How does the calorimeter affect heat released by the system (q)?
Let's take a numerical example. $0.2432$ $g$ magnesium metal ($0.01$ mol) is introduced into a calorimeter containing $50~ g$ HCl $1$ M. The chemical reaction will be $$\ce{Mg + 2 H^+ -&gt;  Mg^{2+} + H2}$$ The composition of the solution is changing between the beginning and the end of the reaction. There is an excess of HCl.  But let's admit in first approximation that the specific heat of the solution does not change and remains as in water ($C_w = 4.185~ J g^{-1} K^{-1}$). The tabulated enthalpy change of the reaction is - $462~ kJ/mol$. The total amount of heat $Q$ produced by the reaction of $0.01 mol$ Mg is : $$Q\ce{= 0.01 mol·462 kJ/mol = 4.620 kJ = 4620 J}$$
If the calorimeter was ideal, without any weight, and with the already mentioned approximation, the change of temperature should be : $$\Delta T = \frac{Q}{C_w·m} =  \frac{4620~ J}{4.185~J g^{-1} K^{-1}·50~ g} = \frac{4620~ J}{209.25~ J K^{-1}} = 22.07 K$$
Now suppose that the calorimeter is made of copper and weighs $m_{Cu} = 20~ g$. The heat capacity of copper is $C_{Cu} = 0.385~ J g^{-1} K^{-1}$. The change of temperature will be reduced and becomes $$\Delta T = \frac{Q}{C_w·m + C_{Cu}·m_{Cu}} =  \frac{4620~ J}{(209.25 ~J K^{-1})~ + (0.385 ~J g^{-1} K^{-1}·20~ g)} = 21.3 K$$
And now if the calorimeter is not well insulated, the final temperature change will be smaller than $21.3 K$.
Sorry for my poor management of the italicized letters. Any help would be appreciated.