how much water in 1000cu.ft.?
494256
7,480.51945 US gallons\n28,316.8466 liters
214638
I am assume stand temperature and pressures at the surface of the earth.\n\nThe volume of the liquid is: \n1 inch x 1 acre \n= 1 in. x 43560 sq. ft.\n= 1 in. x 208.71 ft x 208.71 ft (assume square acre)\n= 1 in. x 2504.52 in x 2504.52 in\n= 6272620.4 cubic inches \n= 27154.2 Gallons (assumung US Gallon)
666448
1 m of H2O = 0.0968 Atm, 1 inch of H2O = 0.00246 Atm\n\n1m of Hg = 1.3158 Atm, 1 inch of Hg = 0.0334 Atm\n\n1 BTU = 251.9958 Calories
616459
Conversion:\n_________________________\n\n1 cu meter = 35.3 cu ft\n_________________________\n\nYour storage room 10 x 10 x 6 = 600 cu ft.\n\n5 cu meter = 5 * 35.3 = 176.5 cu ft.\n\nYour storage room will not fit into the 5 cu meters.\n\nYou are short by some 423.5 cu ft
537035
Formula:  J = K*m*(T2-T1)\n\nSince both K & m are 1 for your problem, J = 125 Joules\n\nK is the specific heat defined as 1 for water.  Varies with other materials.  Also this formula is not strictly true for 1 liter unless you assume the 1 liter weighs 1 kg.  Liters are volume and the formula is for mass.
386344
1067 gallons/minute, approx.\n\nSee reference
657896
32.99 gallons\n39.6 US gallons
951038
Miss Science, you have all the facts but you missed out the most important one, which is that the percentage by mass of water in CuSO4.5H2O is always the same.\n\nthe Mr of CuSO4 is 159.6 and that of 5 water molecules is 90.0. no matter what mass of hydrated copper(ii)sulphate, the masses of CuSO4 and H2O will always be in the ratio of 159.6:90.0\n\nthis means that the percentage mass of water would be 90/(90+159.6)x100% = 36.06%
95239
compressing 2 liters of water into a 1 liter volume\n\nLiquids are not compressible.
1029564
1000 mls = 1 liter\n3.879 liter = 1 gal\nYou are suposed to drink 2 liters of water daily, about 8 normal glasses.\nI know that 1000 ml are 1 lt because I'm Mexican, and in Mexico we use mls instead of oz; you buy 600 ml of soda.\nI know tha 1 gal = 3.879 liter because of the same reason, we also have galons.\nI've heard from doctors that it is safe and good for health to drink around 2 lts of water daily, besides juice, coffee, soda, etc. But no more, well, maybe 2.1 lts is ok, but no 5 lts please
110015
At least 3000 psi.  Recommend at least 180 degree water also.
783798
1 mile = 5,280 ft.\n100 miles = 528,000 ft.\n\nYou'll have a total of:\n528000 x 528000 x 528000 cu. ft. of water.\nAnd a cu. ft. holds 7.48 gallons.\n\nMultiplying it all together it's about 1.1 quintillion gallons!
562654
Okay, I'll try this one -\n\nRun this experiment twice (or more times), each time with a piece of metal having different density.  The metal piece should be of adequate size (a sewing needle isn't going to do it), it needs to be big enough to significantly affect the temperature of the water.  You will need to know the mass of the metal (i.e., weigh it) and have a thermometer capable of reading from room temperature up through boiling or so.  Please note this is not going to give you a highly accurate answer, but can give you a ballpark idea (I'm guessing you're probably not doing this in a 500 million dollar lab).\n\nThe procedure:\n\n1)  Measure and record room temperature as T1.\n\n2)  Weigh the metal, call it M1.\n\n3)  Bring a known quantity of water (weigh it, or measure it and convert to weight, call it M2) to a temperature just short of boiling (say 180F to 200F, use the same temperature for each run of this experiment).  Record the temperature as T2, and remove the heat source (take it off the burner, etc.).\n\n4)  Drop in a metal piece at room temperature.  Allow a couple of minutes for heat to exchange.  Make sure you use \nthe same amount of time for each run of this experiment.\n\n5)  Measure and record the new temperature of the water as T3.\n\n6)  Repeat with your other metal samples.\n\nYou can now estimate the specific heat capacity for each metal.  It will not be super-accurate (heat escapes your system, and temperatures won't have completely equalized), but should give you some idea.  Just assume all temperatures did equalize and that no heat escaped for calculational purposes.\n\nSince (you are assuming) the heat which left the water went into the metal, then the product of mass, specific heat capacity, and temperature change should be the same for both the metal and the water (but in opposite directions).\n\nThis means that:\n\n(M1)(c1)(T3-T1) = (M2)(cw)(T2-T3)\n\nwhere c1 is the specific heat capacity of the metal, and cw is the specific heat capacity of water (look it up on Google to get the units you need).\n\nSolving this, you get:\n\n......(M2)(cw)(T2-T3)\nc1 = -----------------\n........(M1)(T3-T1)\n\n(sorry about the dots, I'm trying to keep everything aligned)\n\nRepeat the calculations for each of the metals.\n\nYou will need to calculate (or look up) the densities of each of the metals.  Then, graph density vs. the specific heat capacities you calculated, and see if your hypothesis is true.\n\nHave fun!
359943
Would need to know the orifice diameter that is flowing air from this 150 psi source.........and if you are asking cu ft per minute.......\n\nThe rate of flow won't change whether there was 6 or 60 cylinders, but more overall delivered cu ft when the cascade was depleted.
941617
The Specific Gravity of water is 1
635522
It depends on the liquid AND the pressure. 100C is for water at atmospheric pressure.
447953
Copper's atomic mass is 63.55 grams/mole.  There are 6.02*10^23 copper atoms in mole of copper.  Thus, 1 atom Cu * 1 mol Cu/6.02*10^23 atoms Cu * 63.55 gCu/1 mol Cu = 1.056 * 10^-22 grams/atom Cu