Question

In an oven, due to insufficient supply of oxygen, $60\%$ of the carbon is converted to carbon dioxide whereas the remaining $40\%$ is converted into carbon monoxide. If the heat of combustion of carbon to $C{O}_2$ is $394kJmo{l}^{-1}$ while that of its oxidation to $CO$ is $111kJmo{l}^{-1}$, calculate the total heat produced in the oven by burning $10kg$ of coal containing $80\%$ carbon by weight. Also calculate the efficiency of the oven.

Answer 1

The reactions taking place in the oven are
$C_{\left(s\right)}+O_2\left(g\right)\to C{O}_2\left(g\right)+394kJmo{l}^{-1}$
$C\left(s\right)+\frac{1}{2}{O}_2\left(g\right)\to CO\left(g\right)+111kJmo{l}^{-1}$
Carbon converted to $C{O}_2$
$=\frac{60}{100}\times 80kg=4.8kg$
Carbon converted to $CO=\frac{40}{100}\times 8kg=3.2kg$
$12g$ i.e., $0.012kg$ of carbon on combustion to $C{O}_2$ produce heat $=394kJ$
$\therefore 4.8kg$ of carbon on combustion to $C{O}_2$ will produced heat $=\frac{394}{0.012}\times 4.8$
$0.012kg$ of carbon on combustion to $CO$ produce heat $=111kJ$
$\therefore 12kg$ of carbon on oxidation to $CO$ will produce heat $=\frac{111}{0.012}\times 3.2kJ$
$=29600kJ$
$\therefore$ Total heat produce $=157600+29600=1,87,200kJ$
If oven were $100\%$ efficient, all carbon would have been converted to $C{O}_2$
Heat produced from $8kg$ carbon would have been $=\frac{394}{0.012}\times 8=262,667kJ$
$\therefore \%$ efficiency $=\frac{187,200}{262,667}\times 100=71.3\%$