# Behold the power of the sun!

**Classical Mechanics**Level 4

*125 points*

Solar collector power generation plants, such as this one in Spain, work by concentrating solar energy into a small region to drive a steam turbine.

Direct sunlight delivers about \(1000~\mbox{Watts/m}^2\) of energy in the infrared, visible, and ultraviolet parts of the spectrum. I want to use mirrors with a total effective area \(A\) to direct this solar energy into a cube \(1~\mbox{m}\) on a side, heating the cube to a steady \(100^\circ\) Celsius so I can start generating steam and running a turbine. What is the necessary effective mirror surface area \(A\) **in \(\mbox{m}^2\)** to do this?

Note that your \(A\) will be smaller than you might expect. This is because we have used components that are 100% efficient - **all** the solar energy gets directed onto the cube. In a real life situation the amount of solar energy redirected by the mirrors is much, much less.

**Details and assumptions**

- The cube absorbs and emits as a perfect blackbody. It's well insulated so you can ignore any conductive or convective heat transfer.
- The air around the cube is \(20^\circ\) Celsius.
- You don't need to worry about any energy lost turning the water into steam. You just want to get the cube up to the specified temperature. (In a real turbine this additional energy loss would need to be taken into account.)
- Effective area takes into account the angle of the mirrors with respect to the sun, so when calculating the effective area you can ignore the orientation of the mirrors. If the mirrors are angled, the true surface area is larger than the effective area.

**Your answer seems reasonable.**

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