How to calculate Solar Pond Evaporation Loss caluculation in an Open system ?
549685
Not sure but these should be of help\ncheck these sites\n\nhttp://scholar.google.com/url?sa=U&q=http://sdl.ae.ntu.edu.tw/~asway/paper/90a_cabs03.pdf\n\nhttp://ag.arizona.edu/azmet/et1.htm\n\nhttp://sti.srs.gov/fulltext/tr2002432/tr2002432.pdf\n\nhttp://www.google.ca/search?hl=en&q=calculation+of++Solar+Pond+Evaporation+Loss&meta=
307035
Using elodea (pond plant) in water (gives the elodea the water it needs for photosynthesis) with a small amount of sodium hydrogen carbonate added (gives the elodea the carbon dioxide it needs for photosynthesis) in a test tube, put it at different distances from a source of light. Then for a minute at each distance, count the amount of bubbles of oxygen being produced by the elodea in the test tube. This shows how the rate of photosynthesis is affected by the light intensity. Remember to use a perspex screen between the light source and test tube so that no warmth is passed from the light source to the elodea, affecting the results.\nI hope this makes some sense!!!\nBest wishes.
897593
If you are doing this in terms of the system,\n\nsince the system is losing energy (delivers heat) then the amount of energy is "-1275 J".  The surrounding add heat (perform work on) to the system, so the amount of energy is "+855".  The internal energy of the system is the sum of the energy loss and the work done to the system, so -1275 J + 855 J = -420 J.\n\nIf you are doing this problem in terms of the surroundings, the all the signs are reversed and you should end up with +420 J.
810276
Well, probably easier to explain empirically -- I know that one of the equations that you would be looking for is Planck's law of black-body radiation.  That, along with other equations involved in transport phenomena could be used to model the system and let you know how much heat you lose from the glass.\n\nOff the top of my head, I can't remember every equation that one would want to use.  Good luck!\n\nThe link is to Wiki's entry for Transport Phenomena -- the first reference there would be an excellent -- if very heavy on the math -- reference for solving this problem.
517654
Need to know the impedance of the solar panel to determine current.
751939
you can either do the math, look it up on a chart or go here:\n\nhttp://leebrewery.com/hydromet.htm\n\nand use the java-calculator.
801823
Quantity of energy lost by the water:\n\nQ = (m)(deltaT)(c)\n    = (50.0g)(2.00 C)(4.18 JK-1g-1)\n    = 418 J\n\nSince the water got colder, the reaction was endothermic, and the enthalpy change for the reaction is +418 J.
726835
A reference to a meterological term: see link\n\nhttp://en.wikipedia.org/wiki/Adiabatic_lapse_rate
631391
let T1 and V1 be velocity and time to lake\nand T2 and V2 be velocity and time to ocean\n \nT1 x V1 = 600\nT2 x V2 = 240\nalso\nT1 = 2T2\nV1= V2 + 6\nhence\n2T2 x V1 = 600\nsubstituting again V1 = V2 + 6\n2T2 x(V2+6) = 600\nT2 x V2     = 240\nor T2       = 240/V2\nsubstituing in equation:\n2 x (240/V2) ( V2 + 6) = 600\nx V2 both sides\n2x 240 (V2+6)= 600 V2\n480 V2 + (6x480) = 600 V2\nsubtract 480 V2 from both sides\n6x 480 = 120 V2\ndivide by 120 each side\n6 x 480/120 = V2\nV2 = 24\nV1 = 30 \nT2 = 10 hours\nt1 = 20 hours \nchecked in top equations
177044
use the fmla.... Q=msT\n\nwhere Q is the quantity of heat required, m is mass, s is specific heat capacity of water, T is the temperature difference. i dont hav a calculator with me now, u do the computations plz...
378935
adiabatic lapse rate?
929275
You would need the specific heat of the material.  The rate of cooling would be the total energy loss (E*A) divided by the specific heat.  This will be non-linear, and your value would be good for only the specific temperature given.\n\nWatts = Energy/time; one watt = 1 joule/sec.
558084
Start with the ideal gas law: PV=nRT.  Since no gas enters or leaves the system, n is constant, and R is the universal gas constant, the product nR has the same value at the beginning and as at the end.  So, collect it to one side, and move the other factors to the other side.\n\nP1 * V1 / T1 = nR = P2 * V2 / T2\n\nSolve for T2\n\nT2 = T1 * (P2 * V2) / (P1 * V1)\n= 300e4 K (.0003e15 atm * .046e-15 cc)/(50e3 atm * .0004e-7 cc)\n=2.07 * 10^7 K
836776
PV=nRT  n=PV/RT   V=nRT/P\n\nR= 0.08206 L atm mol^-1 k^-1\n\n922 torr/760 torr= 1.213 atm\n\nn= (1.213 atm x 35.3 L)/(0.08206 Latm/molK x 315K)\n\nn=1.6565 mol Propane\n\nSTP = 1 atm, 273K\n\nV = (1.6565 mol x 0.08206 Latm/molK x 273K)/(1 atm)\nV = 37.1095 L\nV = 37.1 L (sig figs)
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.
1059332
See in\nhttp://www.solarenergysociety.ca/\nhttp://www.cmhc-schl.gc.ca/en/inpr/su/neze/index.cfm
1119466
I donno ;=)\n\nMay be start with a heat balance equation like\n\nQfinal=Qwater-start  - Qice-start – Qlatent\n\nIn other words the final heat of the system is equal to the original amount of heat minus the heat required to bring ice to the melting point and minus again the amount of heat needed to melt the ice\n\nOr \nQfinal=(Specific-heat of water) (temp in K)(mass-final)\n\nmass-final= mass of water + mass of ice= m1 + m2\n\nNote -273.16 °C =0 degrees on Celsius scale = 273.16 °K (273 I hope is good enough for you?)\n\nQwater-start  =( 4190 J/(kg*K) (t-start K degrees )(m1)( 4190 J/(kg*K)=A(m1)  A= 4190 (76.2 + 273)\n\nQice-start= 2100 J/(kg*K)( - 23.6+273)(m2)= B(m2)\n\n\nQlatent= 334 kJ/kg(m2) =C(m2)\n\nWe have \n\nD(m1+m2)=A(m1)+B(m2)+C(m2)  you know that m1=.280kg  then you figure out m2? \nAnd \nA= 4190 (76.2 + 273)\nB=2100 (- 23.6+273)\nC=334,000\n\nActually Specific heat of ice = 2060 J/(kg*K) not 2100 J/(kg*K)\nAnd Specific heat of water  =4,183  J/(kg*K)