You have been asked to experimentally determine the Calorie (1 Calorie = 1000 ca
ID: 1068800 • Letter: Y
Question
You have been asked to experimentally determine the Calorie (1 Calorie = 1000 calories) values of various desserts. You realize that a combustion reaction is analogous to the digestive reaction occurring during metabolism and the amount of energy released from combustion will tell us how much nutritional energy we can expect from the food. You then need to apply the principles of calorimetry to measure the energy released by the desserts upon combustion. After some research, you discover that calorimetry of combustion reactions is performed in an instrument called a bomb calorimeter, which is shown below. The name “bomb” refers to the inner chamber of the calorimeter that can withstand great increases in pressure.
7. Fill in the following blanks for a true statement regarding the heat exchange in the cotton candy calorimeter.
+qwater = ___q____
+qwater =___________ q ____________
Hint: Look in your lab manual introduction. Use a single word.
8.Your cotton candy experiment was run in a calorimeter with 483 mL of water that started at 20.7°C. After combusting 2.18 g of cotton candy, the temperature of the water was measured at 35.5°C. Find the heat (in J) absorbed by the water, qwater.
9.These numbers will be different from the previous numbers!
Your cotton candy experiment was run in a calorimeter with 316 mL of water that started at 24.0°C. After combusting 1.50 g of cotton candy, the temperature of the water was measured at 41.5°C. How many Calories of energy will be released from metabolizing one 60.0 g serving of cotton candy?
Hint: 1 Calorie = 4184 J. Assume all heat from combustion is transferred with 100% efficiency to the water.
10. For the previous question, your final answer (in Calories/serving) would have________ significant figures.
Explanation / Answer
8) heat absorbed by water(q) = m*s*DT
m = mass of water = 483 g ( density of water = 1 g/cc)
s= specific heat of water = 4.184 j/g.c
DT = 35.5-20.7
q = 483*4.184*(35.5-20.7) = 29.908 kj
9)
heat absorbed by water(q) = m*s*DT
m = mass of water = 316 g ( density of water = 1 g/cc)
s= specific heat of water = 1.0 cal/g.c
DT = 41.5 - 24
q = 316*1.0*(41.5 - 24 ) = 5530 cal = 5.5*10^3 cal
10 . two significant figures.
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