How many grams of ice at 0 degrees C could be melted by the addition of 3.22 kj of heat?
813772
So, heat of fusion ring any bells?
834399
Your answer depends on the nature of your problem, so you'll want to give more info if you want a complete answer.  The specific heat capacity you use depends on which substance is being heated up or cooled off.  The mass will be the mass of the substance that is being heated up or cooled off.\n\nOne exception that frequently occurs is when you are dissolving a substance in water.  Most problems will assume that the solution has the density and specific heat capacity of pure water.  So in other words, if you dissolve 3 grams of ammonium chloride in 100 mL of water and watch the temperature go down, most instructors would want you to use a mass of 100 grams (instead of 103) and the specific heat capacity of water 1.00 cal/g C or 4.184 J/g C if you're working in joules.\n\nHope this helps...this is the best I can do without more info about the problem.
782848
I'm assuming that the ice is at 0°C.\n80°F = 26.666°C.\n79.7cal/g * 15g * 4.186J/cal = 5,004J fron ice to water.\n1cal/g-deg * 15g * 4.186J/cal * 26.666deg = 1,674J from 0°C water to 26.666°C water.\nTotal is 5,004 + 1,647 = 6,651J\n\n\nDoug
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.
377330
Salt, NaCl, turns into Na+ and Cl- ions in water.\n\nThe ions changes the equilbrium of H2O melting and freezing.\nLiquid water is block by the ions, so it has a harder time freezing back onto the ice. Since the ions doenst slow down\nthe melting of ice, but slows down the freeze of water.\nThe ice now melts at below 0 degree.\n\nSo a normal ice water will be at zero degree.\nBut a salt put on the ice/water will melt more of the ice at a lower temp., about 7-10 degree colder, depenting on the ammount of salt and ice avalibale.
775809
I believe the units of your supposed answer are wrong.  They should be kJ/mol and not kJ.\n\nHeat = m * Cp * DeltaT = 60 * 4.186 * (13.2 - 27.2) = - 3516 J\n\nThis heat is absorbed by 28.7 g of KI.  Per mole (MW(KI) = 166 g/mol), the heat will then be:\n\n28.7 g ===> 3516 J\n166 g ===> x\n\nx = 3516 * 166 / 28.7 = 20,340 J/mol = 20.3 kJ/mol
870267
.4084c(80)+.4084(heat of vaporization)\nc is specific heat of water\nthat will give you J/KG C   si units
699816
the first thing you do is figure out how much heat it will take to melt the ice, this is done by multiplying the quantity of ice by the heat of fusion\n\n10g x 335 J/g = 3350J\n\nnow figure out how much the water cools off to give the ice that 3350J that it needs to melt, this is done by multiplying the amount of water by the specific heat and then dividing this into the energy transferred\n\n100g x 4.18J/gC = 418 J/C\n\n3350J / 418 J/C = 8 degrees C\n80-8=72C\n\nso now you have 100g water at 72C and 10g water at 0C, so you can figure out what this mixture will equilibrate to\n\n100 x 72 =7200\n10 x 0 = 0\n7200 / 110 = 65.4C will be the final temp of the mixture\n\nso the answer is D
195189
The heat required is calculated by multiplying the moles of ice and the heat of fusion of water.  Moles of ice equals the mass of ice divided by the molecular weight of ice.  The heat of fusion of water one can get from the book.  Make sure your units cancel appropriately.
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
211914
It depends on the mass. The latent heat of fusion of water (the energy required to freeze it) is 333.7 kJ/kg. That means to freeze one kilogram of water, it requires removing 333.7 kilojoules of energy.\n\nThis assumes that the water is already at freezing temperature. If not, energy must first be removed to get it at freezing temp.
382860
Melting point -11°C (262.15 K) \nBoiling point 150.2°C (423.35 K)
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 ;)!
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
362368
Given:\nm <mass of substance> = 12.0g or 12/1000 kg\nc <specific heat capacity of water> = 4180J per kg per degree celsius\nt <change in temperature> = 20 degrees to 100 degrees i.e. 80 degree celsius\n\nFind:\nHeat req.\n\nSolution:\nHeat required or Q\nQ = m * c * t\n  = 12/1000 * 4180 * 80\n  = 12/10 * 418 * 8 (cancelling 0s of 4180&80)\n  = 12/5 * 209 * 8 (div. 418&10 by 2)\n  = 12/5 * 1672 (mult. 207&8)\n  = 20064/5 (mult. 12&1672)\n  = 4012.8\n\nAnswer = 4012.8 Joules or 4.0128 KJ
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.
971810
If the water is already frozen and you remove more heat, all you will get is colder ice.  Otherwise, we need to know at what point is it at in the latent heat of fusion.