# How much heat is required to raise the temperature of 230 g CH3OH(l) from 22.0 to 30.0 ∘C?

How much heat is required to raise the temperature of $230\ \mathrm g$ $\ce{CH3OH(l)}$ from $22.0\ ^\circ\mathrm C$ to $30.0\ ^\circ\mathrm C$ and then vaporize it at $30.0\ ^\circ\mathrm C$? Use enthalpy of vaporization $\ce{CH3OH(l)} = 38.0\ \mathrm{kJ\ mol^{−1}}$ and a molar heat capacity of $\ce{CH3OH(l)}$ of $81.1\ \mathrm{J\ mol^{−1}\ K^{−1}}$.

I Used the equation $Q_\text{total} = Q_\text{heat} + Q_\text{vap}$. My answer was $88.9\ \mathrm{kJ}$.

However, it is wrong.

• Without seeing your work it's difficult to tell where you went wrong. Your basic method is correct though. Just to check, I get 7.18 moles of methanol. Just considering the enthalpy of vaporization, that's going to be much more than 89 kJ. Sep 17, 2015 at 3:05

The first thing to be sure of is that you're doing all the conversions correctly. Molar mass is the number of grams in a mol. It's found by adding the molar mass of all the elements in a compound times the number in the compound. For example the molar mass of $CH_4$ would be $$\textrm{Molar Mass of }C + 4\times\textrm{ Molar Mass of }H$$ Once you've found the correct molar mass it should have units of mass per mol - most commonly $\textrm{grams mol}^{-1}$. Thus if you have the mass of a substance and the mass required to have a mol, you can divide mass by molar mass to find moles. Then you simply: $$Q_{heat} = \Delta T\times \textrm{ number of moles } \times C_p$$ Where $C_p$ is heat capacity. To find the heat of vaporisation again you just multiply the per mol value by the number of moles: $$Q_{vap} = \textrm{ number of moles } \times \Delta_{vap}H^\circ$$ Where $\Delta_{vap}H^\circ$ is the heat of vaporisation. Be sure you got all these steps right and be aware of the difference between kJ and J! ($1\textrm{ kJ} = 1000\textrm{ J}$) Using this method I got a very different answer to yours. Feel free to post your own working for us to look at!