How can I heat the water within a 150 gal tank that's open to the atmosphere in Feb.?
151759
maybe. in phoenix it would be easy. very difficult and expensive in fairbanks
646959
2H2 + O2 ---- 2 H2O\n\nYou can easily obtain water around the house.
1154531
Just make sure you are swimming slightly b/c if you are not moving you will get a heat shield ora thing around you. Similar to that fume thing when you put gas in your car.
819964
On the vernal equinox, the sun is directly overhead the Earth's equator at local noon, and we are told that the insolation under those conditions is 1368 W/m^2.\n\nAt latitudes other than the equator, the solar energy per unit area is decreased by a factor of cos(L), where L is the latitude.  This is called the "projection effect".  At 40 degrees latitude, the incoming solar energy per unit area is therefore:\n\n1368 W/m^2 * cos(40 deg) = 1368 W/m^2 * 0.766 = 1048 W/m^2.\n\nI'm assuming the question is asking how long it would take to bring the water to a boil, not how long it would take to completely boil away the water.\n\nIn either case, you are missing a key bit of information necessary to answer the question:  the initial temperature of the water.  Let's assume that temperature is T.  I'll also assume that you are located as sea level, so the boiling point of the water is 100 C.\n\nThe specific heat capacity of water is 4.1855 J/(g*degC), and the density of water is ~1.00 gm/cm^3, so to bring one liter of water from temperature T to 100 C requires:\n\nE = 4.1855 J/(g*degC) * 1000 cm^3 * 1 gm/cm^3 * (100 degC - T) \nE = 4.1855*10^3 J/degC  * (100 degC - T)\n\nWe have a collector with a 1 m^2 surface area, so it supplies 1048 W = 1048 J/sec at 40 degrees latitude, all of which we are to assume goes into heating the water.\n\nWith these assumptions, it would take:\n\nt = [4.1855*10^3 J/degC  * (100 degC - T)]/(1048 J/sec)\n\nseconds to bring the water to a boil.\n\nFor instance, if T = 25 C, t = 299.54 sec ~= 5 minutes to bring the water to a boil.
232210
heat coal in an oxygen free atmosphere. without oxygen it will not burn but bond of the carbon and hydrogen atoms will be broken.\n\nCheck out an aquarium store.
848677
I am a HVAC trainer in the USA, and as such I have to explain how to calculate what amounts of energy it takes to do a certain heating jobs.  Looking at your question, I am of the impression that you are asking how much energy it takes to bring the temperature of 100 gallons of water from 74deg F to 90deg F.  I will explain  how to do this calculation, plus ask and answer a question that I feel might be more appropriate to an application. \n\nMost HVAC technicians define a BTU as the amount of heat energy it takes to raise one pound of water one degree Fahrenheit.  Tables will state that there are approximately 8.34 pounds of water in one gallon of water.  Taking this information, we can calculate that there are 834 pounds of water in 100 gallons of water (100 * 8.34 = 834).   So, if we raised 100 gallons of water 1deg F, we will have put 834 BTU’s into that water.  The temperature difference you talk about is equal to the difference between 74deg and 90deg which is 16deg F (90 – 74 = 16).  If one degree requires 834 BTU’s, then 16 degrees will require 16 times that amount, or 13,344 BTU’s (834 * 16 = 13,344).\n\nNow for the real issue, if you have a flow of water entering the slab at 90deg F, and it comes out of the slab at 76deg F, and if the flow rate is …… lets say 8 gallons per minute, than how much energy is the water putting into the slab?  To answer this, we use the same concept as above except we figure the flow rate in pounds per hour and multiply that times the temperature difference.  The process goes like this: (8 gallons per minute) * (8.34 pounds per gallon) * (60 minutes per hour) = 4003.2 pounds per hour.  Now multiply the pounds per hour flow rate times the temperature difference, and we get the energy going into the slab (4003.2 * 16 = 64051.2 BTU’s per hour).  So, the heat pump would have to have a capacity of a little over 5 tons (1 ton is 12,000 BTU per hour) to meet that load. \n\nNow, the energy one puts into a slab should be based on the heat loss of the structure.  It might be that you only have to put 18,000 BTU’s into the slab to offset the heat loss.  The flow rate and temperature difference can be adjusted based on the heat loss.
94072
It depends on how hot you want your water. If you want only moderately heated water, suitable for bathing, you can use a box covered with glass that has a long winding pipe inside it. The pipe should be a small diameter and laid out flat on the bottom of the box with as many turns in it as possible to increas the length. This can go into a reservoir, like a hot water tank, and to make it more sophisticated you can put a circulating pump on the tank that constantly circulates the water through the heating panel. This is how commercial solar water heaters are made, with some sophisticated controls added. \n\nIf you want to boil your water using solar power this can be done with a parabolic mirror that focuses sunlight onto your cooking pot. These can be made from small pieces of mirror laid in a parabolic form (shaped like a satelitte dish) and cemented with plain grout. I have seen these used in some areas and they are quite effective. \n\nOne simple method that I use when camping: Lay a black colored bag of water on the ground or hang it in the sun and wait till it is warm enough for a shower.
776170
1) Raise the temperature from -20°C to 0°C:\nMass of ice * specific heat of ice * temperature rise\n10.0g * 2.1 J/g K * 20° = 420 J\n2) Melt the ice:\nMass of ice * Heat of Fusion\n10g * 333J/g = 3330 J\n3) Raise the temperature from 0°C to 100°C:\nMass of water * specific heat of water * temperature rise\n10.0g * 4.2 J/g K * 100° = 4200 J\n4) Evaporate the water:\nMass of water * Heat of Vaporization\n10.0g * 2260J/g = 22600 J\n5) Raise the temperature from 100°C to 120°C:\nMass of steam * specific heat of steam * temperature rise\n10.0g * 2.0 J/g K * 20° = 400 J\n\n6) Addition: 420 + 3330 + 4200 + 22600 + 400 = 30950 J
1138549
you need to try a science website , i know the temp has to be very high. just put your question into a web search.
693165
You can make a great solar water heater with a small 12 volt water pump which you can purchase from any RV dealer. Put the intake into the pool and connect the out side to a long length of garden hose. Lay the hose in the sun or on the roof and back into the pool. The longer the hose, the warmer the water will be.
1040977
It is done with steam. Just simple steam that you might get from a tea kettle. It doesn't have to be under pressure or any hotter than 212 deg F.
193114
watertemps of 60 and better
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
700946
Interesting question.  I would guess not for long, for the following reason.\n\nThe vapor pressure of liquid water at 32 F or 0 C is 0.08854 psia.  In theory, reducing the pressure above the liquid below this value will cause the water to boil.  However, the escaping steam has a much higher enthalpy than the liquid water.  It draws this heat from the liquid water surrounding the steam bubble as it is formed, thereby cooling the liquid in the process.  Since the liquid is already at the freezing point, the liquid will have to go thru a phase change to ice to supply the heat.  The heat of fusion is much less than the heat of vaporization, so before long you would have a block of ice.\n\nIf you heat the liquid container to match the heat lost in steam, you could keep a boil going, but could not use temperature as a guide since the freeze-up is isothermal.
89807
1. you need firstly to have a total density equivalent to that of fresh or salt water(for whatever purposed required)\nthen bouyancy will be stable/maintained \nthis will make a suitable start\n2. forge a shape suitable to make the sub hydrodynamic, this will enable the prototype to glige through the water \n3. have fun!
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.
967084
I would use a length of nichrome wire (see links below) stretched between a couple copper blocks. You could make some sort of pinch block out of the copper and connect a power supply to the copper. (or drill a hole through it with a set screw, tap a hole and us a pan head screw)  You will have to insulate the copper from the rest of the world so you don't short something out.  G-10 (a glass/epoxy resin) makes a good moderately high temperature insulator.  One of the contacts should be fixed and the other allowed to "float" and allow for changes in the length of the wire due to heating.  A spring could be used to provide tension or an air cylinder.  The power supply will depend on the length and diameter wire you choose.  If there is a problem with the electrical contacts heating up, you could run some cooling water through them.  Just drill a hole through it and 1/8 NPT pipe tape both ends.  You can plumb some water to it and it will keep the copper chilled.