# Concept of Bomb caloriemeters

Considering the example where $0.3212\ \mathrm g$ of glucose (molar mass = 180) being burnt in bomb calorimeter of heat capacity $641\ \mathrm{J\ K^{-1}}$: the temperature rises by $7.793\ \mathrm K$. Is the heat (say '$E$') obtained taking $E=641 \times 7.793 \times 180/0.3212$ the internal energy or heat of combustion? Why? Answer explained in detail will be highly appreciated!

What is the fundamental difference when we completely burn fumaric acid in a bomb calorimeter? Why do we take the heat released as the internal energy? IS it because volume is constant?

Reference: Elements of Physical Chemistry (Paula, Atkins) Q.no:3.21 and 3.22

Let me list some equations first: $$w = -p\Delta V,\qquad\Delta U = q + w,\qquad \Delta H = \Delta U + \Delta (pV)$$ where $w$ is work, $U$ is internal energy, $q$ is heat transfer and $H$ is enthalpy

Now, when you are trying to calculate the internal energy, enthalpy or anything else, you should first note what values are constant and what values change through out the reaction. For example, if I burnt a sample of glucose in an open beaker, the volume would obviously change, but the pressure would remain constant. This is because the beaker is open, which means that pressure exerted on the products and reactants is the external atmospheric pressure which is always remains a constant value. Noting the fact that pressure is constant is important as it allows us to simply the equations that I have listed above. $$q = \Delta U - w = \Delta U + p\Delta V$$ $$\Delta H = \Delta U + p\Delta V = q$$ Therefore, by realising that pressure is constant, we are able to state the $\Delta H = q$

Now lets consider the case of burning glucose in a bomb calorimeter. Here is a diagram of one below:

So lets consider what values remain constant and what values change throughout the combustion. As you should be able to see, the bomb calorimeter has a fixed volume (it can't expand or contract), hence $\Delta V = 0$. Since the bomb calorimeter is a closed object, the pressure on the system is the internal pressure, not the atmospheric pressure which is a constant. This internal pressure will change as the reaction proceeds (in this case it will increase as gases are produced from the combustion). Therefore $\Delta p \neq 0$.

Now, using the data from above, we can obtain the following expressions: $$w = 0,\qquad\Delta U = q,\qquad\Delta H = q + V\Delta p$$

So when you do the calculation: $641\ \mathrm{J\ K^{-1}}\times7.793\ \mathrm K$ you are finding the total heat transfer ($q$) from the system to the surroundings (the bomb calorimeter). Now form the above information we know that the total heat transfer is also equal to change in internal energy, therefore: $$641\ \mathrm{J\ K^{-1}}\times7.793\ \mathrm K = \Delta U$$

I think you were trying to find the molar internal energy for the reaction, which the calculation will be then: $$\Delta U_\mathrm m = \frac{641\ \mathrm{J\ K^{-1}}\times7.793\ \mathrm K\times180\ \mathrm{g\ mol^{-1}}}{0.3212\ \mathrm g}$$