# 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} &= -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?

• 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?
– Klik
Jul 3, 2014 at 12:20
• yes i meant that.. 2.46 is the specific heat of ethanol I got it from the book Jul 3, 2014 at 12:24
• 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?
– Klik
Jul 3, 2014 at 12:50

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".
$$\begin{multline} \pu{25.95 g} \cdot (\pu{35.6 ^\circ C} - \pu{28.5 ^\circ C}) \cdot S \\ = \pu{38.65 g} \cdot (\pu{28.5 ^\circ C} - \pu{24.7 ^\circ C}) \cdot \pu{2.46 J/(g* ^\circ C)} + \pu{19.3J/^\circ C} \cdot \pu{3.8 ^\circ C} \end{multline}$$
After rearranging, we get $$S = \pu{2.36 J/(g * ^\circ C)}$$.