At what speed Ice melts in 20C environment. Grams / Minute.?
931428
The speed at which it melts it dependent on the amount of heat that is transferred to the ice. The speed at which heat is transferred is dependent on the environment. \n\ne.g. ice will melt a lot faster when it is placed in water of 20 degrees Celcius, than when it is placed in air of 20 degrees Celcius.
602468
Absolute Zero or around 20 K.\n\nMelting point 14.01 K  (−259.14 °C, −434.45 °F) \nBoiling point 20.28 K  (−252.87 °C, −423.17 °F)
721040
I get 45.13 degree Celsius.\nReasoning:\nChanging 23 gram ice in water of 0 degree requires 7.66e3 Joule, according to part 1. This amount of energy taken from the 70 degree water decreases the temperature of the latter by 7.66e3 x (70-10) / 2.9e4 = 15.85 degrees (using part 1 and 2).\n\nSo now we have a mixture of 23 gram water at 0 degree and 115 gram (5 times as much!) at 54.15 degrees. This will average out to (5/6) x 54.15 = 45.13 degree Celcius.
512231
20 C is about room temp... it's a gas, clear and odorless.
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
820184
I think you should break things down and think of the question like this.\n\nThe heat for a process is 250.3J, Its doubled with the same temperrature difference, the temperature difference is 5C.
784706
Work must:\n be converted into heat.\n\nTo determine the amount of heat needed to melt the snow-\ntake 4000g H20 change to moles by dividing by 18 then multiply by the molar heat of fusion\n\n(4000 g H2O)(1mol/18) = 222 moles(6.03 kJ/mol) = 1300 kJ or \n1,300,000J = Work done to melt snow\n\n1,300,000 = 0.1 x 75 kg x 9.8 m/s2 x d\nThe answer is 18 km
726471
Look at your units for latent heat. Joules per g. That means it takes 334 J to convert each gram of ice to liquid water. You have 250 g of ice. That should tell you what the formula is.
366921
Q = mc delta T\nHeat = mass times specific heat times change in temperature\nspecific heat for water is 4.186 kJ/kg C\n-41840 J (negative because absorbed) = .200 kg x 4.186 x 10^3 J/ kg C x (Tf - 20)\nTf = final Temp.\nsolving the equation gives you Tf = -30.0 degrees Celsius
730140
The ice would gain 124.3 kJ in moving from -10 °C to zero.\nThe water would lose 252.0 kJ in moving from +10°C to zero.The difference in the two energies is sufficient to melt 383 g of ice.\nSo the mixture will be at 0 °C and will consist of 6383 g of water and 5617 g of ice.
216514
That all depends on the altitude. At 1 atmosphere of pressure (sea-level), the melting point is 32F and 0C. At higher pressures, this temperature lowers. This is why ice skating works. The pressure of the blade causes the ice to melt, letting the skate glide.
862439
The steam.  It would have to go through a phase change and give up a bunch of energy to turn back into water at 100.  Assuming you mean equal mass of water and steam. \n\nSame question for ice.  How much energy does it take to melt 1 Kg of ice at 0 to water at 0?
328951
The latent heat of fusion for water is 335 Joules per gram. so thats exactly how many joules of energy you need to melt one gram of ice
767956
Carbon dioxide liquefies/boiles at -57°C and solidifies/melts at -78 °C . People use these carbon dioxide properties, to test other properties. One such  test is impact testing of metals, at low temperature.  \n\nSolidified/Solid carbon dioxide is called dry-ice.  The  test using solid carbon dioxide is called  dry-ice test.
377359
What volume of water in what period of time?\nAs soon as you introduce steam to the ice, it will start melting and if the water runs off, it will probably achieve 20 degrees celsius fairly quickly.\n\nIt also depends not only on the temperature of the ice, but the heat content. "When ice melts, it absorbs as much heat energy (the heat of fusion) as it would take to heat an equivalent mass of water by 80 °C, while its temperature remains a constant 0 °C." Wikipedia, frozen water.
773883
Stop for a minute... Look over your question. Note 2 things\na. 77.2 AU\nb. An icy object\n\nNow start over slowly and find your mistake...YOU CAN DO THIS!!!
629709
She will need enough heat to raise the temperature of the ice 20 degrees and then to fuse it all at a constant temperature.  Water's heat of fusion would not have a temperature term as it is a constant temperature process.\n\nSo, it would be:\n2100 J/kgC * 2.0 kg * 20 C + 335,000 J/kg * 2 = 754,000 J or 754 kJ