calorimetry que...........just checkit out?
143592
Heat required to rise the temperature of  ice from \n-10 degree centigrade to 0 degree centigrade \n= ms(t2-t1)\n=10 * 2.1 * 10  g* {joule/(g*degree)} (degree)\n=210 joule\n\nHeat required to convert ice to water  \n=mass x specific latent heat fusion of ice \n=10 * 336  g*joules per gram\n=3360 joule\n\nHeat required to rise the temperature of  water from \n0 degree centigrade to 100 degree centigrade\n=ms(t2-t1)\n=10 * 4.2 * 100\n=4200 joule\n\nHeat required to convert water to steam\nmass x specific latent heat of vaporisation of water\n10 * 2264\n22640 joule\n \nHeat required to rise the temperature of steam from \n100 degree centigrade to 110 degree centigrade \n= ms(t2-t1)\n=10 * 2 *  (110-100)\n=200 joule.\n\ntotal of all is  30610 joule.
701285
In ideal conditions, yes.\n\nIn your error analysis, you could talk about how styrofoam calorimeters are not the best ones to use and some heat could have been lost to the environment.
1128546
specific heat Q = mc(t2-t1)\nput everything in SI units..\nthen \n\n40,000= 500/1000 * c(68-20)\n40000=0.5* 48*c \nhence c= 40000/24= 1666.67 joules/kg oC.
550455
http://www.calorielab.com/
793636
shawty's NOT good at stoichiometry...has made a fundamental error in the final line so end result is rubbish...\nhow can you end up with MORE Cu than you started with when it's 2 moles : 2moles? \n\n(8/159)x 127 = 6.4 g
174163
thermodynamics?  heat transfer?  physics?
944636
I just did a bomb calorimeter experiment in my physical chemistry lab, I am sure most universities with chemistry programs will have them.
187241
See reference. Calorimetric constant is the heat capacity of the inside of a calorimeter.
501446
http://www.caloriesperhour.com/index_burn.html\nhttp://www.healthstatus.com/cbc.html
155062
Quantity of heat (Q), measurement = joule (J)\nSpecific heat capacity in liquid, measurement = c/(kJ kg^-1)
952146
The first part of the question gives you the information you need to determine the "calorimeter constant", or the amount of heat required to change the temperature of the whole calorimeter set-up by 1 degree.\n\nThe calcium carbonate/HCl stuff is the reaction whose enthalpy change you are trying to determine, using the same calorimeter as before.  When you combine calcium carbonate and HCl, you'll get a reaction that produces CO2 and leaves you with a solution of calcium chloride.  So the overall reaction that you're looking at is:\n\nCaCO3 + 2 HCl --> CO2(g) + H2O + CaCl2\n\nCalcium carbonate will only decompose to CO2 and CaO when you heat it--not when you react it with an acid.\n\nHope this helps....I'll let you do the calculations.
477699
http://www.caloriesperhour.com/index.html\n\ni use it all the time.
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!
622985
yes it is 294cal... first you go 150 - 10 degrees to find out how much the temperature change was  (140).  multiply 0.21 cal by 10 g to find out how many cal is given off for each C if the aluminum is 10 g (2.1), and then you take that number and times it by the temperature change (140 x 2.1). and voila. you get 294 cal. which is one of the choices :)
163823
Q=mc(Tf- Ti)\nWhere Q is the heat added\nm is the mass\nc is the specific heat capacity\nTf is the final temperature\nTi is the initial temperature
315183
watch your calories as much as possable ?
577335
The SI unit: calorie (cal)