Asked by chloe
In an experiment to determine heat loss to a calorimeter, 100.0 mL of cool water was measured into the calorimeter and allowed to reach room temperature before 100.0 mL of heated water was added. The results of this experiment produced the data below. What is the absolute value of the calorimeter constant in J / ºC?
ΔTheated water = -6.2 ºC
ΔTcool water= 11.2 ºC
Specific heat of water = 4.184 J deg-1 g-1
Density of water = 1 g / mL
Mass of calorimeter = 40.91 g
ΔTheated water = -6.2 ºC
ΔTcool water= 11.2 ºC
Specific heat of water = 4.184 J deg-1 g-1
Density of water = 1 g / mL
Mass of calorimeter = 40.91 g
Answers
Answered by
DrBob222
I am confused by the wording in the problem. For example, let's say room T is 10 C so we put 100 mL of cool water in the calorimeter and the final T is 3.8 to make dT = -6.2. Then we allow the system to warm back up to 10 C? Is that right?
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.