8.25

[Remi Lathrop 1G]

This question is essentially a 2 part question, the first which asks us to find the specific heat capacity of the calorimeter which I understand how to do. The C(calorimeter) = 0.478 kJ/˚C
The second part says that 100.0 mL of 0.200 M HBr(aq) was mixed with 100.0 mL of 0.200 M KOH(aq) in the same calorimeter and that the temperature rose by 2.49˚C and then asks for the change in the internal temp.

the solution manual says q= (2.49) (0.478) but does not include the 200mL (200g) of solution. Why is this?

[Sean Monji 2B]

In the first part of the problem, you find the heat capacity (the kilojoules required to raise the whole calorimeter by a degree kelvin), which is a ratio that can be applied to all reactions in that specific calorimeter where there is a change in temperature. This unlike specific heat capacity where you need to know the amount of the substance in the reactions. Therefore, the amount of the substance isn't needed for this problem, and you can multiply the ratio (kilojoules / change in kelvin) by the change in kelvin of the calorimeter created by the reaction to find the amount of kilojoules produced by the reaction.
Since the volume is constant, w = 0, and therefore the change in internal energy is equal to the kilojoules produced by the reaction.

Btw, the insides of the calorimeter must be a closed, not isolated system, for a change in internal energy to be possible. Because the reaction itself is losing energy to the surroundings (seen in the rise of temperature of the calorimeter), the change in internal energy must be negative.
Hope that makes sense