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In a Bomb Calorimeter, we measure the change in temperature of water $\Delta T$. But when it comes to calculating the change in internal energy we use $$\Delta U = C_{cal}\Delta T$$ Where $C_{cal}$ is the specific heat capacity of the whole calorimeter. Why is that? What I mean is that we're measuring the change in temperature of the water and assuming that heat from the reaction transferred completely to water, so shouldn't we apply $$\Delta U = C_{water}\Delta T$$ since we're calculating the change in temperature of the water?

Or are we assuming that the change in temperature of the water is the change in temperature of every element in the calorimeter?

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  • $\begingroup$ Can you give a reference i.e where you learned this? $\endgroup$
    – Buraian
    Oct 6 '20 at 10:35
  • $\begingroup$ This isn't where I learnt but it says the same thing chem.libretexts.org/Bookshelves/… $\endgroup$
    – Eyy boss
    Oct 6 '20 at 10:38
  • $\begingroup$ Oh btw I think you should change your title because right now it's really broad $\endgroup$
    – Buraian
    Oct 6 '20 at 11:52
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Basically, a bomb calorimeter consists of a small cup to contain the sample, oxygen, a stainless steel bomb, water, a stirrer, a thermometer, the dewar or insulating container (to prevent heat flow from the calorimeter to the surroundings), and ignition circuit connected to the bomb. (*)

The whole device known as calorimeter consists of a bomb and a water bath, they are not two separate entities. So, $ C_{cal}$ is actually referring to the heat capacity required of the steel vessel and also the water. The construction of the calorimeter is such that the thermal mass of the calorimeter system completely absorbs the heat of reaction (**).


*: This Wikipedia

**: Refer 29:27 of this video

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  • $\begingroup$ So the change in temperature is actually tge change in temperature of whole apparatus? $\endgroup$
    – Eyy boss
    Oct 6 '20 at 11:07
  • $\begingroup$ Yes, the calorimeter is the whole thing not just the bomb $\endgroup$
    – Buraian
    Oct 6 '20 at 11:08
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    $\begingroup$ It's even said so in ncert, see page-163 under (a) $ \Delta U$ measurements [ I'm assuming you study in cbse] $\endgroup$
    – Buraian
    Oct 6 '20 at 11:14
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In your first equation, the $\Delta U$ is not the change in internal energy of the gas reaction. It is the change in internal energy of the surroundings (the calorimeter and water), which is the heat transferred from the reaction mixture to the calorimeter and water. Since the calorimeter/water and the reaction chamber are both of constant volume, the $\Delta U$ in this equation is equal to minus the change in internal energy due to the reaction.

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  • $\begingroup$ But that means that change in temperature of water and the calorimeter is same. Is that true? Also what about the second equation $\Delta H = \Delta U + \Delta n_g RT$ $\endgroup$
    – Eyy boss
    Oct 6 '20 at 4:55
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    $\begingroup$ Yes, the change in temperature of water and calorimeter are the same at final equilibrium. The second equation is used for a reaction involving ideal gases to get $\Delta H$ once $\Delta U$ is known. It follows from $$\Delta H=\Delta U+\Delta PV=\Delta U+(\Delta n_g)RT$$ $\endgroup$ Oct 6 '20 at 11:15

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