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ANSWERS_TO_THE_QUESTIONS 6 - ANSWERS TO THE QUESTIONS 1 If...

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ANSWERS TO THE QUESTIONS 1) If our value were off by 1% from the accepted value of the calorimeter constant, it would be seen the same amount of the in the calculations. The change is directly proportional to the heat of reaction: Q = - W x ΔT W = 2420 cal => W’’ = 2420 x 99 / 100 = 2395.8 cal = 10.03 kJ Q = - (10.03 kJ) x (2.15) = -21.56KJ The value we found before was (-21.78+21.56)/-21.78 is nearly 0.01. 2)By measuring heat capacities by a calorimeter, we determine the heat capacities experimentally by measuring heat of reaction before and after the reaction when known amount of heat is transferred into calorimeter. The heat capacity of the calorimeter is actually the water equivalent. The sum of the heat capacities of the parts of the calorimeter gives us the water equivalent of the calorimeter. So water equivalent is the heat capacity of the calorimeter.And we kow that   water equivalent is the heat capacity of the calorimeter. 3)In the first part in the graph (0-5 min) is, which is the time need to make temperature of all parts of the calorimeter nearly same; therefore, there is no significant temperature change.
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