Question

Consider the combustion of propane:

${\mathit{C}}_{\mathbf{3}}{\mathit{H}}_{\mathbf{8}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{5}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathbf{}\mathbf{3}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{l}\mathbf{\right)}\mathit{\Delta }\mathit{H}\mathbf{=}\mathbf{-}\mathbf{2221}\mathit{k}\mathit{J}\mathbf{}$

What mass of propane must be burned to furnish $\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathit{k}\mathit{J}$of energy, assuming the heat transfer process is $\mathbf{60}\mathbf{\%}$efficient.

Short answer

Mass of propane that burns with $8\times {10}^{7}kJ$of $heat=1091.4\times {10}^{4}kg$

Step-by-step solution

Total heat

Form the statement that the heat transfer process is 60 percent efficient

It means, 60 percent of heat = 2221 kJ

100 percent of heat$=\frac{2221}{60}\times 100=3701kJ\approx 3700kJ$

Calculation of Heat

Mass of propane = 44 g

Now, 44 g of propane burned to furnish 2221 kJ of energy.

So, 1 g of propane furnish heat$=\frac{3700}{2221}\times 44=73.3kJ$

Or we can say 73.3 kJ of heat is required to burn 1 g of propane.

Calculation of Mass

Mass of propane that burns with 73.3 kJ of heat = 1 g

Mass of propane that burns with $8\times {10}^{7}kJ$of heat

$$\begin{gathered} =\frac{1}{73.3}\times 8\times {10}^7=0.10914\times {10}^{7}g \\ =1091.4kg \end{gathered}$$