[ramchip]

Disclaimer: not a physicist.

My intuitive understanding is that, as you heat the absorber, it becomes hotter and hotter, until it reaches the same temperature as the sun's surface. At this point, the absorber (also a blackbody) radiates just as much energy as it receives from the sun; thus there is no more heat exchange. Otherwise, you would be taking two objects at the same temperature and making one hotter and the other one colder with no work from outside the system, violating thermodynamics.

There is a more formal study in solar energy books like (http://books.google.com/books?id=81WI2LwrpkcC&lpg=PA25&#...) or (http://www.physics.drexel.edu/~jenks/Solar%20Energy.pdf) (PDF, page 21 onwards is quite relevant).

[anthonyb]

Your intuition seems wrong to me. The two major problems are that thermodynamics refers to energy, not temperature. Thus, as long as total energy is conserved, there's no problem. The second issue is that solar heating is due to radiation, not conduction/convection, so there's no heat exchange as such.

Think of it in terms of photons - if you're collecting and concentrating photons, the upper limit is the total number of photons which the sun produces. I'm pretty sure that if you focused all of those into a square meter of the earth's surface, then it'd reach more than 5760K.

[ramchip]

> Thus, as long as total energy is conserved, there's no problem.

Where did the other laws go?

> The second issue is that solar heating is due to radiation, not conduction/convection, so there's no heat exchange as such.

I don't understand how there could be "solar heating" but no "heat exchange".

Unless I misunderstood the papers I cited, the efficiency equations show that there is a limit (efficiency drops to 0% with Ts=Ta), but I'll still try to address the "intuition" part of your post.

Let's say we are in a black room with no windows, everything being at room temperature. I use lenses to concentrate the infrared light from the walls onto a black cube. Would the cube heat up from the concentration of photons?

[anthonyb]

> I don't understand how there could be "solar heating" but no "heat exchange".

Hmm, I thought that "heat exchange" referred to direct transfer of heat energy, perhaps I'm misremembering. http://en.wikipedia.org/wiki/Heat_transfer#Radiation doesn't seem to differentiate, but bear in mind that there's no way to focus molecular vibration with a mirror.

And if more photons hit the cube with your lens than without it, then yes, the cube will heat up.

[ramchip]

Then focus the infrared light on a thermoelectric generator instead of a cube. The generator powers a widget. The dissipated heat from the generator and the friction on the widget goes back to the walls and is later reemitted as infrared.

That's a perpetual motion machine.

[anthonyb]

...until it hits a new equilibrium, at which point you're back to square one.

btw, from what I've been reading, it looks like you're right (ie. you can't heat anything directly to > the sun's temperature) but I can't find any explanation which makes any intuitive sense at all, just lots of stuff like http://en.allexperts.com/q/Physics-1358/Black-body-radiation... (note the hand waving) or else a bunch of hard core entropy equations.

[gbhn]

You can't do that (at least not with passive components). There's a famous optics result called the 'brightness theorem' that explains why not.

[anthonyb]

I think you mean "Conservation of radiance". It corresponds pretty much directly with "Power per unit area". Obviously, mirrors and lenses can reduce the unit area, while keeping power constant (at best).

http://en.wikipedia.org/wiki/Radiance#Intensity

[forgottenpaswrd]

My intuition is that you "cold" your absorber when you extract work from the electrons. One of the big changes of recent years had been power electronics that let you convert the charge-voltage pair into anything you need(like the voltage to charge a battery).