This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
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 datas 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....
View Full Document