Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
Assume you are performing the calibration step of Experiment 8 and you begin with 80 g of water at 20 oC and 80 g of water at 8...Asked by Sam
Assume you are performing the calibration step of Experiment 8 and you begin with 50 g of water at 20 oC and 50 g of water at 80 oC. After adding the two portions of water into your calorimeter setup and following the procedure outlined in the experiment, you determine the temperature of the mixed portions of water to be 45 oC. What is the heat capacity of the calorimeter?
Assume room temperature is 25 oC.
Assume room temperature is 25 oC.
Answers
Answered by
DrBob222
Some of the water gained heat. The calorimeter gained heat. Some of the water lost heat.
heat gained + heat lost = 0
I would calculate heat lost by water, subtract heat gained by water. The difference is the heat gained by the calorimeter.
heat gained = mass water x specific heat water x (Tfinal-Tinitial).
heat lost = mass water x specific heat water x (Tfinal-Tinitial).
Take the difference. That's the heat gained by the calorimeter = heat capacity x delta (Tfinal-Tinitial).
heat gained + heat lost = 0
I would calculate heat lost by water, subtract heat gained by water. The difference is the heat gained by the calorimeter.
heat gained = mass water x specific heat water x (Tfinal-Tinitial).
heat lost = mass water x specific heat water x (Tfinal-Tinitial).
Take the difference. That's the heat gained by the calorimeter = heat capacity x delta (Tfinal-Tinitial).
Answered by
elizabeth
1008 cCal
Answered by
Anonymous
25 cCal
Answered by
Addison
15 cal/C
Answered by
Cameron
25cal/C
There are no AI answers yet. The ability to request AI answers is coming soon!