Problem 5 – Hint 5

Just to sum up all the hints: for maximal efficiency of heat conversion into electrical energy, entropy must not increase. Hence, the entropy lost by the satellite at its cold side must be equal to the entropy obtained by the satellite at its hot side. Then, the charging power is found from the energy conservation law (the 1st law of thermodynamics). When expressing the entropies, keep in mind that while the heat is lost at the cold side only due to thermal radiation at the temperature $T_c$, the heat obtained at the hot side is the difference of the thermal radiation from Sun, and the thermal radiation of the satellite at temperature $T_h$. NB! the thermal radiation from Sun is absorbed by the satellite at the temperature of its hot side. (This means that in a big picture, there is an unavoidable increase of entropy when the Sun’s radiation is absorbed: Sun’s thermal radiation is emitted at the Sun’s temperature $T_\odot$, and absorbed at the temperature $T_h$. However, further increase of entropy can be avoided if an ideal heat engine is used inside the satellite.) Once an expression for the charging power is obtained, it will be quite easy to figure out what are the best emissivities for both the cold and hot sides of the satellite. Finding the optimal ratio of the temperatures $T_c/T_h$ requires a little more work: if everything goes well, you’ll obtain an equation which needs to be solved numerically.

Please submit the solution to this problem via e-mail to physcs.cup@gmail.com. Problem 5 will be kept open for submissions until 13:00 GMT, 15th May 2022.