Ah, wait. The light dawns. As Peter Gluck said, this only makes sense if
you are talking about the electric bill. I see. Jones Beene wrote:
There are two P-in points, one for the whole system and one for the
calorimetry.
If by "system" you mean the cost of running the experiment, including
the pump, the overhead lights, the building HVAC, the power meters and
computers, the coffee pot, and so on, then yes, we must take this into
account. Arguably, it does take all of the electricity in the room to do
the experiment. You might also toss in the gas it takes to drive to
work, and the energy from the food that the researchers eat. However, in
a typical physics experiment, they do not count this other energy. It
makes no difference to the conclusion, because it is expended outside
the cell.
The pump energy is also unpredictable. Some pumps are noisy and
inefficient. Others are highly efficient and run cool. (I know a lot
about pumps in this range because I go through many of them with my
outdoor pond.) If you spend more and get a good one, the pump will
consume less power. No one would say the Rossi device is more impressive
because someone paid $500 for a good pump, instead of $75 for an
el-cheapo model, saving a little money on the lab power bill. In any
case this pump power will not show up in the calorimetry. No one would
include the amount of power the pump consumes in a performance report on
this device. For that matter an HVAC guy testing an ordinary boiler will
not include pump power in his tests either; he only measures the gas or
electric heating power.
If you happen to have 50 Italian people watching the experiment, you are
likely to expend a great deal of electricity in the espresso machine
there in the corner of the lab. With 50 reporters, you will have to turn
on more overhead lights. These are random variations and they have no
connection to the experiment itself.
- Jed
