DATA: Calorimteter: Trial 1 Mass of cold water = 50.0g, mass of hot water = 50.0
ID: 1003456 • Letter: D
Question
DATA: Calorimteter:
Trial 1 Mass of cold water = 50.0g, mass of hot water = 50.0g, temperature of hot water = 64.8 degree celsius, temperature of cold water = 22.8 degrees celsius, final temperature of mixed water = 22.7 degrees celsius.
Trial 2 Mass of cold water = 50.0g, mass of hot water = 50.0g, temperature of hot water = 45.6 degrees celsius, temperature of cold water = 23.2 degrees celsius, final temperature of mixed water = 33.6 degrees celsius
Data: Reactions
Reaction 1: Mg + 2HCl --> MgCl2 + H2 Mass of Mg = 0.116g, volume of HCl = 100.0mL, temperature of HCl = 22.7 degrees celsius, Final temperature = 27.8 degree celsius
Reaction 2: MgO + HCl --> MgCl2 + H2O Mass of MgO = 0.731, volume of HCl = 100.0mL, temperature of HCl = 22.5 degree Celsisus, Final temp = 28.7 degrees celsius
Questions: Calculate the calorimeter constant, Calculate q of calorimeter, Calculate q of contents (assume density of solution is 1.00g/mL, mass of content is solution mass plus solid mass), calculate q of reaction
Explanation / Answer
From Trial 2 data:
Heat lost by hot water = mCpdT = 50 x 4.184 x (45.6 - 336) = 2510.4 J
Heat gained by cold water = mCpdT = 50 x 4.184 x (33.6 - 23.2) = 2175.68 J
Heat gained by calorimeter = 2510.4 - 2175.68 = 334.72 J
Calorimeter constant = 334.72/(33.6 - 23.2) = 32.184 J/oC
q(calorimeter) = 334.72 J
From reaction 1: q(reaction) = mCpdT = 100 ml x 1.00 g/ml x 4.184 J/oC x (27.8 - 22.7) = 2133.84 J
q(contents) = mCpdT = 100.116 x 4.184 x (27.8 - 22.7) = 2136.31 J
similarly the calculations for other Trial and second reaction can be done.
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