Unformatted text preview: y both the
beaker (remember it had HCl in it) and the coffee cup calorimeter. Get 20.00 mL of HCl
in the beaker and 20.00 mL of NaOH in the coffee cup calorimeter. Once again,
determine the temperature of both solutions, being careful to rinse off the temperature
probe and to dry it between solutions. Begin taking temperature measurements of the
NaOH, and quickly, but carefully (to avoid splashing) pour the HCl into the NaOH.
Cover the calorimeter and swirl carefully. Continue taking temperature readings until the
temperature levels off, or until it drops slightly. All solutions can go down the drain with
running water; clean your equipment. Return the calorimeter.
Calculations:
For any heat transfer which does not involve phase changes, we have q=mc∆T, where
∆T=TfinalTinitial. Here, q is heat (measured today in “calories”), m is mass (measured in
grams), T is temperature (measured in degrees Celsius), and c is the specific heat. For
water, we have cwater =1cal/go C. Since qobject =  qcalorimeter, we have mobject cobject ∆Tobject = mwatercwater∆Twater . Thus, cobject = mwatercwater∆Twater/mobject ∆Tobject .
The mass of the object was measured directly. To determine the mass of the water,
subtract the mass of the calorimeter from the mass of the water and calorimeter.
Plot a graph of temperature versus time (by convention, we always mean plot “y
versus x”, so to say “plot temperature versus time” implies that temperature belongs on
the y axis, while time belongs on the x axis). From the graph, determine ∆Twater, and the
final temperature. Since heat is being absorbed by the water, ∆Twater should be a positive
number.
The initial temperature of the object is taken to be 100o C (since, after all, it started in
boiling water. The final temperature of the object should be the same as the final
temperature of the calorimeter if you were patient enough. Thus, ∆ Tobject =Tfinal100, and
should be a negative number.
Calculate the specific heat of each object and report them. From a table of specific
heat values, can you predict what the objects were made of?
Heat of Reactions:
You will need to calculate the heat of dilution for both the HCl and the NaOH. For
NaOH, determine the initial temperature by the initial temperature of the NaOH and the
water; if they are different, use the average as your initial temperature. The final Dakota State University page 180 of 232 Experiment 17: Calorimetry General Chemistry I and II Lab Manual temperature is the maximum temperature reached on mixing. The heat of mixing, then, is
simply
∆HNaOH,mix =m∆ T
where ∆ T is the change in temperature and m is the mass of the total solution (take it to
be 40 grams; we are assuming the density is 1.00 g/mol). In an analogous fashion,
determine ∆HHCl,mix .
The total energy change, ∆ Htotal is determined the same way as above, but use the run
where you mixed the HCl and NaOH. To find the heat of reaction, note that
∆Htotal=∆Hrxn +∆HNaOH,mix +∆HHCl,mix
Use the values you’ve calculated to find ∆Hrxn; this is the heat of reaction. Divide this
value by the number of moles of acid (or base, whichever is less) as calculated by the
concentration of the acid or base and the volumes we’ve...
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 Fall '11
 Kumar
 Chemistry, pH, The Crucible, Dakota State University

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