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110-102 lab #10

110-102 lab #10 - T hot = 54.7 – 78.1 =-23.4 C M cool =(d...

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Kate Stonesifer Chem 110-102 November 16, 2009 Team: John C. General Chemistry 1 Experiment # 10: Calorimetry: Heat of Reaction Part: A, Determination of the Heat Capacity of the Calorimeter
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Introduction : In this experiment we used a calorimeter to determine the heat of deionizer water. To do this we heated water and took the temperature of it before and after heating, then we got more water and recorded its temperature. By placing the heated water and the room temperature water into the calorimeter, we were able to record the exact heat in degrees Celsius. With this data we calculated C cal and found that our range was a little above the 15/50 J/degrees C, we expected to be in. Method: See Lab manual Data: Temp ( C) Volume (mL) Initial Temp 22.1 84 Hot Temp 78.1 Cool Temp 23.0 42 When Mixed Temp 54.7 Density of water at 22.1 C is .99775 Calculations: M hot = (d x v = m) so (.99775) x (84) = 83.811
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Unformatted text preview: T hot = 54.7 – 78.1 = -23.4 C, M cool = (d x v=m) so (.99775) x (42) = 41.9055 T cal = T cool so 54.7- 23 = 31.7 C, S hot = 4.18, S cool = 4.18 Questions: Part A 1) No because our data for mass was recorded before we heated anything. 2) T hot would increase in size and therefore decrease the number we used in our calculations when subtracting and then dividing. This would make C cal , in our data’s case, smaller and more accurate to the 15/50 J/degrees C. Conclusion: Through the use of a calorimeter, we were able to determine the exact heat loss in the reaction when hot water was mixed with cold water. Our data was not perfect and we got 83.43 J/ C, which was slightly out of the range expected of 15/50 J/ C, yet this could be due to incorrect measurements in the data or the calculations of the data themselves....
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110-102 lab #10 - T hot = 54.7 – 78.1 =-23.4 C M cool =(d...

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