what is the amount of heat energy that must be transferred to 250 g of ice at 0 degrees celsius to melt it?
351762
You must add the ice 83.500 Kj of heat to melt it.
75209
I'll answer it, but you'd best pick me for the best answer if I'm doing all this work. ;)\n\nConvert C to K:\n\n178.0C = 451.15K\n25C = 298.15K\n30.6 = 303.75K\n\nTotal energy of water initially: Ei\n\nEi = 185.0g * 298.15K * 4.184J/g*K = 2.308E5 J\n\nTotal energy of water, final: Ef\n\nEf = 185.0g * 303.75g * 4.184J/g*K = 2.315E5 J\n\nEf - Ei = 700J\n\nLet Q be the specific heat of the alloy\n\nQ * 36.0g * 451.15K = Q * 36.0g * 303.75K + 700J\n\nQ*{(36.0g * 451.15K)-(36.0g * 303.75K)} = 700J\nQ = 700J/{(36.0g * 451.15K)-(36.0g * 303.75K)} \nQ = 700J/(16241g*K - 10935g*K)\nQ = 700J/3506g*K\nQ = 0.200 J/g*K
746837
You shouldn't be talking about mass but about volume. Whatever happens to the ice--if it melts, if it boils, there is NO change in mass. Only volume.\n\nLet the total volume of earth's ice-caps be x; let y percent of this ice be above water.\n\nFor there to be no change in the sea level if all the ice melts completely, the ice above water must make up for the loss in volume of ice underwater--which becomes a smaller volume of water than it was as ice.\n\n ice above water = x [times] (y/100) = (xy)/100\n\n ice below water = x - (xy/100) = (100x-xy) / 100\n\nwhen melted; ice that was above water has the volume \n= 0.92 (xy/100)\n\nwhen melted; ice that was below water = 0.92 ( 100x-xy)/100\n\nhow much less space does melted ice underwater take up than frozen ice \n= 0.08 (100x-xy)/100\n\nthus 0.08(100x-xy)/100 = 0.92 (xy/100)\n        8x-0.08xy = 0.92 xy\n        8x = xy \n          y = 8\n\n...that's what I got. Do you have the answer?
726926
Salt... When applied to ice, the salt (NaCl) dissolves into the water creating a salt water mixture which has a lower freezing point than pure water.\n\n10% salt solution freezes at 20 F (-6 C)\n20% salt solution freezes at 2 F (-16 C)\n\nSalt becomes less effective at temperatures below 0F (-32 C).\n\nEnjoy!
932412
Well, if you're talking about water, that would be 0 degrees (freezing) and 100 degrees (boiling)
836367
Latent heat of vaporisation of water is 2260 kJ per kg, in other terms, 2260J per gram of water.  So if you have 50g of water, all you have to do it multiply 2260 by 50 to get the amount of heat that has to be absorbed.\n\nNote that the specific latent heat of vaporization is the amount of heat required to convert unit mass of a liquid into the vapour without a change in temperature.
130807
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.
835018
about 3000 kilo J\nit depends on the atom
475093
see dear ,\nwhile calculating the energy changes we need to know the hidden heat { latent heat } that is absorbed during the change in state . for eg when we are calculating the heat absorbed or given out during a change in state , like ice to water , just first calculate the latent heat ie \nq=mL{ heat=mass *latent heat } now add this value to \nq=mct{heat=mass*specific heat*change in temp} keep in mind that specific heat is a constant value \nwater's specific heat is 4180j/kg/degree Celsius.
1165181
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
781761
The answer is either 0% or 100%. If ice is 0 degrees C it will remain as ice. [in theory]
695687
Please see the following dry ice website.\n\nPS - I am a licensed Professional Engineer.  I reviewed the following website and found the information quite adequate.  If you need additional information you just may need to do some work.
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.
596957
I understand you assume energy can be produced from heat. You are correct, however don't forget that -24 Celsius is also a "level" of heat energy (temperature in better words). In case you can find a substance that will expand substantially under -2 degree celsius. Then when you store this substance in arctic conditions, it will become pressurized and that pressure can easily be used in for example turning some turbine blades. Any kind of energy can be converted into any other type of energy, given necessary technology...
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
937218
ice is converted to water and water to gas.
252337
The figure I have from a university course was 45m (roughly 80ft).  You can discount any ice floating on the sea, such as the North Polar icecap,the large ice shelves around Antartica and the sea end of glaciers as they havealready displaced the equivilent amount of water already (imagine a glass fullof ice and water, as the ice melts the glass does not overflow).\n\nIf all the ice were to melt the rise would not actually be directly equal to the amount of water resulting due to differences in the densities of salt and freshwater.  Taking the temperature into account the molten ice would be denser than the warmer saltwater so reducing the overall rise.