How to determine the specific heat of methanol from calorimeter data? 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} &amp;= -q_\mathrm{ethanol}\\
  msT &amp;= -msT\\
  \pu{25.95 g} \cdot  (\pu{28.5 ^\circ C} - \pu{35.6 ^\circ C}) \cdot  s
    &amp;= -\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 &amp;= \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?
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} = $ &quot;Joules absorbed by the calorimeter&quot;.
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 $S = \pu{2.36 J/(g * ^\circ C)}$.