What is the ratio of 1tsp of water to 2 gallons steam, how would I figure this out?
115995
You <can> measure steam. It is merely water in a gaseous state; gases occupy volume just like liquids.\n\nI'm not totally sure what you're asking though. Do you mean that during the boiling process, 1tsp (0.005L) of liquid water generates 2 gallons (7.5708L) of water vapour? \n\nThat's a pretty massive expansion, but if it is the case, the ratio of liquid:steam would be\n0.005:7.5708\nor\n0.001:1.51416\nor \n1:1514.16
580754
i haven't done this in a while.\n\nThe volume of the reservoir is equal to (1/2)*(height)*("base")*length.  the reservoir is a rectangular prism with the "base" being the distance across the surface of the water.  We are given the shape of the full volume so we can scale down to the level of the water to find what the volume in the reservoir at its current level.\n\nif you think about the cross-section of the volume, and use half of that forming a right triangle, we can determine the ratio of the surface (the "base") to the height (water depth).  In this case, it is (40/2)/(20*sqrt(3)) or 1/sqrt(3).  so ("base"/2)/(height) = 1/sqrt(3)  and then "base" = (2*height)/sqrt(3).\n\nsubstitute that into the volume equation so that\nvolume = [(height^2)*length]/sqrt(3) or literally,\nV = [(h^2)*L]/sqrt(3).  differentiate with respect to time,\n(dV/dt) = [2*h*(dh/dt)*L]/sqrt(3) and substitute the numbers in:\ndV/dt = 200 - 50 = 150\nh = 10*sqrt(3)\nL = 450\nand solve for dh/dt which is the rate of change of the water level in yds/hour.  these are odd units, you might want to change to inches/hour.\n\ni did it this way and got 1/60 yds/hr or 0.6 inches/hr.\n\nbut again it has been a long time....
336893
You've used 258.3kJ to boil the water.\n\nIf you use 2400 kJ/kg to turn water to steam, and you waste 50g of water then;\n\n=> 2400 / 1000 * 50 = 120 kJ\n\nTotal Energy used = 258.3 + 120 = 378.3 kJ
920577
actualy it depends on the pressure because the formula is:\nQ=m*Cp*Delta(T)\nAnd Cp for water is dependant on Temprature you are giving heat to the water. you can find the analytical method for finding Cp in Thermodynamics  van wylan ed 6
361907
If the tank is a cylinder, the Volume is πr²h.  If you get the volume in cubic feet, multiply by 7.5 to get gallons.
149069
since water seeks its own level, just connect a tube at the bottom of two 5 empty gal jugs so that water can flow in one another then plug the tube opening with cotton . fill one with 5gal of water and the other with 3gal of water.Now remove the plugs GRAVITY will do the rest dividing the two volume of water into   two eqaul 4gal jug of water . No water is wasted.
1004492
The optimum ratio of H2 to O2 is given by the formula for water, which is the only product of combustion.  Two molecules of water contain 2 molecules of Hyrdrogen and 1 of Oxygen, so the ratio is 2 to 1 (by volume).\nIf you change the ratio, you get a little less energy because one of the reactants isn't being completely used up and you'll have either an excess of H2 or O2 depending on whether the mix was rich or lean to start.
780627
WATER BOIL AT:  27*C at 26.739 (pressure(torr)  thru 30*C at 31.824\n pressure(torr)
354060
0.2 gal. or 25.6 fluid oz.
798267
This is a Hess' Law problem.\n\nThe strategy I like to teach for these is to get the biggest thing you need first then get rid of the biggest thing you don't need next.  So we want to get the 2 N2O5 first then get rid of the resulting 4 HNO3.\n\nUse -2 times the first eq to get your 2 N2O5, -4 times the third to get rid of your 4 HNO3, finally -2 times the second to get rid of the water:\n\n2 N2O5 + 2H2O > 4 HNO3   -152.4 kJ\n4 HNO3              >  2 N2 + 6 O2 + 2 H2  +696.0 kJ\nO2 + 2H2           >   2 H2O  -572 kJ\n\nEverything cancels properly to get the desired equation, and delta H is -28.4 kJ.
141221
The original water temp is 25C, as was given in the problem... I'm assuming you meant to ask what was the initial temp of the water that was added, which isn't too hard. Here goes.\nQ(heat)= m(mass) x c(specific heat) x delta T(change in temp)\nso Q = (150g)(4.something J/gC-look it up)(25C-14C)\nBecause of conservation of energy and assuming ideal insulation, the heat LOST by the 150g of water during the mixing is equal to the heat GAINED by the 200g of water. So, once you have Q, rearrange the equation to look like [delta T] = Q/mc = Q/(200g)(c)\nThat will give you the temperature difference between 14C and the 200g of water's initial temp. Delta T should come out as being positive, since the temp is going up. \nWhen you have more complex mixtures of things with different specific heats and add processes and changes, the math becomes a lot more of a headache but the same basic principles apply just the same.
772330
You just need a few basics to estimate things.\n\n1.  6000 watts is 6 BTU/sec.\n2. 1 BTU heats 1 lb of water 1 degree F\n3. 1 gallon of water weighs 8.3 lbs.\n4. A nominal natural convection heat transfer coefficient is 1 BTU/(ft^2 * hr * deg F)\n\nIf you barrel is well insulated, just use the above (except 4) and you will see that 25 gallons will heat at 0.027 deg F per second.  \n\nIf you arent insulated, you need to use the heat transfer to see what portion of the 6000 watts is lost as the temperature rises.  You can readily do this in general form and the result should be an exponentially asymptotic temperature.
984937
The equation is: 2H2 + O2 ---> 2H2O\nSo you would form 2L of water VAPOR\nIf you want this in terms of LIQUID water, remember that one mol of a gas occupies 22.4L, so you have 2/22.4 mols of water (MW=18) formed, so...\n2/22.4 x 18 = 1.6 grams or 0.0016L of LIQUID water.\nThis assumes that all three gasses act as 'ideal gasses' (which is only close) and conditions are reduced to STP...so answer is not exactly correct, but 'close enough for government work'  :-))
1165181
It's been awhile since i've done this but i'll give it my best\n\nFirst off, the first energy you need to calculate is the energy needed to raise the temperature of ice to 0 degrees C. so...\n\n(12.37 g)(.48 cal/deg/g)(35 deg) = +207.816 cal\n\nWhen the ice changes to water you have a positive heat of fusion. So...\n\n(12.37 g)(80 cal/g) = 989.6 cal\n\nNext you have to raise the water temperature from 0 degrees C to 100 degrees C. So...\n\n(12.37 g)(1.0 cal/deg/g)(100 deg) = 1237 cal\n\nNext you have the positive heat of vaporization.\n\n(12.37 g)(540 cal/g) = 6679.9 cal\n\nNext once it is steam, you must raise the temperature from 100 degrees C to 146 degrees C. So...\n\n(12.37g)(.48 cal/deg/g)(46 deg) = 273.1296 cal.\n\nNow that you have the energy for each step add them together.\n\n207.816 + 989.6 + 1237 + 6679.9 + 273.1296 = 9387.45 cal\n\nI'm pretty confident in my breakdown but it defintely wouldn't hurt to have someone else check it
1003334
Assuming D is diameter, calculate the volume of the tank, subtract the volume of the cylinder and convert the remaing volume to gallons.  There are 128 cubic inches (0.0741 cubic feet) per gallon.\n\nyou do the math, that is obviously the intent of your assignment
1107016
To find amount of water produced, first , we need to write a balanced equation for the reaction:  \n\n2H2 + O2 = 2H2O\n\nTo do any Stoechiometry calculations we need to change the grams to moles.\n\n1 g H2 * 1 mol/2.016 g = .496 moles of H2\n10 g O2 * 1 mol /32 g =.3125 moles \n\nThe ratio between H2 and H2O in the balance equation is 1:1\nOn the other hand, the ratio between O2 and H2O in the balanced equation is 1:2.  \nThat means we can make .496 moles of H2O from the given H2 and 03125*2 = .625 moles of H2O from the given O2.  In this case H2 is the limiting reactant since H2 will run out first. That means we can only make .496 moles of H2O from the given H2 and O2. \nNow we need to change the moles of H2O to grams.  \n.496 moles * 18.016 g/1 mol H2O = 8.936 grams.
804606
For cubic measure:  L x W x H = CI / 231 = G\nL = Length\nW = Width\nH = Height\nCI = Cubic Inches (Volume)\nG = Gallons\nMeasure the Length, Width, and Height in Inches.\n231 Cubic Inches to one Gallon.\nConvert Volume to Cubic Inches\nLength x Width x Height  = Cubic Inches  / 231 = Gallons\n\nFor Right cylinder measure: Pi x r'2 x H = CI / 231 = G\nPi = 3.14159\nr = radius (half of diameter) '2 = squared\nH = Height (or length)\nCI = Cubic Inches (Volume)\nConvert Volume to Cubic Inches. \nPi x r squared x H = CI / 231 = gallons