heat the house if the efficiency in generation and distribution of electrical energy is 40%.
 Is this answer correct?

[Win,odd Dhamnekar]

Since there are no replies  about the correctness or otherwise of the answers given to these questions a, b, c, d, e, f, and g, I assume all the above answers are correct. (Do you want to see my workings? )

[mjc123]

Yes, show your working. If you just say "this is the number I get, is it right?" we have to work it out ourselves, and evidently nobody wanted to do that. If you show your working, we can see quickly whether you got the method right, even if we don't bother to check your arithmetic.
But one thing I can say at once is that your answers have too many significant figures.

[Win,odd Dhamnekar]

 Answer a)
 ΔH_C° of CH₄ (Methane)= CH₄(g) + 2O₂(g) CO₂(g) + 2H₂O (l) ΔH=-890 kJ/mol at 25°C

 We need 3500 kWh of heat by combustion of CH₄= 12.6 million kJ , [itex]\frac{12600000 kJ}{0.89}=14000000 kJ[/itex] (with 100% efficiency of gas furnace) are required.

[itex]\frac{14000000.0 kJ}{890.8 kJ}=16000[/itex] moles of CH₄ are required.

Mass of 16000 moles of CH₄=260000 grams .

 Density of CH₄ is 0.681 g L^-1, so the volime of CH₄ required in liters = 380000 liters = 13419.57 ft³ of CH₄ will be required to keep warm the small house.

Answers to other questions with workings will be provided very soon.

[mjc123]

Your method is correct. I get 13193 ft³.
You seem to be rounding off intermediate numbers to 2 sig figs, and then quoting your final answer to 7 sig figs! That's the opposite of what you should be doing. You should keep the intermediate values to several sig figs (not as many as a calculator might give you, but say 4 or 5), then round off your final answer to an appropriate number of figures (I would quote my answer as 13200 ft³). If you need to report an intermediate value as an answer, or to use your answer in further calculations, give your answer to the appropriate sig figs, but use the more precise value in further calculations.

[Win,odd Dhamnekar]

Answer to b)
ΔH_C° of C₃H₈= C₃H₈ + 5O₂ 3CO₂ + 4H₂O ΔH=-2219 kJ mol^-1 at 25°C

[itex] \frac{14000000 kJ}{2219 kJ}=6309.15[/itex] moles of C₃H₈ are required.

Mass of 6309.15 moles of LPG = 278214.51 g.

Given, Density of LPG =0.5318 kg L^-1,volume of LPG required in liters = 523.2 liters of LPG = 138.21 U.S.gallons are required to keep warm the small house.

Answer to c) 16000 moles of CO₂ will be produced in the combustion reaction of CH₄ , the mass of which would be 16000 moles * 44.009 g mol^-1of CO₂= 700 kg of CO₂

Answer to d) 32000 moles of H₂O will be produced in the reaction of combustion of CH₄, the mass of which would be 32000 moles * 18.015 g mol_-1 of H₂O = 580 kg of H₂O


Answer to e) Average molar mass of  dry air is 28.97 g/mol. The mass of Oxygen containing in one mole of dry air is 0.23 * 28.97 g =6.6631 grams. Combustion of 16000 moles of CH₄  requires 32000 moles of O₂, the mass of which is 32000 moles * 31.998 g /mol = 1023936 grams.

One mole of dry air contains 6.6631 g of oxygen, so [itex]\frac{1023936 grams}{6.6631 grams} = 153672.6148[/itex] moles of dry air is required, the mass of which is  4451895.65218 grams.

 The density of air is 1.22 g/L , so[itex]\frac{4451895.65218 grams}{1.22 grams} = 3650000[/itex] liters of dry air is necessary to keep small house warm.

Answer to f) As electricity is 100% efficient , 3500 kWh of electricity is required to keep small house warm. 

Answer to g) As efficiency in the production and distribution of electricity produced in coal-fired power plant is 40%, we need 3500 kWh /0.4 = 8750 kWh of heat to keep small house warm.  8750 kwh × [itex] \frac{0.45359237 kg}{1 lb} \times \frac{1 lb}{2.26 kwh}= 1760 kg[/itex]

[mjc123]

I haven't checked all your calculations, but the methods are correct and the magnitude of the answers looks about right.