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A $\pu{25.95 g}$ sample of methanol at $\pu{35.6 ^\circ C}$ is added to a $\pu{38.65 g}$ sample of ethanol at $\pu{24.7 ^\circ C}$ in a constant pressure calorimeter. If the final temperature of the combined liquids is $\pu{28.5 ^\circ C}$ and heat capacity of the calorimeter is $\pu{19.3 J/^\circ C}$ determine the specific heat of methanol. The heat capacity of ethanol is $\pu{2.46 J/(g * ^\circ C)}$.

\begin{align} q_\mathrm{methanol} &= -q_\mathrm{ethanol}\\ msT &= -msT\\ \pu{25.95 g} \cdot (\pu{28.5 ^\circ C} - \pu{35.6 ^\circ C}) \cdot s &= -\pu{38.65 g} \cdot (\pu{28.5 ^\circ C} - \pu{24.7 ^\circ C}) \cdot \pu{2.46 J/(g* ^\circ C)}\\ \therefore s &= \pu{1.96 J/(g * ^\circ C)} \end{align}

The correct answer, however, is $\pu{2.36 J/(g * ^\circ C)}$. I think that there is a mistake in my solution, because the heat capacity of the calorimeter is given, and I didn't use it.

Could anyone explain where I have been wrong?

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  • $\begingroup$ It's a good idea to include units in your equation. Where does the number 38.5 come from? (I think you mean 28.5). Also, why are you multiplying the right side by 2.64? $\endgroup$ – Klik Jul 3 '14 at 12:20
  • $\begingroup$ yes i meant that.. 2.46 is the specific heat of ethanol I got it from the book $\endgroup$ – Maher Jul 3 '14 at 12:24
  • $\begingroup$ You don't mention ethanol in your problem, you only mention methanol. Could you double check the question you've written. Is it that you are given a sample of methanol and it is added to a sample of ethanol? $\endgroup$ – Klik Jul 3 '14 at 12:50
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You should have included the heat capacitance of the calorimeter. Think about it, if the calorimeter was at $\pu{24.7 ^\circ C}$ before the methanol was added and then it rose $\pu{3.8 ^\circ C}$ along with the liquids, where did the energy come from to increase its temperature? Since you know that the heat capacitance of the calorimeter is $\pu{19.3 J/^\circ C}$ then you can find out how much energy the calorimeter absorbed to increase in temperature by multiplying its heat capacitance by its temperature increase. I.e. $\pu{19.3J/^\circ C} \cdot \pu{3.8 ^\circ C} = $ "Joules absorbed by the calorimeter".

Then we simply apply the heat lost equals heat gained concept:

$$\begin{multline} \pu{25.95 g} \cdot (\pu{35.6 ^\circ C} - \pu{28.5 ^\circ C}) \cdot S \\ = \pu{38.65 g} \cdot (\pu{28.5 ^\circ C} - \pu{24.7 ^\circ C}) \cdot \pu{2.46 J/(g* ^\circ C)} + \pu{19.3J/^\circ C} \cdot \pu{3.8 ^\circ C} \end{multline}$$

After rearranging, we get $S = \pu{2.36 J/(g * ^\circ C)}$.

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