Asked by Lance
Calculate the calorimeter constant if 25g of water at 50C was added to 25g of water at 25C with a resulting temperature of 35C?
The answer is 52.3 but I do not know how to get to this. I hope someone can help. Thanks!
The answer is 52.3 but I do not know how to get to this. I hope someone can help. Thanks!
Answers
Answered by
DrBob222
heat lost by 50C H2O + heat gained by 25C H2O + heat gained by calorimeter = 0
[mass H2O x sp.h. x (Tfinal-Tintial)] + [mass H2O x sp.h. x (Tfinal-Tinital)] + Qcal((Tfinal-Tinitial) = 0
[25 x 4.18 x (35-50)] + [25 x 4.18 x (35-25)] + Qcal(35-25) = 0
Solve for Qcal. I get 52.25 which rounds to 52.3.
[mass H2O x sp.h. x (Tfinal-Tintial)] + [mass H2O x sp.h. x (Tfinal-Tinital)] + Qcal((Tfinal-Tinitial) = 0
[25 x 4.18 x (35-50)] + [25 x 4.18 x (35-25)] + Qcal(35-25) = 0
Solve for Qcal. I get 52.25 which rounds to 52.3.
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