I wrote:
> In any case, even if 38 gigawatts had been input before the event, that > would make no difference if all of that heat came out as soon as it went in. > Other people here have confused this issue. For example, Robert Leguillon wrote: "I take an old blacksmith's anvil. I warm it in a kiln over two day to roughly orange-hot (it is going to hold this heat for a LONG time, especially if well-insulated). . . . The energy expended in getting the anvil up to "operating temperature" would more than balance this equation, and is necessary beyond a doubt. Think of it as potential energy, just like a coiled spring or a raised weight." The second statement is correct, but as I pointed out before, the first statement is a misunderstanding. After 10 minutes in the kiln, the anvil reaches the highest temperature it will reach. It does not get any hotter after that, so it does not store any more heat. You can leave it for two days or 20 years, it will not store any more energy. The "potential energy" is limited. To extend the analogy, once you raise the weight as high as it can, to the top pulley, you can hold it up there for two days but you will not add any more potential energy to the system. In the case of the Oct. 6 Rossi demonstration, as soon as the heat out balances the heat in (taking into account the low calorimetric recovery rate), no more heat energy is stored in the system. As I said, whether the heat added amounts to 33 MJ or 33 GJ makes no difference whatever; it all comes out, except for a little added at the beginning to bring the cell up to the terminal temperature. You can determine how much heat a substance stores at a give temperature by looking up the specific heat. Assuming that the thermal mass of the Rossi cell is 30 L of water (which has by highest specific heat), then it is easy to determine what temperature it would have to reach in order to store up 33 MJ. The temperature would rise 263 deg C, ignoring pressure and boiling. As I mentioned, if it stored up this heat instead of dissipating it, the cell would get hot but the cooling water loop would magically refuse to diffuse any of this heat (since it is storing up) and the cell would appear to remain at room temperature. If the cooling water loop warmed up and a Delta T appeared, then it would not be storing heat. Needless to say, that's impossible. - Jed
