Question

The earth receives $\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{14}}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{s}$of solar energy. What mass of solar material is converted to energy over a 24-h period to provide the daily amount of solar energy to the earth? What mass of coal would have to be burned to provide the same amount of energy? Coal releases 32kJof energy per gram when burned.

Short answer

$1.72\times {10}^5\mathrm{kg}$of solar material was used to create energy.

$4.84\times {10}^{17}\mathrm{g}$of coal would be the mass.

Step-by-step solution

Einstein’s mass energy

The Einstein mass energy will be used to calculate the rate of change in energy and this equation can be represented as:

${\mathbf{\text{ΔE=Δm·c}}}^{\mathbf{2}}$

Where,

• $\mathbf{\text{ΔE}}$=change in energy
• $\mathbf{\text{Δm}}$=change in mass
• ${\mathbf{\text{c}}}^{\mathbf{2}}$=velocity of light

Find the mass of coal which is required to be burned to provide the same amount of energy.

The planet receives$1.8\times {10}^{14}\mathrm{kJ}/\mathrm{s}$of solar energy; thus, we need to figure out how much of that energy is converted to mass in a 24-hour period.

Earth receives the following amount of energy per hours:

$\mathrm{E}=1.8\times {10}^{14}\mathrm{kJ}/\mathrm{s}\times 24\mathrm{h}\times \frac{60\min }{1\mathrm{h}}\times \frac{60\mathrm{s}}{1\min }\times \frac{1000\mathrm{J}}{1\mathrm{kJ}}=1.55\times {10}^{22}\mathrm{J}$

Let's now determine how much solar material is turned into electricity.

$\begin{array}{rcl}\mathrm{E}& =& \mathrm{m}\times {\mathrm{c}}^2\\ \mathrm{m}& =& \frac{\mathrm{E}}{{\mathrm{c}}^2}\\ & =& \frac{1.55\times {10}^{22}\mathrm{J}}{{\left(3.0\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2}\\ & =& \frac{1.55\times {10}^{22}\frac{\mathrm{kg}\times {\mathrm{m}}^2}{{\mathrm{s}}^2}}{9.0\times {10}^{16}\frac{{\mathrm{m}}^2}{{\mathrm{s}}^2}}\\ & =& 1.72\times {10}^5\mathrm{kg}\end{array}$

Each gramme of coal releases 32 kJ (32000 J/g) of energy.

Find the mass of coal that, if burned for 24 hours, would provide the same amount of energy as the sun.

${\mathrm{m}}_{\mathrm{coal}}=\frac{1.55\times {10}^{22}\mathrm{J}}{32000\mathrm{J}/\mathrm{g}}=4.84\times {10}^{17}\mathrm{g}$