how much heat is required to covert 0.3 kilograms of ice at 0' C to water at the same temperture?
88715
Latent heat of ice is 334kJ/kg.\nThis means, to convert 1kg of ice at 0c to 1kg of water at 0c it would require 334kJ (or 334000 Joules). \nSo, to convert 0.3kg of ice to water at the same temperature would require 0.3*334 or about 100kJ.\nThe answer would be (A) above.
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
781761
The answer is either 0% or 100%. If ice is 0 degrees C it will remain as ice. [in theory]
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
1164067
Third Law Of Thermodynamics (I am studying that too)\n\nIt is impossible to reduce the temperature of any system to absolute zero by a finite number of operations.\n\nIf you are asking how much internal energy ice has at -273K, then the answer is zero.\n\nGood luck with this.
137352
No, it isn't 33 degrees F.  At 32 degrees (definition 0 degrees Celsius, 273 degrees K) and at one atmosphere of pressure, water and ice coexist; this is the melting point of ice.  \nIf the water is not pure, the melting point is lower than previously stated (freezing point lowering is a colligative property).
251863
At 20 degrees C, it's 0.6g/100mL. You can convert that to a Ksp.
860053
measure the cold water temp and take a large amout of cold water than hot water and mix them .\nuse the realation mhot (Thot - T2) = mcold(T2 - Tcold)\nwhere:\nmhot = amount of hot water , Thot = hot water temp\nmcold = amount of cold water , Tcold = cold water temp.\nT2 = the temp. of water after mix.\n(considering the specific heat of water constant at this range of temp.)
142627
First we must use the Kelvin scale as is the only absolute scale starting from 0. You have quantified the amount of cold so you must use an asbolute scale. Kelvin measures the absolute temperature with 0 representing NO kinetic motion whatsoever of the molecules. Kinetic motion is another way of defining temperature. That said here we go:\n\n0 degrees F is ~255K so half of that is 122.5K which is -233.57 degrees F. \n\n0 degrees C is ~273K and half of that is 136.5K which is -136.5 degrees C. \n\nNumbers are all approximate.\n\nAlternatively, you can set an arbitrary definition for cold say 40 degrees F which is 278K. If its 255K (0 deg f)today, then you take the difference between 278 and 255 and multiply it by 2 and you get 46K. Substract the difference from 255 and you get 204K which is -92 deg farenheit.\n\nI hope I didnt lose you yet but I actually had fun doing this ;)!
106190
This seems like a homework problem...I'll give you a clue.\n\nsince, the ice is not changing temperature, all you need to do is figure out how much energy(per kilogram) it takes to ice into water...then multiply that number by 0.3.\n\nI'm guessing there will be a number in your book somewhere stateing the required energy to change 1 kilo of ice to water.
288276
ρwater ~ 1g/ml\n\nq = 1000g x 4.18J/ºC.g x (37ºC-0ºC) = 155.660kJ (1.6x10^5J to correct Sig Figs)\n\nsecond part. The amount of heat required in part 1 will melt 155.660kJ / 6.01kJ/mole = 25.9 moles of ice or 468g
811304
let thetemperature of the mix be 't'*Cfind the heat lost by coffee\nfind the heat gained by ice in three stages\na)from -2.2*C to 0*C\nb)for melting and becoming water at 0*C\nc)for getting to the equilibrium temperature of the mixture\nequate and solve
1066111
let final temp be x\nHeat lost by 0.5gm h20 = mass*specific heat(a constant)*(98 -x)\n= 0.5 *4.2*98-x\n= 2.1(98-x)\nheat gained by 0.235kg water = 235*4.2*(x-20)\n=987(x-20)\nheat lost = heat gained\n2.1(98-x) = 987(x-20)\n205.8 - 2.1x = 987x - 19740\n989.1x  = 19945.8\nx = 20.165 degrees C\nfinal temp = 20.165 deg C
1183481
Pure water.  The ice will have to melt before it could boil.  Water doesn't go through sublimation
613974
You need to use the combined gas law:\nP1V1/T1 = P2V2/T2, but your temps need to be in Kelvins, not Celsius.\n\nIt also seems that the volume is being held constant, so we can cancel volume out of the problem.\n\nSo, convertine the temperature first, we get:\n10C + 273 = 383K for our temp1\n\nSo, subbing into the formula we get:\n100mmHg/283K = 900mmHg/x\n\nSolve for x and get your answer.  The temperature will be in Kelvins.
55129
The heat required can be divided into 5 parts:\n1. Heat required to race the temperature to the melting point.\n2. Heat required to melt the ice.\n3. Heat required to raise the water to the boiling point.\n4. Heat required to change the water to steam.\n5. Heat required to raise the temperature of the steam to the final value.\n\nCompute part 1 using the specific heat of ice.\nCompute part 2 using the heat of fusion.\nCompute part 3 using the specific heat of water.\nCompute part 4 using the heat of vaporization.\nCompute part 5 using the specific heat of steam.\n\nYou can do the math, after all it is your homework assignment.
934498
This is a three step problem.  Since you're losing heat temp is going down.  First calculate the amount of heat needed to go from 25.0 C to 0.00 C.\n\nq1 =  1 g X 4.18 J/gC X -25 = -104.5 J\n\nNow you have water at 0 C which needs to freeze.  You need to remove the heat of fusion to make ice.  deltaHf is -335.5 J/g.\n\nSo q = m X deltaHf = 1.0 g X 335.5 J/g = -335.5 J.\n\nNow we've used up 440 J so far, and the remaining 24 J will lower the temperature of our 1.0 gram of ice.\n\nThe heat capacity of ice is 2.1 J/gC and we'll need to solve for delta T.\n\n-24 J = 1.0 g X 2.1 J/gC X delta T\n\ndelta T = -24 J/ 2.1 JC= -11.4 C.\n\nSo the final state and temperature would be ice at - 11.4 C.\n\nSo the answer of -12 C should be correct with more care paid to significant figures and proper rounding.