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Question:

When $\pu{60 g}$ of $\ce{C60}$ is combusted in a bomb calorimeter that has a water jacket containing $\pu{300.0 g}$ of water, the temperature of the water increases by $\pu{10 ^\circ C}$. Assuming that the specific heat of water is $\pu{4.18 J g^{-1} K^{-1}}$, estimate $\Delta E$ of combustion per mole of $\ce{C60}$.

Answer: $\pu{-150.5 kJ/mol}$

What I have tried: $$ \begin{align} Q &= m \times c \times \Delta T \\ Q &= 300 \times 4.18 \times 10 \end{align}$$

That's the only equation I know and it doesn't seem to work for this problem.

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    $\begingroup$ 1) chemistry help question and more chemistry help? are not titles that would be in any way likely to help future visitors of the site. You have a rather specific problem in both your questions, you should have titled them appropriately. 2) I don’t believe that that be the only equation you know. $\endgroup$ – Jan Oct 27 '15 at 13:02
  • $\begingroup$ @Jan Ok, I changed the title and obliviously I know more equations but in this regard I don't. I don't know any other equation that could be applied to solve this problem, so for the sake of other future visitors to this site can you at least point me in the right direction because i'm completely lost $\endgroup$ – user3882522 Oct 27 '15 at 13:07
  • $\begingroup$ Suggestion 1: Add units. (That should be a given for all equations.) Then you should realise what you have just calculated, what that is equivalent to physically and maybe then you can see which step is remaining. $\endgroup$ – Jan Oct 27 '15 at 13:09
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    $\begingroup$ One last time: Please use units, it gets you so much further. $\endgroup$ – Jan Oct 27 '15 at 13:26
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    $\begingroup$ Ah, we’re almost there. Why are you using $5~\mathrm{mol}$ of carbon, because in the question there clearly are not $5~\mathrm{mol}$ of carbon. Check out which type of carbon you are using. $\endgroup$ – Jan Oct 27 '15 at 13:39
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You have 60 g of $\ce{C60}$ (which has a molecular weight of 720.6), so that means you'll need to multiply your answer by 720.6/60 which (cleverly) turns out to be 12.01. Another way of saying it is that you have calculated the heat of combustion for 1/12.01 moles of $\ce{C60}$, so you'll need to multiply your answer by the inverse in order to get the desired (per mole) quantity.

Using your notation, numbers, and noting that a change in one degree Celsius is equivalent to a change in one kelvin:

$$\Delta E_{\rm molar} = -12.01\;\mathrm{mol} \cdot 300\;{\mathrm g} \cdot 4.18\;{\mathrm J} \cdot {\mathrm g}^{-1}\, \cdot {\mathrm K}^{-1}\, \cdot 10\;{\mathrm K} = -150.6054\;\mathrm{kJ}\cdot\mathrm{mol}^{-1}$$

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  • $\begingroup$ thank you for your answer, I've been trying to figure this out for hours $\endgroup$ – user3882522 Oct 27 '15 at 14:11

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