What is the heating value of propane in BTU/GAL?
598451
90,962 BTUs per gallon gross heating value
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Usually the composition f natural gas is as follows:\nComponent   Typical Analysis(mole %)  Range(mole %)\nMethane          94.9                  87.0 - 96.0\nEthane           2.5                   1.8 - 5.1\nPropane          0.2                   0.1 - 1.5 \niso - Butane     0.03                  0.01 - 0.3\nnormal - Butane  0.03                  0.01 - 0.3\niso - Pentane    0.01                  trace - 0.14\nnormal - Pentane 0.01                  trace - 0.04\nHexanes plus     0.01                  trace - 0.06\nNitrogen         1.6                   1.3 - 5.6\nCarbon Dioxide   0.7                   0.1 - 1.0\nOxygen           0.02                  0.01 - 0.1\nHydrogen         trace                 trace - 0.02\n  \nSpecific Gravity 0.585                 0.57 - 0.62\nGHV(MJ/m3), dry* 37.8                  36.0 - 40.2
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Table 1. Fuel properties.\n\n__________________________________________________________________________\n                              Fuel      Heat of     Cetan     Viscosity \n                             Weight   Combustion    Number   Centistokes \n___________________________________________________________________________\n                            Lbs./gal.   BTU/gal.\n \nNo. 2 diesel                  7.05      140,000        48        3.0 \n100% Biodiesel (B100)\nMethyl or ethyl ester          7.3      130,000        55        5.7 \nB20 mix (20/80)                7.1      138,000        50        3.3 \nRaw vegetable oil              7.5      130,000     35 to 45   40 to 50\nthe btu is 130,000 per gallon
569208
You need the specific heat of water, which is 4.184 J/g*C.  Assuming a density of water of 1 g/ml (not always the safest assumption, but close enough), then your 500 ml = 500 g.  You're raising the temperature 35 degrees, so the heat you will need will be:\n\n4.184 * 500 * 35 = 73,220 J = 73.22 kJ.\n\nThe combustion of decane provides 6.78 MJ/mol, or 6780 kJ/mol.  In order to give 73.22 kJ, you would need to burn 73.22/6780 = 0.0108 mol of decane.  Given decane's molecular weight of 132 g/mol, you would have to burn 1.425 grams.
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You want "energy density:"\n\nEthanol (Ethyl Alcohol):\n26.8 MJ / kg\n= 2905 kCal / lb.\n\nCrude Oil:\n45 MJ / kg\n= 4879 kCal / lb\n\nAviation Kerosene:\n43 MJ / kg\n= 4662 kCal / lb\n\nI don't understand the "If measured with a calorimeter, what degree celsius would it reach." Are you talking about the combustion temperature?
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Propane is a by-product from two sources: natural gas processing and crude oil refining. Most of the LPG used in the United States is produced domestically. When natural gas is produced, it contains methane and other light hydrocarbons that are separated in a gas processing plant. Because propane boils at -44°F and ethane boils at -127°F, it is separated from methane by combining increasing pressure and decreasing temperature.\n\nThe natural gas liquid components recovered during processing include ethane, propane, and butane, as well as heavier hydrocarbons.\n\nPropane and butane, along with other gases, are also produced during crude refining as by-products of the processes that rearrange or break down molecular structure to obtain more desirable petroleum compounds.
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http://energy21.freeservers.com/bookrep.html  Want to know something else? I've found a way to convert gasoline into Propane! And it can never be marketed! All Engines from 1996 on up must run at an Air to Fuel Ratio of 14.7 Parts of Air to 1 Part of Fuel! My System goes 100 to 1 Air to Fuel Ratio, but it will never happen! You want something interesting to Research ? This is it!
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PV = nRT\nn = PV/RT\n= ([2.5 * 101.325] * 30 ) / (8.31 * [273 + 327])\n= 1.5 mol of CH4 (2 Sig Fig)\n\n1 CH4 = 1 CO2 Therefore 1.5 mol of CO2\nm(CO2) = 1.5 * (12 + 32)\n= 66 g
209800
Dry natural gas can contain other hydrocarbon components such as ethane propane, butane, pentane, hexane and so on. Each of the molecules have a higher heating value that pure natural gas (methane)\n\nDry natural gas can also contain inert materials such as some CO2 or N2. These can take up space in each volume of natural gas and not contribute any caloric value.
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Take table for R134a refrigerant:\n\na) delta s = entropy change\n--inlet for evaporator assuming 50%moisture= interpolate to get S \n@200kpa for R134a\n0       Sg   \n0.5     S\n1       Sf\nget Sg and Sf from table at 200kpa for R134a. \n \n--out for evaporator = Sg at 200Kpa (get from table) since it is sat vap.\n\nthe difference of both Sfg and Sg is the answer.\n\nb)  delta S = Q / T  where Q = 120 KJ and T = -5+273 = 268K\n\nc) add both a) and b) is the total entropy
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barrel (petroleum) bl ≡ 42 gal (US) = 158.987 294 928 L \n\nThere are 42 gal(US) ! US gal is diffrent from UK gal and imperial gal
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You have not stated where these vapors are coming from or what their purity is, but in general propane vapors can be compressed, cooled and thereby liquified. At approximately 200 psi and 100 degree F. propane can be condensed from a vapor to a liquid. I say approximately, because propane seldom exists in the pure form but is usually mixed with small amounts of methane, ethane and butane.
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What teacher still gives exercises in such mixed units? Degrees Fahrenheit and Kelvins. Distance in feet....but the results is required in grams?\n\nThe volume of air is 16000 ft^3 or 453 m^3. The density of air is 1.293 gram/liter (I looked that up) or 1293 gram/m^3. We have therefore 5.86 x 10^5 grams of air to heat.\nThe temperature change is 30 deg F or 16.7 deg K.\nSo the total heat needed is 1.01 x 16.7 x 5.86 x 10^5 -9.88 x 10^7 Joule. \nThe combustion of methane provides 890.8 x 10^3 J/mol, which means that we need 111 mol of methane or 111 x (12+4x1) = 1775 gram of CH4\n\nThe reaction equation is CH4 + 2 O2 -> CO2 + 2 H2O.\nSo one mol of methane will produce one mol of CO2. A total of 111 mol are produced, or 111 x (12+2x16) = 4884 gram of CO2.\n\n(And all that stuff goes in the atmosphere warming up the greenhouse a little more.)
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Straight chain hydrocarbons (e.g., methane, butane, propane, gasoline, deisel) run about 20,000 BTU/lb.  High explosives typically are 2,000 BTU/lb because they must include the heavy oxidizer.  Most other combustibles (wood, paper, methanol, ethanol) are in the range of 5000-10000 BTU/lb.\n\nIt is interesting that high explosives have less energy than firewood.  For explosives, its not the energy, its how fast it is relaesed.
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One BTU. Or British thermal unit
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Standard enthalpy of formation 227 KJ/mol\nStandard enthalpy of combustion -1300 KJ/mol\nStandard entropy 201 J/K/mol
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The specific heat of liquefied methane is 3.50 J/g/K. The specific heat capacity of LNG is likely to be close to this as LNG is about 90% methane. If you need a more precise value, you will need to find out how much methane, ethane etc is in the LNG and look up the appropriate specific heat values and calculate a weighted average. See page 2 of the URL below.