How to compute the heater power of a swimming pool?
963007
multiply the mass of the water by the specific heat and the (temperature required minus the ambient temperature).that will give you the energy in joules.if you know in how much time you want this temperature to be achieved you can calculate the power as power is nothing but the rate at which the energy is dispensed
1043560
easy if its circular: square the diameter (multiply the diameter by itself) then multiply the diameter  by 3.142, divide the answer by 4, then multiply this answer by the height of the pool.
929436
You have 8 rungs that are nine inches apart with the last being 9 inches from the bottom of the pool so I would draw it on a piece of paper I I I I I I I I I as nine rungs and count the spaces in between the rungs (lines). I get 8 spaces of nine inches. So then 8 spaces times nine inches is 72 inches. Now we need to convert inches to feet so we divide 72 by 12 inches per foot. That works out to 6 feet. if the pool fills at 1 foot per hour then it will take 6 hours to fill the pool to the top rung. There is an algebraic approach but that isn't what they'd be looking for in 5'th grade. Teach you son to draw a image and label it when working with word problems. Its a great habit and eventually he'll learn to visualize without the paper. Cheers:0
209317
The straight calculation is this:\nVolts X Amps = Watts(power)\n\n4.5 watts would be the answer.
1187482
Kilowatt\nCost depends on method of generation and location
812469
I'll give you some data and you can work out the details.  Total mass of the atmosphere, 5.136E18 kg.  Assumed average temperature of the atmosphere, 15 C. Enthalpy of vaporization of water at 15 C, 44.4 kilojoules per mole (by interpolation).  That should be enough data to do the trick -- except that I have no idea how much electricity Ottawa uses.  Call the power company and ask..\n\nPostscript: I tried to reply to your private note, but you have to do an address verification protocol with Yahoo before it will accept e-mails posted through your avatar.\n\nPost-postcript (11/21):  Yahoo is still claiming that your e-mail address is unconfirmed.  A pox on them!
697703
The heater part will work fine, the problem will be with the pump motor. If the motor is actually 220 volt it will run ok but 12.5% slower (less water pressure) A lot of things in the US that are called 220 volt will only be using 220 for the high power part. I.E. a 220 window air conditioner often has a 110 volt fan, it only uses one hot leg and neutral. If your hot tub has anything (pump motor, lights, air compressor) that uses 110 volts, that part won't work without being modified. There is no neutral in European 220. In some cases you will get something to work anyway but it is using earth leakage on other wiring and will not last for long, the voltage isn't steady. It will also get you killed in a worst case. You will not have any trouble getting/finding a good electrician there to make the changes needed and it won't cost nearly what it would here.
949133
OMG!\ni have the same question!\nmine is:\nA piece of an unknown metal has a mass of 20.0 g and at 100.0 C is placed in a styrofoam cup calorimeter containing 50.7 g of water at 22 C. If the final temp is 27.2 C, calculate the specific heat capacity of the metal. Assume that all the energy lost by the metal is gained by the water.\n\nok, haha fine, not the SAME question, but close enough. But, i'm uber confused. ><
945879
The first method would be to compare you utility bills to see if there is a drastic change (or any) between last year at this time and the most recent bill with the insulation on.  \n\nFor an exact calculation, you need the following:\n\nThe R value of the insulation\nLength of pipe insulated\nTemperature of the water inside the pipe\n\nThen plug them into this equation:\n\nQ=UA(Tinside - Toutside)\nwhere Q is the amout of Btus \nU is the thermal insulation (or lack there of) expressed as =1/R\nTinside = water temperature\nToutside = air temperature (average) where the pipe is installed\nA = the surface area of the pipe\n\nOnce you get that for the insulation, do it once with out the insulation.  For the second R value without the insulation, use a R-value of 0.5.  The two different Q's you get will be the amount of Btu's you saved.  \n\nTo convert that to $, you need to convert the Btu's (Q) to kW.  Divide the Btu's by 3.412 to get watts.  That will be the amount watts you saved at your hot water heater/boiler or whatever is heating the water (assuming it is electric).  You then look at your electric bill to determine the cost $/W or $/kWh.  Take the number you got for watts and convert to kilo Watts by dividing by 1000.  Then multiply it by the cost off of you utility bill for $/kWh.  \n\nThis is all for a one hour period.  You need to calculate that for a month, day, or year (720, 24, 8765 hours respectively).  You now have what your savings for this project was.
765230
EER = Q in (BTU/hr) divided by W in (watts)\n\n600/500= 1.2
83134
you really should have given numbers \n\nkinetic energy will be 1/2mv^2\n\nIts linear speed can be determined by distance over time  (d/t)\n\nthe next 3 answers need me to know what data you have in order calculate their values
1015451
Power = work done per unit of time.\n\nP=W/T
939720
It will vary upon the size of the pump and how much it runs.  The pump will give ratings for kwh (kilowatt hours) and the pump's capacity in either gallons or liters per hour. Simply divide your estimated sump by the pump's rate per gallon and then multiply by the pump's kw rate.  If you want to get tricky, you can then look at your electric bill and multiply the kwh you calculated by your average billing rate and you'll then know how much juice you use and how much it costs.
794988
Energy radiated per second is given by\nE = e*sigma*A*T^4   where \ne = emissivity; \nsigma = Stefan's constant = 5.67 * 10^-8 W/m^2/K^4\nA = surface area\nT = temperature in kelvin.\nSubstituting the various values,\nE = 249 * 10^-8 Watts.
587092
Insulate the rod with a good insulator. (Styrofoam for instance). Get a source of measured heat.(electric heater with known wattage).Get two temperature measuring instruments.(  cu-constantan thermocouples).Get a stop watch.Measure the cross sectional area of the rod by appropriate scales.Measure the length of the rod.\n\nUse the equation: Q=k*(T1-T2)/dt, where\nQ=Btu, T=temperature at one end of the rod, t=time\n\nbe careful with the units.\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nElectric
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.
87578
P = EI (Power equals voltage E times current I)\n\nMany electrical properties can be explained as analogous to water flow.  For example, a transformer "transforms" high voltage and low current into low voltage and high current, keeping the power constant (except for minor losses) at full load.\n\nThis would be equivalent to a high pressure water pump supplying low water flow to the primary winding and taking out high water flow at low pressure from the secondary winding to drive a low pressure water turbine.\n\nYou could represent the transformer core as a hollow square of wood or even thick cardboard.  On one side wrap say ten turns of thin clear plastic tube around the core to represent the primary (first) winding.  On the other side wrap say five turns of larger diameter plastic tube around the core to represent the secondary (second) winding.  Fill the primary tubing with water and red food coloring to represent "hot" high voltage and the secondary tubing with water and blue food coloring to represent "cooler" low voltage.\n\nThe explanation of course is that the high voltage primary winding will have more windings of thin tube (similar to more thinner copper wire windings) because the current will be less than in the secondary winding.\n\nThere will be less windings for the secondary coil using larger tubing because the current will be higher at low pressure.\n\nSupply the secondary tube with a shut off valve (or clip) to represent the fact that there will be no current unless the circuit is open.  In the formula, if current is zero power is also zero (on both sides of the transformer).\n\nYou can represent the pump and turbine with pictures (or drawings) of same but it would not be impossible to rig an actual pump and turbine, just not necessary to "see" what is happening.\n\nThe turbine can use up more and more power until the maximum power of the primary winding is reached but not more.  The magnet of the core will be "saturated at full power."\n\nThe rest is explaining magnetic flux in the core and how it is produced by the primary winding and cuts through the wires of the secondary to produce voltage and current if there is a load.  Each winding of a coil adds to the effect.  A transformer with equal primary and secondary windings would have the same voltage and current under load.  I believe that the phase would shift 180 degrees because of lead/lag.   Good luck.