Textbook 4A.13

[005778617]

Hello everyone,

I don't quite know what to do first in this problem. Can someone explain?

A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solution in the calorimeter ,(q = -3.50 kJ) resulting in a temperature rise of 7.32°C. In a subsequent experiment, 100.0 mL of 0.200 M HBr(aq) and 100.0 mL of 0.200 M KOH (aq) were mixed in the same calorimeter and the temperature rose by 2.49 °C. What is the change in the internal energy of the reaction mixture as a result of the neutralization reaction?

[Allison Peng 1D]

When the calorimeter is calibrated, it compares a known heat change with a measured change in temperature to determine Ccal (in this case, it is Ccal for a calorimeter with 200mL of solution in it). Therefore, by using the first sentence of the question, you can determine Ccal.

In the second part of the problem, you can use the Ccal you found and the new measured temperature change to determine the heat that the second reaction releases (using q=Ccal*(deltaT)). It's important to check that the volume of solution in the second reaction is the same as the volume that you calibrated the calorimeter with (200mL, in this case), otherwise the Ccal value will not be accurate.

[MaiVyDang2I]

For the first part, the q given is of the system, and we can use that to find q of calorimeter to find C. q of the calorimeter would be the same magnitude but opposite sign of q of system. And C can be found using qcal = C delta T.
Then for the second part, you can use that C to find q of the calorimeter (qcal = Ccal delta T). Then the opposite sign of that would be q of the system.