Why would you have to subtract the heat capacity of the calorimeter when calculating the heat of the reaction?

Question

aaronq · Answer

you wouldnt subtract it, you would add it to account for the thermal energy absorbed by the calorimeter itself. The set-up looks something like this:|dw:1405199730169:dw| The water and the calorimeter absorb the heat being given off by the (exothermic) reaction.

                         $\sf \large \underbrace{-q}_{reaction}=\underbrace{q}_{water+calorimeter}$

Look at the equation below that accounts for the heat absorbed,

       $\large \sf q_{absorbed}=\underbrace{m_{water}*C_p*\Delta T}_{water}+\underbrace{C_p\Delta T}_{calorimeter}$

The first part accounts for the thermal energy absorbed by the water and, the second, for the thermal energy absorbed by calorimeter. These 2 together are the total thermal energy  released by a reaction.