{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–2. Sign up to view the full content.

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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern