I saw this basic calorimetry question and it included the heat capacity of the calorimeter (1550 J/°C) but it also included that there were 150 g of water inside the calorimeter as well (4.184 J/(G°C). I saw that in order to calculate the total q, you add q of calorimeter to q of water. I'm having a hard time envisioning why you have to add the two instead of taking only the q of the calorimeter (isn't q of water included in the q of calorimeter?). The question stated that ∆T was equal to +1.211°C.

  • $\begingroup$ No, the heat capacity of the specific calorimeter being used must be determined experimentally. You can look up the heat capacity of water in a table. $\endgroup$
    – MaxW
    Apr 11, 2020 at 2:03
  • $\begingroup$ Think of the calorimeter as the container. Its heat capacity is fixed. Then you add to this the heat capacity of the water, a figure that can vary depending on how much water you've poured into the calorimeter. $\endgroup$
    – theorist
    Apr 11, 2020 at 6:42
  • $\begingroup$ Oh ok, so let’s say that there’s a temperature change, would that temperature change be equal to both the water and the calorimeter? $\endgroup$
    – biryaniboi
    Apr 11, 2020 at 10:09


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