Coal burns according to the reaction $\mathit{C}\mathbf{+}{\mathit{O}}_{\mathbf{2}}\mathbf{\to }\mathit{C}{\mathit{O}}_{\mathbf{2}}$. The heat of combustion is $\mathbf{3}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathit{j}\mathbf{/}\mathit{k}\mathit{g}$of atomic carbon consumed. (a) Express this in terms of energy per carbon atom. (b) Express it in terms of energy per kilogram of the initial reactants, carbon and oxygen. (c) Suppose that the Sun ($\mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{30}}\mathit{k}\mathit{g}$) were made of carbon and oxygen in combustible proportions and that it continued to radiate energy at its present rate of $\mathbf{3}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{26}}\mathbf{W}$. How long would the Sun last?

The combustion heat is,$E=3.3x{10}^{7}J/\mathrm{kg}$ .

The mass of the Sun is,$m=2.0x{10}^{30}\mathrm{kg}$ .

The power is, $P=3.9x{10}^{26}W$.

The expression for the number of atom in mass of mis given by,

$\mathbf{N}\mathbf{=}\frac{{\mathbf{mN}}_{\mathbf{A}}}{\mathbf{M}}$

Here, $\mathit{m}$is mass of Sun, ${\mathit{N}}_{\mathbf{A}}$is Avogadro number, and $\mathit{M}$is molar mass of carbon.

The expression for the heat of combustion per atom in terms of energy is given by,

$\begin{array}{l}{\mathbf{E}}_{\mathbf{N}}\mathbf{=}\frac{\mathbf{E}}{\mathbf{N}}\\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{E}\mathbf{M}}{\mathbf{m}{\mathbf{N}}_{\mathbf{A}}}\end{array}$ ...(i)

Here, $\mathit{E}$is energy.

Substitute all the known values in equation (i).

$$\begin{gathered} E_N=\frac{\left(3.3\times {10}^{7}J\right)\left(12.0g/mol\right)}{\left(1000g\right)\left(6.022\times {10}^{23}mo{l}^{-1}\right)} \\ =6.576\times {10}^{-19}J \\ =\left(6.576\times {10}^{-19}J\right)\left(\frac{1eV}{1.6\times {10}^{-19}J}\right) \\ =4.1eV/atom \end{gathered}$$

Therefore, the heat of combustion per atom in terms of energy is$4.1eV/atom$ .

In each event, two oxygen atoms combine with one carbon atom to form carbon dioxide, so the mass of the reactants is,

$2\left(16.0g\right)+12u=44u$

The number of atoms in of reactants is,

$$\begin{gathered} N_r=\frac{\left(1000g\right)\left(6.022\times {10}^{23}mo{l}^{-1}\right)}{44g/mol} \\ =1.37\times {10}^{25} \end{gathered}$$

The energy per kilogram of the initial reactants is given by,

$E_r=N_r{E}_N$ ...(ii)

Substitute all the known values in equation (ii).

$$\begin{gathered} E_r=\left(1.37\times {10}^{25}\right)\left(6.576\times {10}^{-19}J\right) \\ =9.00\times {10}^{6}J \\ =9.00MJ/kg \end{gathered}$$

Therefore, the required energy is $9.00MJ/kg$.

Let $P$be the power output of the Sun.

The burn time of the sun is given by,

$t=\frac{E_s}{P}$ ...(iii).

Here, $E_s$is the energy produced by the mass of the sun.

Find the number of reactants atoms in the mass of sun.

$$\begin{gathered} N_s=\frac{\left(2\times {10}^{33}g\right)\left(6.22\times {10}^{23}mo{l}^{-1}\right)}{44g/mol} \\ =2.74\times {10}^{55} \end{gathered}$$

Find the energy produced by the mass of the sun.

$$\begin{gathered} E_s=N_s{E}_N \\ =\left(2.74\times {10}^{55}\right)\left(6.576\times {10}^{-19}J\right) \\ =1.802\times {10}^{37}J \end{gathered}$$

Substitute all the known values in equation (3).

$$\begin{gathered} t=\frac{1.802\times {10}^{37}J}{3.9\times {10}^{26}W} \\ =4.62\times {10}^{10}s \\ =\left(4.62\times {10}^{10}s\right)\left(\frac{1y}{12\times 30\times 24\times 3600s}\right) \\ \approx 1.5\times {10}^{3}y \end{gathered}$$

Therefore, the required time is $1.5\times {10}^{3}y$.