50.0 mL of water at 40.5 °C is added to a calorimeter containing 50.0 mL of water at 1 After waiting for the system to equilibrate, the final temperature reached is 28.3 °C. Calcu heat capacity of the calorimeter. (sp_heat of water = 4.184 J/gx°C) Athot Atcold qhot qcold = qcal = 28.3 °C -40.5 °C -12.2 °C 28.3 °C -17.4 °C 10.9 °C mxsp_heat At (50.0 g)x(4.184 J/gx°C)x(-12.2 °C) -2550 J mxsp_heat At (50.0 g)x(4.184 J/gX°C)x(10.9 °C) 2280 J By rearranging the equation, qhot ' IV - (q cold +qcal), we can solve for q cal: qhot - qcold 2550 J - 2280 J 270 J 2701
50.0 mL of water at 40.5 °C is added to a calorimeter containing 50.0 mL of water at 1 After waiting for the system to equilibrate, the final temperature reached is 28.3 °C. Calcu heat capacity of the calorimeter. (sp_heat of water = 4.184 J/gx°C) Athot Atcold qhot qcold = qcal = 28.3 °C -40.5 °C -12.2 °C 28.3 °C -17.4 °C 10.9 °C mxsp_heat At (50.0 g)x(4.184 J/gx°C)x(-12.2 °C) -2550 J mxsp_heat At (50.0 g)x(4.184 J/gX°C)x(10.9 °C) 2280 J By rearranging the equation, qhot ' IV - (q cold +qcal), we can solve for q cal: qhot - qcold 2550 J - 2280 J 270 J 2701
Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter12: Thermodynamic Processes And Thermochemistry
Section: Chapter Questions
Problem 14P
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Hello, I just want to know if the work i attached for a calorimetry lab is correct, is the qcal supposed to be positive in the reaction or is teh work shown incorrect
Transcribed Image Text:50.0 mL of water at 40.5 °C is added to a calorimeter containing 50.0 mL of water at 1
After waiting for the system to equilibrate, the final temperature reached is 28.3 °C. Calcu
heat capacity of the calorimeter. (sp_heat of water = 4.184 J/gx°C)
Athot
Atcold
qhot
qcold
=
qcal
=
28.3 °C -40.5 °C
-12.2 °C
28.3 °C -17.4 °C
10.9 °C
mxsp_heat At
(50.0 g)x(4.184 J/gx°C)x(-12.2 °C)
-2550 J
mxsp_heat At
(50.0 g)x(4.184 J/gX°C)x(10.9 °C)
2280 J
By rearranging the equation, qhot '
IV
- (q cold
+qcal), we can solve for q cal:
qhot - qcold
2550 J - 2280 J
270 J
2701
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