Calorimeter Question!?
958888
Use the formula:\nQ (heat) = Cp (Heat Capacity) * delta T (Change in Temperature)
750041
.You need to refer to your text or on-line to find the instructions.  A bomb calorimeter uses electricity to burn the material in the presence of oxygen.  You would need a sealed metal container (and you need to know the metal to know its specific heat), a source of electric current with exposed wires to provide the spark (be careful with the live wire!) a second larger container for water in which the first container is submerged, and a thermometer.  Good luck and be careful
337103
Temperature measurement
929275
You would need the specific heat of the material.  The rate of cooling would be the total energy loss (E*A) divided by the specific heat.  This will be non-linear, and your value would be good for only the specific temperature given.\n\nWatts = Energy/time; one watt = 1 joule/sec.
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. ><
336045
Go to google n type in calorie counter n there is a site that has the words caloriecounter---- < that is another word that i dont remember anyways n it will have a HUGE list of stuff or you can type it in n find out the calories in them.\nWe had to do sum like this for school n thats how i found out. Try it.
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.
393560
Go to www.freecaloriecounter.com
673582
This information should enable you to solve the problem. \n\nLatent Heat of Freezing or Solidification \n\nThe amount of heat required per unit mass of a substance at freezing point (constant) and under 1 Atmosphere pressure to convert it from the liquid to the solid state is called Latent Heat of Freezing.\n\n \ne.g. At 00C & 1 Atmosphere pressure 80 K-calorie of heat supplied to 1 Kg of ice melts it into water and 80 K-Calorie of heat taken away from 1 Kg of water converts it into ice. \n\n&#92; L = 80 Kcal / Kg = Latent heat of fusion of ice\n\n\n    Latent heat of freezing of water \n\nso if 80 Kcal/Kg= latent heat of fusion of ice\nthen you can cross out the K and you have 80 cal/gram is the heat of fusion of water. but you hve 25 grams of water to freeze, not just 1 gram. so what comes next. what do you have to multiply by 25?\n80 right? \nthats right. see it wasnt so hard was it?\nso the answer is 2000calories or 2 Kcal.
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
64647
2x (G2 + M2 ---> 2GM)      DH=-280.0*2= -560.0 kJ\n2x (L + 2M2 ---> LM4)      DH=-350.0*2= -700.0 kJ\n   (L2G4 ---> 2L + 2 G2)   DH=122.0*(-1)=-122.0 kJ (change sign when flip over)\n\nAdd them together, we'll have:\n2G2 + 2L + 6M2 + L2G4 ---> 4GM + 2LM4 + 2L + 2G2\nthen cancel 2G2 and 2L on both side, we end up with:\nL2G4 + 6M2 ---> 2LM4 + 4GM\nthe enthalpy of this reaction is, therefore the sum of each enthalpies calculated above.\nDH=(-560.0) + (-700.0) + (-122.0)= -1382.0 kJ (exothermic process)\nhope that'll help!
643566
In the Environmental Lab, we used the Parr Oxygen Bomb Calorimeter to measure the change of heat or capacity of heat for different substances.  What I basically mean is that we put the substance into a reaction vessel and then into the reaction container(bomb).  The bomb is pressurized,  filled with Oxygen, ignited and the data from the reaction is extrapolated to give us the Caloric Value for the substance.  \n\nWe took in Hazardous Waste for incineration in an industrial Kiln.  It was important to know the Caloric Value for these substances as it would determine how much other fuel (Usually natural gas) we would have to use to maintain the optimum temperature for the Kiln to operate.
512100
Well, it's 335 Joules.\n\n1 Joule = 0.238845896627496 Calories, so 335 Joules = 80 calories.
889824
okay.  here's the formula to solving this problem.  heat=mass*calories*change in temperature.\n\n1)5 grams\n2) the change in temperature is 100 degrees\n3)1 calorie per gram\n\nso the answer is 5*100*1=500
915069
The food is "burned" in a bomb calorimeter.  The energy given off by the food causes the water to change temperature.  The change in temperature of the water allows us to calculate the calorie value of the food sample.
309325
Since m*C*dT=Q;\n\nHeat capacity=135000 J/(5100 g*(31.5K - 18K))= 1.96 J/gK\n\n(K for Kelvin)
681590
Three sources of error of using foam as a calorimeter vessel involve: \n1. Pressure   \n2. Ignition\n3. Material interference.\n\n1.  Limited Pressure in the cup\nFoam used in the construction of drinking cups is very limited in its ability to be pressurized.  It would be a poor calorimeter vessel because you couldn't hold any pressure.  To effectively measure the caloric value of a substance, you need a much wider pressure gradient than foam will allow. \n\n2.  Ignition of substance being measured\nFoam doesn't withstand exposure to heat very well.  Since a difference in heat is an integral part of measuring a substances' caloric value, and the fact that foam won't take a very big difference n heat, then foam isn't a good choice. \n\n3.  Material interference with reaction \nThe Caloric measurement of a substance is a chemical reaction.  In a Calorimeter, this reaction is conducted inside a nonreactive, non-interfering vessel.  If foam were used, the potential of interference via byproducts from the foam heating up could be substantial.