The most intelligent way to answer the question, "Why does ice melt in a warm room?"\n\nObviously you would say it is because the room is warmer than the ice, but think deeper than that. Think about diffusion of energy and equalization. I can answer this question with 7 words or 1000 words. How do you answer it?
128768
Through entropy of heat, the movement of the electrons in the warm surrounding air is transferred to the electrons in the solid/liquid/gas as the electrons accelerate in their orbit around the nucleus the state of the matter changes through the three states I have mentioned.  All of these parameters are leading to the eventual "heat death" of the universe.  Entropy as observed in all systems.
175314
I had written a detailed answer here, and Answers lost it for me, so I'll try once more and make it a bit shorter.\n\nThe snowflake phenomenon occurs because of the interaction of a microscopic unstable dynamic system (the formation of dendrites on a free-floating ice crystal) and a macroscopic one (turbulent motion in the atmosphere).\n\nThe microscopic system balances surface tension against heat diffusion; the combination of forces amplifies the tendency of water molecules to organize themselves in six-sided crystals to the point that it is visible at the snowflake scale (around a millimeter).\n\nIt takes about an hour for the typical snowflake to form. During this time, the branches of the flake go through various temperatures and zones of humidity. The variations of freezing caused by these differences don't vary enough across the scale of the forming snowflake to make the branches grow differently, so we get a hexagonally-symmetric crystal at the end, whether a flake, a needle, or a plate.\n\nEach snowflake is being blown around and up and down in the macro-scale turbulence of the atmosphere while the snow is falling; because no two snowflakes are at all likely to follow the same path through the exact same temperature and humidity gradients, the likelihood of duplicate snowflakes is pretty much zero.
51667
The temperature where the substance turns from a solid to a liquid. The melting temperature of ice is 31.9999 degrees Fahrenheit.\n\nThe melting point for gold would be much higher as it must be very hot for gold to turn to a liquid.
40041
>>>>>at 4 degrees above its freezing point, water sinks.\n\nThis is true.\n\n>>>>any lower, and it begins to freeze, forms crystals, and floats.\n\nThis is not true. Water begins to freeze at 0oC, not at 3.9oC or even 1oC.\n\n>>>>any warmer, and the molucular action makes it less >>>>dense.\n\nThis is not true. Water becomes less dense at warmer temperatures due to kinetic motion impeding the formation of hydrogen BETWEEN molecules.
8626
Water gives up far more (latent) heat when cooled than soil does.  Therefore the ground around the puddle may drop to well below freezing before the water in the puddle cools and freezes solid.  This could cause the puddle to freeze at the edges first then work inward toward the middle although a thin skim of ice may cover the entire puddle.  As the freezing water advances toward the center of the puddle it pushes any excess air before it and air can collect in the middle from all sides.  This can form a hollow cavity.  Incidently, ice can freeze in large grains and since a mixture of minerals and water freezes at a lower temperature than pure water, minerals are left at the boundaries of the grains.  Later when the ice melts, the boundaries melt first and can produce weak "rotten" ice that easily breaks through when walked on.  In short, an advancing wall of freezing water or a freezing crystal can exclude foreign matter.
193538
It depends on the temperature of the water and the air but if they are the same, then the ice in the water will melt faster because water is a better conductor for heat.
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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
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Well,  Hot & Cold are not only subjective (What's Hot for he is cold for me) and relative.  Better to think of temperature.\n\nTemperature is the measure of average kinetic energy.  The hotter, the more energy and vice versa.  \n\nAlso, as temperature is an average, some of the "hot" water may be cold.  In fact, there could be a several molecules that are frozen - we'd never know - but the average temperature is "hot."\n\nWhen you mix them heat is transfered from the high average kinetic energy water, "Hot," to the low average kinetic energy water, "Cold."  The "Cold" water absorbs the heat in the transfer and the results is as follows...\n\nThe "Hot" water is cooled, and the "Cold" water is warmed.  The results is 24 oz's of water that has an average kinetic energy, or temperature, higher than that of the "Cold" water, and lower than that of the "Hot."
20713
1) We need to raise the temperature of the ice from -16.73°C to 0°C, which is a temperature rise of 16.73°C.\n\nMass of ice * specific heat of ice * temperature rise\n38.80g * 2.1 J/g K * 16.73° = 1,292.89 J\n\n\n2) We now need to add heat until all the ice is melted:\n\nMass of ice * Heat of Fusion\n36.80g * 333J/g = 12,254.40 J\n\n\n3) Now that the ice is all melted, it's all water, but the temperature is still 0°C. We now have to raise the temperature of the water to 100°C, which is a temperature rise of 100°.\n\nMass of water * specific heat of water * temperature rise\n36.80g * 4.2 J/g K * 100° = 15,456.00 J\n\n\n4) Now we have to change all the water into steam.\n\nMass of water * Heat of Vaporization\n36.80g * 2260J/g = 83,168.00 J\n\n\n5) All the water is now steam, but the temperature of the steam is still 100°C. We now have to raise the temperature of the steam to 227.6°C, which is a temperature rise of 127.6°.\n\nMass of steam * specific heat of steam * temperature rise\n36.80g * 2.0 J/g K * 127.6° = 9,391.36 J\n\n\n\nNow add up all the Joules\n1,292.89 J + 12,254.40 J + 15,456.00 J + 83,168.00 J\n + 9,391.36 J = 121,562.65 J\n\nConvert to kJ = 121.56kJ
391270
I would say "Please turn the A/C down" if I wanted the room to be warmer, since the A/C cools the air, and the more it's turned up, the cooler it makes the air.\n\nAnd to answer your last question ... ALL THE TIME! :)  Whether it's A/C or something else, often two people will have different ways to say or ask the same thing, and one or the other of us will have to clarify.  I don't think that's just you or me, but happens to everybody at one time or another!  \n\n.
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Note: I'm going to convert all the intermediate steps into kJ by dividing by 1000. I won't show that step.  \n\nFirst, let's find out how much heat it'll take to get the ice from -17.87 degrees to zero degrees, where it has to be in order to melt.  This is a change of 17.87 degrees.  (Because one degree Celsius and one Kelvin is the same size degree, they are interchangable.) We'll use the specific heat capacity of ice.  \n59.90 g ice * (2.1 J/g*K)*(17.87 K)=2248 J = 2.248 kJ\n\nNext, let's find out how much energy it would take to melt that ice at 0 degrees to water at 0 degrees using the heat of fusion:\n59.90 g ice * (333 J/g)=19950 J = 19.95 kJ\n\nNext, let's find out how much energy it would take to get the water from 0 to 100 degrees (so it can become steam) using the specific heat of water:\n59.90 g water * (4.2 J/g*K)*(100. K)=25160 J = 25.16 kJ\n\nNext, let's find out how much energy it would take to evaporate the water using the heat of vaporization: \n59.90 g water* (2260 J/g)=135400J = 135.4 kJ\n\nFinally,  let's find out how much energy it would take to get the steam from 100 to 250.7 degrees for a total change of 150.7; we will use the specific heat capacity of steam:\n59.90 g steam* (2.0 J/g*K)*(150.7 K)=18050 J= 18.05 kJ.  \n\nNow, just add up all of the heats required at each of these steps: \n2.248 kJ+19.95 kJ+25.16 kJ+135.4 kJ+18.05 kJ=200.8 kJ, with correct significant figures.
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The latent heat of fusion -- once you cool water to 0°C, you have to remove more heat to actually crystallize it.\n\nIce at 0°C requires the addition of heat to melt into water at 0°C.  That heat comes from whatever you are cooling.\n\nThe latent heat of fusion for ice is 333.6 joules/gram
327620
Heat is transferred by conduction, convection and radiation. In medium-temperature applications radiation would only be a small part of the total, and an opaque material will prevent radiative escape of heat in any case. Conduction and convection are minimized by the structure of a typical insulator in addition to the material properties mentioned in the first answer. Styrofoam, for instance, consists of very thin partitions of styrene plastic which enclose gas-filled bubbles. This means each convection cell or mixing space is small and essentially a small proportion of the total insulation thickness, so you don't have the coldest internal gas mixing with the hottest gas. Conduction is low because gas conducts heat more slowly than a solid. The conduction path through the solid styrene is convoluted and of small thickness, both of which minimize conduction. In houses, fiberglass insulation uses similar principles except that the air spaces are connected. This means a larger convection cell potentially exists, but it is baffled by all the glass strands, so circulation and mixing are much slower and smaller-scale.
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I've heard that its the boiling water that freezes first. \n\nI think because the water molecules are hot (i.e. they are moving fast) the energy (or heat) will escape (move to equilibrium with other items in the freezer) at a faster rate than the water molecules in colder cup of water. \n\nBut really I'm mostly guessing with my explanation.
168406
An icecube doesn't melt at room temperature, it will stay at 0 degreescelsius (33 degrees Fahrenheit) untill the ice is completely melted.Only when the ice is completely melted will the temperature of thesystem rise. This is because of all the energy put into the ice isconsumed in converting the ice from solid water to liquid water. Thelatent heat of fusion/melting of water is 80 Joules per gram (it takes\n80 joules to melt or freeze one gram of water).\n\nAn ice cube melts at room temperature because it tries to\ncome intothermal equilibrium with it's enviroment. Heat flows into the ice,which increases the total energy of the ice. The molecules in thesolid water start to vibrate faster and they get enough energy tobreak free of the electromagnetic attraction that holds the solid\nwater molecules together.\n\nAfter these electromagnetic bonds break, the water becomes fluid.\n\nIf you heat water to it's boiling point the same process occurs. Themolecules vibrate enough and get so much energy that they can overcomethe force of atmospheric pressure and the water molecules escape into the atmosphere.\n\nTo that end... i dont think there is an actual melting point of ice.  There are many variables to account for when you consider the melting point of ice. \n\nRoom temperature is taken to be roughly 21–23 degrees Celsius (69-73 degrees Fahrenheit).\n\nBoiling Point, 100 C
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The entire world moves toward entropy, it's not in a state of entropy.  There are small scale effects that counteract entropy because of randomness and because the sun feeds us with energy constantly.\n\nWhen a snowflake forms, that is against entropy (it is becoming more "organized").  When a living thing grows up, that is against entropy.
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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