# How does the calorimeter affect heat released by the system (q)? [closed]

Here are the 2 questions I am pondering about

1. How would the calculated quantity of heat (q) released by a system differ if the calorimeter's heat capacity was taken into account?

My thought process is that the calculated quantity of heat (q) released by a system would increase if the calorimeter's heat capacity was taken into account, b/c the calorimeter would aid in absorbing the heat along with the surroundings, so change in temperature would increase ($$q=m c \Delta T$$)

1. How would quantity of heat (q) released by a system differ if the calorimeter was not a good insulator?

Here, I am not sure how to word the answer, but if the calorimeter wasnt a good insulator, heat released by a system would be lost to the environment, and thus less heat released would be calculated via $$q=m c \Delta T$$ due to a smaller change in $$\Delta T$$.

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 $$\ce{Mg + 2 H^+ -> Mg^{2+} + H2}$$ 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 $$21.3 K$$.