Why is water an excellent material to use in a calorimeter?
141010
Because calorie is the energy needed to increase the temperature of 1 g of WATER by 1 °C.
76194
Thermal energy is a measure of molecular motion, not collision, and the more motion there is in the water, the higher the thermal energy (kinetic), and the higher the temperature.
762941
The change in heat = specific heat * mass * change in temperature (Q=C*g*ΔT).\n\nAssuming that the Aluminum also ends at 28.3 °C, then the change in heat for the Aluminum is\n\n0.0902 J/g°K * 35.7 g * (28.3°-81.9°)\n\nThe above will be negative as heat was lost from the Aluminum.  The change in heat for the water is\n\n4.184 J/g°K * 75.0 g * (28.3°-24.9°)\n\nThe above will be positive as the water gained heat.\n\nThe Heat Capacity of the calorimeter is the difference between the heat lost by the Aluminum and that gained by the water divided by the change in temperature (which is the same as for the water (28.3°-24.9°).
291253
Because water increases your body's electrical conductivity.
83439
Because water has a high SPECIFIC HEAT VALUE (C=J/gC) which just means that it can absorb a realitivy high ammount of heat (J=jouls) and keep its overall mass (g=grams) about the same temperature (C=celsius)\n\nits like 4.186 J/gC or somin i think, which is very high, while steel is only .113 J/gC
562654
Okay, I'll try this one -\n\nRun this experiment twice (or more times), each time with a piece of metal having different density.  The metal piece should be of adequate size (a sewing needle isn't going to do it), it needs to be big enough to significantly affect the temperature of the water.  You will need to know the mass of the metal (i.e., weigh it) and have a thermometer capable of reading from room temperature up through boiling or so.  Please note this is not going to give you a highly accurate answer, but can give you a ballpark idea (I'm guessing you're probably not doing this in a 500 million dollar lab).\n\nThe procedure:\n\n1)  Measure and record room temperature as T1.\n\n2)  Weigh the metal, call it M1.\n\n3)  Bring a known quantity of water (weigh it, or measure it and convert to weight, call it M2) to a temperature just short of boiling (say 180F to 200F, use the same temperature for each run of this experiment).  Record the temperature as T2, and remove the heat source (take it off the burner, etc.).\n\n4)  Drop in a metal piece at room temperature.  Allow a couple of minutes for heat to exchange.  Make sure you use \nthe same amount of time for each run of this experiment.\n\n5)  Measure and record the new temperature of the water as T3.\n\n6)  Repeat with your other metal samples.\n\nYou can now estimate the specific heat capacity for each metal.  It will not be super-accurate (heat escapes your system, and temperatures won't have completely equalized), but should give you some idea.  Just assume all temperatures did equalize and that no heat escaped for calculational purposes.\n\nSince (you are assuming) the heat which left the water went into the metal, then the product of mass, specific heat capacity, and temperature change should be the same for both the metal and the water (but in opposite directions).\n\nThis means that:\n\n(M1)(c1)(T3-T1) = (M2)(cw)(T2-T3)\n\nwhere c1 is the specific heat capacity of the metal, and cw is the specific heat capacity of water (look it up on Google to get the units you need).\n\nSolving this, you get:\n\n......(M2)(cw)(T2-T3)\nc1 = -----------------\n........(M1)(T3-T1)\n\n(sorry about the dots, I'm trying to keep everything aligned)\n\nRepeat the calculations for each of the metals.\n\nYou will need to calculate (or look up) the densities of each of the metals.  Then, graph density vs. the specific heat capacities you calculated, and see if your hypothesis is true.\n\nHave fun!
855413
Latent heat of fusion of ice is only 80 calories/gm whereas the latent heat of vaporisation of water is as high s 536 cal/gm.Therefore C is correct
187241
See reference. Calorimetric constant is the heat capacity of the inside of a calorimeter.
1136153
many materials absorb better at the infrared wavelengths than the shorter wavelengths so the absorbed energy is conserved as heat.  There are materials that absorb shorter wavelengths and convert the energy to heat, such as a microwave oven???  Water absorbs at that frequency (1-2 GHz) and the water molecule resonates.  The vibrational energy is re-emitted as heat due to a Stokes shift.\n\nThe shorter wavelengths are more energetic and more capable of depositing energy that might show up as heat but it depends on the absorption of the material itself.
621304
If you are dealing with a reaction involving ammonium carbonate in solution, and are then measuring the temperature rise, you need the heat capacity of water, not ammonium carbonate!\n\nPlease post the whole problem if that is not enough to get you started.
1159854
I didn't know they used calcium until i checked it out.  Magnesium makes since as all of those metals in that column are real nice and bright when dropped in water.  I think calcium does something similar, something about it is preferable under water, though I don't know my periodic table too well.  \n\nFrancium is the craziest in water.\n\n"It burns with a yellow-red flame and forms a white nitride coating when exposed to air. It reacts with water, displacing hydrogen and forming calcium hydroxide."
731763
I would say that as 1 calorie raises the temp of 1 cubic cm or 1g of water by 1 degree that you would need 25 calories to raise 25g of water to 1 degree c therefore melting it.
295266
water\nits specific heat is 4.18 kJ/kg(degree)C\nits much more highthan any substance available.\nthats why its used in car radiators
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. ><
274980
Since there is no temperature change in the substance, only a phase change (solid --> liquid), all you need to use is,\nQ = Lm\nwhere Q is the heat energy, L is the latent heat of fusion, and m is the mass of the substance.\nI dont happen to know the latent heat of fusion for water (I think it is like 2260 J/g, maybe), but I am sure it is in your book somewhere.\nJust plug the values into the formula and you are set.\n\nEDIT:\nThe Latent heat of fusion of water is 333 J/g.\nThe latent heat of vaporization is 2260 J/g.
805636
Unfortunately the answer above is not entirely correct for this context. (although the statement is correct).\n\nThe reason why water is used is because of its large specific heat capacity.  \n\nThat is also the reason why when we run (and because our bodies are mainly water), the the body temperature does not rise excessively because of the body’s high heat capacity.
39424
heat is the measurement of thermal energy