chemistry 101
Assume you are performing the calibration step of Experiment 8 and you begin with 50 g of water at 20 oC and 50 g of water at 80 oC. After adding the two portions of water into your calorimeter setup and following the procedure outlined in the experiment, you determine the temperature of the mixed portions of water to be 45 oC. What is the heat capacity of the calorimeter?
Assume room temperature is 25 oC.
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Answers
The heat capacity of a calorimeter can be determined using the following equation:
Heat Capacity (c) = q/(ΔT)
Where q is the heat transfer (in Joules) and ΔT is the change in temperature (in oC).
For this example, q is the sum of the heat lost by the water at 80 oC and the heat gained by the water at 20 oC, or:
q=(50g x 4.184 J/g/oC x (80 oC - 45 oC)) + (50g x 4.184 J/g/oC x (45 oC - 20 oC)).
q = 12,276 Joules
Therefore, the heat capacity of the calorimeter is:
c = q/(ΔT) = 12,276 J / (45 oC - 25 oC) = 541 Joules/oC
