Question

The sun supplies electromagnetic energy to the earth. It appears to have an effective temperature of approximately \(5800 \mathrm{K}\). On a clear summer day in North America, the energy incident on a surface facing the sun is approximately \(0.95 \mathrm{kW} / \mathrm{m}^{2} .\) The electromagnetic solar energy can be converted into thermal energy by being absorbed on a darkened surface. How might you characterize the work potential of the sun's energy when it is to be used to produce work?

Short answer

Characterize the work potential of the sun's energy when it is to be used to produce work. The work potential of the sun's energy when it is used to produce work can be characterized as approximately \(0.901\mathrm{kW}/\mathrm{m}^2\). This value represents the maximum power that can be converted into work considering the Carnot efficiency, which represents the maximum conversion efficiency of any heat engine.

Step-by-step solution

Find the Carnot efficiency

The maximum efficiency of any heat engine, also called Carnot efficiency, is given by the formula: $$ \eta_\text{Carnot} = 1 - \frac{T_\text{cold}}{T_\text{hot}} $$ where \(T_\text{cold}\) is the colder system temperature, and \(T_\text{hot}\) is the hotter system temperature. In our case, the sun can be considered the hot reservoir, with a given temperature of \(5800\mathrm{K}\), and Earth is the cold reservoir, which we will assume a temperature of around \(300\mathrm{K}\) for simplicity.

Calculate Carnot efficiency

Now that we have the temperatures for both the hot and cold reservoirs, we can calculate the Carnot efficiency: $$ \eta_\text{Carnot} = 1 - \frac{300}{5800} \approx 0.948 $$ The calculated value represents the maximum efficiency at which a heat engine can convert the sun's energy into work.

Determine the incident power

The problem states that the power of the sun's energy incident on a surface in North America is approximately \(0.95\mathrm{kW}/\mathrm{m}^2\). We will use this value to find the work potential for the sun's energy.

Calculate the work potential

Finally, to calculate the work potential, we can multiply the incident power by the Carnot efficiency: $$ W_\text{potential} = P_\text{incident} \times \eta_\text{Carnot} $$ $$ W_\text{potential} = 0.95\mathrm{kW}/\mathrm{m}^2 \times 0.948 \approx 0.901\mathrm{kW}/\mathrm{m}^2 $$ So, the work potential of the sun's energy when it is to be used to produce work can be characterized as approximately \(0.901\mathrm{kW}/\mathrm{m}^2\).