how many BTUs would be required to raise 45lbs of water 18F?
93911
From ?. You have to know the starting temperature.\n\n By definition, 1 BTU will raise the temperature of 1 pound of water 1F degree.\n\nSo subtract the starting temperature from 18 and then multiply by 45.
132159
Open up your thermodynamics book and do your own homework.\nHint.. add the energy required to raise the temp from 100C to 112C plus the energy used to change state, plus the energy used to go from 76C to 100C.  You should have all the info you need in your problem, and the energy required to do the change is in your book.
748326
The fuel should increase to take the load. My experience is on turbine generators, one was 9 mega. WATS.
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.
182655
water 8000 pounds \nair not even 1\nhelium (-1)\nlead - really heavy\n\nhope that helps - im guessing you ment water - salt water weights a bit more
904003
Temperature rise of water = 2 deg C = 275 K\nq=mc(change in temperature)\n  = (100)(4.18)(275)\n  =114950J\n  = 114.95 kJ\nHmm. Either I'm wrong or the question is incomplete. Did the question perhaps give more information on the reaction?
57870
You would want at least 50,000 BTU, Hope this will answer your question. I had an 1800 sq. ft. home once so I know. It costed me $5,000.00 installed with A/C and ducts.
753895
PV=nRT\nn=PV/RT\n\nR=10.7316 ft³ · psi · °R-1 · lbmol-1\nP=14.7 psi\nT=530 R, 540 R\nV=100ft^3\n@70F=530R\nn=14.7*100/(10.7316*530)=.2585\nmol wt=28.8\nm=28.8*.2584=7.44 lb\n@ 80f=540R\nn=14.7*100/(10.7316*540)=.2537\nm=.2537*28.8=7.31 lb
941617
The Specific Gravity of water is 1
865811
30.5 m of water.\n\nConvert units to desired units.\n\n1 atm is 14.?? psi is about 32 ft of water.\n\nLook in reference for more detailed unit conversions.\n\nOr use P= &#92;rho g H.
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.
98169
at least half a gallon.
143234
I think you at least need to know the density of pure acetic acid to solve your problem. We can assume density of water at 1.0 g/mL.\nIf we assume that the density of your 40% w/v solution is 1.0 g/mL then:\n\n    V = volume of 40% solution required would be: \n\n        V x .40  = 0.0225 x 250 \n\n  V = 14.1 mL \n\nYou place the 14.1 mL in a flask and add water to complete 250 mL and voila... You are done. \n\nLet me know if this is wrong.
525961
Peat yields 8.000 Million Btu/Ton according to the Department of Energy figures.  Dividing by 2,000 gives us 8 thousand BTU per pound.\n\nHope this helps!
898274
Yes, you can do it, but it will require iterations.  \n\n1) Convert all units to standard engineering units (feet, pounds, seconds).\n\n2) Set up the mechanical energy equation:\n(p2 - p1)/gamma = z2 - z1 + (f*L*v^2)/(2*D*g)\nwhere p2 -p1 is the pressure across the pipe (20 psi, but convert to lb/ft^2), z2 - z1 is the rise in the pipe (if any), gamma is the specific weight of water (62.4 lb/ft^3), L is the length of the pipe (50 ft), D is the I.D. of the pipe (ft), and g is the acceleration of gravity.  \n\n3) Now you have a problem: v is the velocity of the water, and f is the friction factor, which depends on v.  So, you guess a value for v, and calculate f from it.  You look up f on a chart (it's called a "Moody chart").  But you'll need the Reynolds number and relative roughness of the pipe surface to do that.  Then you try another v, find f, etc.  Or, you can try this:\n\nhttp://www.fluidmech.net/jscalc/vmd03.htm\n\n4) Just keep plugging in different velocities until the head loss (right side of the equation) equals the left side.  Then convert to volumetric flow rate.\n\nHey, I get 14.2 ft/s, 0.0774 ft^3/s = 34.7 gpm
985910
You have 2 38.0 mL volumes, so the total volume is 76.0 mL.\nThe reaction is H2SO4 + 2 KOH --> 2 H2O + K2SO4\nThe rise in temperature (delta T) is 30.17°C - 23.50°C = 6.67°C\n\nThe heat capacity of water is 1.0 cal/(g*°C)\nThe density of water is 1.0 g/mL\n\nThe H of the reaction is 1.0 cal/(g*°C) * 1.0 g/mL * 76.0 mL * 6.67°C = 506.9 cal or (506.9 cal * 4.186 J/cal * 0.001 kJ/J) = 2.122 kJ\n\nThere were 0.82 mol/L of KOH, and 38.0 mL, so there were .82*.038 mol = 0.03116 mol of KOH\n\nDeltaH per mole of KOH is 2.122 kJ / 0.03116 mol = 68.1 kJ/mol
342773
Want to see something neat??\n\nType "19 gallons in cubic feet." into google.\n\nWorks for all kinds of conversions.