Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Cyberconvict on March 23, 2006, 09:07:51 PM
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I know how to figure out the calorimeter's constant most of the time, but this worksheet I have is calling for a method I've never used. It says:
A calorimeter is to be calibrated: 51.203 g of water at 55.2C is added to a colorimeter containing 49.783 g of water at 23.5C. After reaching equilibrium , the final temperature reached is 37.6C. Calculate the calorimeter constant.
How do I go about solving this? I know the specific heat of water is 4.18 J/g-K and that there are 100.986g water, but I don't know what the temperature change is.
EDIT: I found the formula (mC?T)warm water = -[(mC?T)cold water + (Ccalorimeter ?T)]. I'm assuming that the change in temperature is based on the cold to the equilibrium point and the warm to the equilibrium? Thus ?T for warm would be -17.6C and ?T for cold water would be 14.1C? Please tell me if I'm correct.
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http://www.chemicalforums.com/index.php?board=4;action=display;threadid=5954;start=msg26410#msg26410
Try this link and see if it helps at all :)
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Yea that helped me, but I don't see how I can use that for this, because I am unable to take values for a worksheet problem. Maybe I'm just looking at this wrong.
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Don't think in terms of calorimeter, think in terms of simple heat balance. Your only unknown in the heat balance for the system is calorimeter constant.