i want to know about water freezing-melting curve.?
482741
You need a few different constants to figure this out.\n\nWater has a Specific Heat of 1 kcal/kg/°C\nThat means it needs 1 kcal per kilogram to heat the water 1 degree celsius\n\nIt has a density of 1000 kg/m3\n\nIt has a Latent heat of Fusion of 80 kcal/kg (melting) and 65 J/kg (freezing)\n\nSo if you know the volume you can work out the weight of the water. Once you have the weight (or you already know it), you can work out the rate of the drop (or increase) of water temperature.\n\nOnce you get to the point of freezing, or melting, simply add on the latent heat, and you can then get the total amount of energy required to change the water from one temperature to the next.\n\nSo for example;\n\n1) How much energy to heat 2 kg water from 18°C to 100°C is =>\n\n2kg * 1 * (100-18) = 164 kcal (The heat up curve is fairly straight)\n------------------------------->\n\n2) 2kg of water from 5°C to -10°C is as follows;\n\n(2 x 1 * (5 - -10)) + (65 * 2) = 160 kcal\n\nCurve drops down fairly constantly, then flats out until the water turns to ice, then continues to drop until you are at your temperature.\n\nI hope that helps a little.\n\nIf you want to figure out the times for all this to happen, you need to either know what heat source you are using (for example the wattage of the kettle element), and also you need to figure out the heat losses from the container (e.g. kettle or pipe, etc.)\n\nThese can all then be converted to an equivalent kcal number.
364087
sea water = S\ndistilled water = D\nfreezin point = fp\nboiling point = bp\n\nS fp < D fp ~ 0 deg C < 100 deg C ~ D bp < S bp\n\nRemember freezing point depression and boiling point elevation.  If you have problems remember which drops just remember that they salt the roads in winter to help the roads not freeze.
338330
The answer to this is made tougher by the fact that the water will be freezing.  I mean how can you answer a question where the water is ice cold and use the data provided at STP?  Think you need to either re-read your question or provide a little more info.
963637
Gold's freezing point is 1064.58 degrees celsius.\n(This is the melting point and melting points are the same as freezing points.  Water, for example, melts at 0 degrees celsius but also freezes at 0 degrees celsius.)
869302
1) Calculate the heat needed to melt 100 grams of ice completely: \nMass of ice * Heat of Fusion\n100g * 333J/g = 33300 J\n\n2) Calculate the cooling we can obtain \nMass of water * specific heat of water * temperature loss\n180g * 4.2 J/g K * dT = 33300 J => dT = 44.0 degrees\n\nConclusion we need only to melt 100g * 30/44 = 68.1 gram of ice to cool down the water to zero degree. 31.9 grams of ice remains.\n\n3) 1) Calculate the heat needed to melt 50 grams of ice completely: \nMass of ice * Heat of Fusion\n50g * 333J/g = 16650 J\n\n4) Calculate the cooling we can obtain \nMass of water * specific heat of water * temperature loss\n180g * 4.2 J/g K * dT = 16650 J => dT = 22.0 degrees\n\n5) Calculate the average temperature of the water.\nAfter melting of the ice we have 50 g of water at 0C and 180 g of water at 30-22=8C. This average out to 6.3C
741460
Two things are unusual about water.\n\n1. Frozen water (ice) is less dense than liquid water, as a few answerers have already said.  For every other liquid, the frozen form is more dense.\n\n2. As it cools, water gets steadily denser until about 4 degrees C, then it gets slightly less dense as it cools to its freezing point at zero degrees C.  Every other liquid just gets steadily denser and denser until it freezes.\n\nBoth of these unusual things are quite important for life on earth, but the second one is much more important than the first one.
587656
So anti-freeze and salt causes a freezing point depression in the water which means as you add solvent such as salt the freezing point is lowered (hence when you add salt to your windshield the ice melts).  There is a simple mathematic equation that will relate the change of Temp freezing to the amout of solvent added.  (ΔTf = Kfcm).  Kf is the freezing point constant of the solvent and cm is the molal concentration of the solvent. So as you can see from the equation the ΔTf will be greater as you increase cm so the effects of adding a solvent will depend on how much solvent you add.  \n\nSo a simple experiment would be to add increasing amounts of salt and antifreeze to constant volumes of water and determine the freezing temperatures of each one.  You should also want to calculate the molal concentration of each one and try to determine before hand what the freezing points should be (hypothesis).
1083390
The water around an iceberg or floe becomes colder than freezing (like putting ice cubes in a drink), but won't turn to ice since it is salty.  And waters in the polar regions are down near freezing to begin with.  When the berg gets sunlight some of the surface will melt off.  When the berg finally drifts to warmer waters, it will melt until it is gone.\n\nFor an example: Take a sink and fill it with water.  Put in some rock salts and stir until dissolved.  Now put in several trays' worth of ice cubes.  Stir.  Some of the cubes will melt since the tap water was not freezing, but eventually the sink water will get to a temperature below 32 degrees F or 0 degrees C without freezing solid.  At that point the coldness of the water keeps the remaining ice cubes frozen and the ice cubes keep the water at a below freezing temp.  Eventually it will melt cause there will be heat absorption into the system raising the water temperature, but it will take awhile.
530330
Ice melting\nWater freezing\nWater boiling\nAmmonium chloride subliming
819865
Conservation of energy.\n\nCalculate the amount of heat stored in the 28ºC water using 0ºC as the starting point.  This energy must first melt the ice, for which you'll need to use the heat of fusion of water.  The remainder of the energy will warm up the entire amount of water.
1019201
First you have to know the pressure and the exact salinity to answer this question.\nBut generally :\nAt a salinity of 35PPT and a pressure of 1.5 PSI\nthe freezing point would be -1.930 degrees C or 28.5 F
174433
273K = 0 degrees celsius therefore water freezes\n\nre jackster: freezing point and melting point are the same\nre kay: absolute zero is 0K at which all particles stop moving .. 273K is not absolute zero
162986
We typed "how does salt melt ice" into the Yahoo! Search box and were soon traveling the well-worn path to How Stuff Works. \nThere we learned that the answer has to do with melting points. Most people know that water freezes at a temperature of 32°Fahrenheit (0°Celsius). Salt dissolves fairly rapidly in ice, lowering its freezing point in an effect known as freezing point depression, and melting it in the process. \n\nBut what's going on at the molecular level? Basically, salt molecules are bonding with water molecules, making it more difficult for ice to form. The result is a solution with a lower freezing temperature. For example, a 10% salt solution freezes at 20°F (-6°C), while a 20% salt solution freezes at 2°F (-16°C). \n\nYou've also probably noticed the effect of salt water on freezing points if you've tried to make ice cream or if you've wondered why the ocean doesn't freeze at the same time as bodies of fresh water, such as rivers, lakes, and ponds.
271556
In addition to the fact that at 4 degrees celcius, water has an absolute density of EXACTLY 1 g/cc, at 0 degrees, it is in equilibrium with its solid phase (ice) which has a density of LESS than 1 g/cc due to its open cage like crystal structure. So, its exact density at 0 degrees is impossible to standardise.
5352
This is a simple and complex question all in one.  I was a chemical engineer before I became a registered nurse and believe me, this could have a very detailed and complex answer to it.  However, I will simply say that it is due to the other compounds in tap water that affect the melting of ice.  Water and ice have been studied extensively and just about any website dealing with chemistry would be able to help you with this question.  If you want something very detailed, let me know.
845517
One way is to use the Clausius-Clayperon Equation.  See the link below and use the boiling point of water for one of the pressure and temperature pairs.  \n\nP.S.  It's equation 22 in the linked document.
874149
Yes.  Adding salt to frozen water lowers the melting point.  \n\nConsider that snow and ice are really the same thing, in different mass quantities - water in solid state due to ambient temperature.\n\nBest to you.