I wrote:
As I mentioned yesterday, a calorimeter can measure an endothermic reaction as easily and as accurately as an exothermic reaction. In your hypothetical example with 2 kW going into the system for two hours, you will definitely see 1.98 kW emerge from the system during the entire two hours.
On average, I mean. When you first start heating there is a large gap between input and output. You can observe this when you heat a large, covered pot of water on the stove. You can hold your hand 10 cm above the pot comfortably for a while. After the water comes to boil you cannot do this.
. . . you will see all stored energy released during this phase, and you can add it into the total energy balance. You will see that total input energy is far exceeded by total output.
Please note I mean the total energy balance for the entire run. Not the power balance at any given moment. You need to concentrate on the _energy balance_.
During a test with a joule heater and a good calorimeter the energy balance is about 0.9 units of energy per 1 unit of energy produced. ("Produced" means either input, or generated internally in a chemical or nuclear reaction.) With a superb calorimeter the recovery rate rises to 0.95 or 0.98. I expect this is a lousy calorimeter because of the heat transfer to the secondary loop, and I expect it will recover ~0.7 or ~0.8 of the heat produced by the cell. Since the thing was running for hours with no input, and the heat balance includes all of the energy originally input to heat up the water (which must come out after the reaction is quenched) obviously the output heat will far exceed input energy.
Leguillon seems to have notion that heat originally stored as the water is warmed up somehow vanishes and is never accounted for. That is not how a calorimeter works.
- Jed
