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The question I'm trying to answer is:

Calculate the mass of butane gas that would be needed to heat $724\ \mathrm{cm^3}$ of water from an initial temperature of $7.44\ \mathrm{^\circ C}$ to $50.7\ \mathrm{^\circ C}$. The thermochemical equation for the combustion of butane is:

$$\ce{2C4H10(g) + 13O2 -> 8CO2(g) + 10H2O(l)}\quad\Delta H=-5748\ \mathrm{kJ/mol}$$

I know that the equation for enthalpy is $$Q=m\times c\times\Delta T$$ and I calculated this for water:

$$Q=724\times4.18\times43.26$$

$$Q=130918.6032$$

I don't know where to go from here in order to figure out the mass of butane needed. Please tell me if I have done something wrong or if you can help me to solve the rest of the problem.

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  • $\begingroup$ If you want to be sure that your result isn't off by some orders of magnitude please check the units. Unit analysis looks boring but is helpful. $\endgroup$ – Klaus-Dieter Warzecha Feb 10 '14 at 9:36
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You're on the right track.


\begin{eqnarray} Q & = & m\cdot c\cdot \Delta T\\ \mathbf{unit\ analysis} \qquad \textrm{kJ} & = & \textrm{kg}\cdot \frac{\textrm{kJ}}{\textrm{kg} \cdot \textrm{K}} \cdot \textrm{K} \end{eqnarray}
For water, the specific heat capacity $c = 4.18 \large{\frac{\textrm{kJ}}{\textrm{kg}\cdot \textrm{K}}}$. You have that right.

The temperature difference $\Delta T = (50.7 - 7.44) = 43.26\ \textrm{K}$. We're talking about differences, so we can use kelvin as the unit. You have that right too.

Now, what is the mass of $724\ \textrm{cm}^3$ water?

Choose any unit you want, but adjust the outcome of your calculation according to the unit analysis above!

Now you know the kJ you need.

From the combustion equation, you know the amount of heat delivered by $n$ moles of butane. Note the $n$!

Now you can calculate how many moles of butane you will need.

From there, the calculation of the mass is not difficult either.

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