how much heat is required to convert 0.3 kilogram of ice at 0'C to water at the same temperture?
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.
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.
27210
im not sure of the exact curve.  but i know you have take in account the heat of fusion (the energy it takes to convert ice to liquid water, where the temperature stays the same, 0 degrees celsius, even though energy is being added)  and the heat of vaporization (the energy it takes to convert liquid water into steam, where the temperature remains the same, 100 degrees celsius, even though energy is being added)
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.
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 ;)!
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.
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
530952
It depends on the temperature of the ice. 1 calorie is the amount of energy needed to raise the temperature of 1 gram of water by 1 degree Celsius. 1 calorie = 4.2 joules.\n\nTherfore, 1 gram of ice at -4C would require just over 16.8 joules to warm it up to just above the freezing mark.
501191
There is a formula...but this is much easier\n\nhttp://www.onlineconversion.com/temperature.htm
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).
725751
steam to water \n15.2 x 2.03 KJ x (158-100) = 1789.648\n15.2 x 40.6 = 617.12\n\nwater to ice\n15.2 x 4.18 x 100C = 6353.6\n15.2 x  6.02 = 19.5\n\nice at 0degree C to -50\n15.2 x 2.06 x 50 = 1565.6\n\ntotal quantity of heat\n1789.648+617.12+6353.6+19.5+1565.6 = 10345.468\n\n\ni hope it is right.. nt vry sure..
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.
1119466
I donno ;=)\n\nMay be start with a heat balance equation like\n\nQfinal=Qwater-start  - Qice-start – Qlatent\n\nIn other words the final heat of the system is equal to the original amount of heat minus the heat required to bring ice to the melting point and minus again the amount of heat needed to melt the ice\n\nOr \nQfinal=(Specific-heat of water) (temp in K)(mass-final)\n\nmass-final= mass of water + mass of ice= m1 + m2\n\nNote -273.16 °C =0 degrees on Celsius scale = 273.16 °K (273 I hope is good enough for you?)\n\nQwater-start  =( 4190 J/(kg*K) (t-start K degrees )(m1)( 4190 J/(kg*K)=A(m1)  A= 4190 (76.2 + 273)\n\nQice-start= 2100 J/(kg*K)( - 23.6+273)(m2)= B(m2)\n\n\nQlatent= 334 kJ/kg(m2) =C(m2)\n\nWe have \n\nD(m1+m2)=A(m1)+B(m2)+C(m2)  you know that m1=.280kg  then you figure out m2? \nAnd \nA= 4190 (76.2 + 273)\nB=2100 (- 23.6+273)\nC=334,000\n\nActually Specific heat of ice = 2060 J/(kg*K) not 2100 J/(kg*K)\nAnd Specific heat of water  =4,183  J/(kg*K)
41651
The formula for calculating specific heat is q=mC(T2-T1). Where q=heat exchange, m=mass, C=specific heat, T2=final temp, and T1=initial temp.\n\nSince you have two materials inside your container, your formula would be q=[mC(T2-T1)]for the water + [mC(T2-T1)] for the metal.\n\nThe container is insulated so there is no heat exchange between the outside environment and the materials in the container. So q=0.\n\nThe formula reduces to 0=[mC(T2-T1)]for the water + [mC(T2-T1)] for the metal. \n\nOr [mC(T2-T1)]for the water = -[mC(T2-T1)] for the metal\n\nC of metal= [mC(T2-T1)]for the water/[-m(T2-T1)]for the metal.\n\nC of metal= [(185g)(4.184 J/gK)(303.75K-298.15K)]/\n            [-(36g)(303.75K-451.15K)]\n\nC of metal = 7775.313J=7.775kJ
170755
Specific heat 0.13 J/gK
62689
specific heat of steam is 2.02 J/g*K\n\nU steam= 215000J - 75.0g*2.02J/g*K(25K)= 211213J\n\nU liquified = 211213 - 40,700J/mol*mol/18g*75g = 41629J\n\nU water @ 0 C = 41629J - (4.18J/g*K)*75g*100K = 10279J\n\nU freezing = 10279J - 6000J/mol * mol/18g * 75g = -14721\n\nso you have some water and some ice, or liquid and solid phases.
128148
It would depend on the amounts (mass) of the water.\n\nedit:\nThe specific heat of water is not 1 [artless_man], it is 1 calorie per gram degress celcius, the units make all the difference.  Notice the lower case c in calorie.\nA "better" specific heat of water, using metric units, is 4.186 J / (gram * degree celsius).\n\nHowever, for this problem, you dont even need to know the specifc heats since both substances are water.