What  Is Specific Heat?Give Its Unit.?
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The specific heat of a substance is the radiation heat contained  by each atom or molecule per unit volume and temperature for a specific material.\nIt is calculated as follows; K(Bolzman's constant) multiplied by one kilogram mass of the material and divided by the atomic mass of the specific material.\n\nIts Unit is Joule(energy)per kilogram for each degree of absolute temperature.\n\nThis means that different materials dont have the same heat energy content for the same volume and atomic energy level.\n\nI hope if you find this explanation helpfull.
885641
Assuming that they have equal masses, it would be whichever material has the highest specific heat, because that defines how much heat energy is required to increase a unit mass of material by a unit temperature.  Water is known for having a very high specific heat, so it could be that, but you should really look up the specific heat (or "specific heat capacity") for all of these materials in your text book.
705505
The specific heat of water is 1 calorie/gram °C = 4.186 joule/gram-°C.  If you want to use 4.18, that is okay.\n\nThe heat added = the specific heat of water (c) times the mass of the water (m) times the change in temperature.\n\nQ = cmT or\n\n= 4.18 joule/g-°C * 125 g * 2.44 °C\n= 4.18 * 125 * 2.44 joules (you can do the math)\n\nThis is for 1.75 g of CaCl2 which has a molecular mass of 111.0 g/mol, so the number of moles is:\n\n1.75 g / 111.0 g/mol = 0.01577 moles of CaCl2.\n\nDivide the heat change (in joules) above by 0.01577 moles to get the heat per mole.
190244
Specific Heat Capacity\nThe specific heat capacity of a solid or liquid is defined as the heat required to raise unit mass of substance by one degree of temperature. This can be stated by the following equation: \n\nQ= mc(changes of T)\nwhere, \nQ= Heat supplied to substance, \nm= Mass of the substance, \nc= Specific heat capacity, \nT= Temperature rise.
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Acetone       2.13kJ/kg K or  0.51  btu/lb F\nWater, fresh 4.19kJ/kg K or  1.0   btu/lb F\n\nsee also http://www.engineeringtoolbox.com/specific-heat-fluids-d_151.html
827524
Easier compared to one with a higher heat capacity:\n\nH/(m*Cp) = (delta T), where m is the mass of the object, H is the heat that must be added (or removed) to raise (or lower) its temperature, delta T is its change in temperature, and Cp is its heat capacity. From the above formula, you can see that larger values of Cp result in small temp. changes, for the same heat amount.
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.
216745
One unit kelvin, or 5/9 degree farenheit
891159
use the equation\nE(enthalpy change) = m(mass) c(specific heat) DeltaT(change in temperature)\n\ntemperature change is initial temperature minus final\nsince temperature increased then the temp change is negative\n\n-2.00C
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Remember your equation for specific heat capacity::\n\nHeat content = mass x spec heat capacity x delta T\n\nq = m*C*dT\n\n225 J = (15 g) * (0.80 J/gram-deg C)* dT \n\nTherefore\n\ndT (in C) = 225 J  /  (15 g)(0.8 J/g-C) = 18.75 deg C\n\nMind your units and make sure they cancel.  I did.
734516
Easy.  \n\nYou have q = mass * specific heat * change in temp.\n\nFind the specific heat of water and your set.  Q is joules so you need to convert.
170755
Specific heat 0.13 J/gK
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You could start with these sites and go from there.\n\nhttp://www.sciencebyjones.com/specific_heat1.htm\nhttp://en.wikipedia.org/wiki/Specific_heat_capacity\nhttp://www.rwc.uc.edu/koehler/biophys.2ed/heat.html
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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.
949133
OMG!\ni have the same question!\nmine is:\nA piece of an unknown metal has a mass of 20.0 g and at 100.0 C is placed in a styrofoam cup calorimeter containing 50.7 g of water at 22 C. If the final temp is 27.2 C, calculate the specific heat capacity of the metal. Assume that all the energy lost by the metal is gained by the water.\n\nok, haha fine, not the SAME question, but close enough. But, i'm uber confused. ><
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Q (heat energy) = m(mass in gram)*c(specific heat in J/gm/K)*t (change in temp)\nQ = 25 * 1 * (36-24)\nQ = 300 J
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.065kg * 35 * specific heat of Aluminum\n\njust make sure your units check out; you may need to convert this equation into calories as specific heat is normally defined in joules I think.