Dawson H.

asked • 03/02/21

This one I am stuck....High-purity benzoic acid (C6H5COOH; ΔHrxn for combustion = −3227 kJ/mol) is used as a standard for calibrating bomb calorimeters. A 1.221-g sample burns in a calorimeter...

(heat capacity = 1365 J/°C) that contains exactly 1.370 kg of water. What temperature change is observed?

My steps:

-qreaction = - (1.221g HBZ) X 1 mol HBZ over 1221.2g HBZ X -3227 kJ over 1 mol HBZ X 103 J over 1 kJ = 3.22647 X 104 J

3.22647 X 104 J = 4.184 J over g 0C X (1.370 X 103 g) X ΔT + 1365 J over 0C X ΔT

3.22647 X 104 = 5732 X ΔT) + (1365 X ΔT)

ΔT =

1 Expert Answer


J.R. S.

What about the heat absorbed by the water in the calorimeter? Shouldn't that be included in the calculation. q = mC∆T + Ccal∆T? 319 70 = (1370 g)(4.184 J/gº)(∆T) + (1365)(∆T)


Ramesh V.

Thank you so much Mr. JRS


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