Excuse me Jed, but I think that is very simple for you to say that I do not understand calorimetry if you reply to a question that I did not ask.
The refrigerator example is quite evident, but is unfit to our situation, by various causes. The main one is that there you have an abrupt *change *of air temperature, while in the 18h test the air temperature is falling at a modest rate of 0,36 °C/h that is very simple to follow for the calorimeter. But the main question is that in your report you made a statement that is not true. This is you: *The temperature rose for 1.5 hours until it stabilized 0.6°C above room temperature (Fig. 19.) It stabilized because heat losses equal the power from the pump.* If from now on the losses are equal to the pump power, since you have Loss = K * deltaT and PumpPower = loss = constant and since K is valid over a broad range of deltaT you should have a constant deltaT. Instead, deltaT keeps increasing for a few hours when considering the missing file. I fully agree with your words, howewer. So in the case we both do not understand anything of calorimetry. So going back to the plots in the missing file you considered only the first 1.5 hour only because just after the ambient temperature starts decreasing. What have it happened if the ambient did not change for 5-6 hours? Can you answer this question? Where in the data do you see that 0.6 °C is the maximum? The true is that you simply stop there and are happy with your data, but there is no theoretycal reason. Please don't be contemptuous and dismissive; it is not the case. If someone does't understand calorimetry it is not me. By the way we performed a full simulation of this calibration: we got exactly the same curve but at a much higher pump power. We shall show you and Mizuno the results, hopefully at ICCF-19. Will you be there? Regards 2015-01-13 16:52 GMT+01:00 Jed Rothwell <[email protected]>: > Gigi DiMarco <[email protected]> wrote: > > >> I could say that this is false but I will be fair and I will say that >> this is not true. From the missing file (Mizuno's data) we get the >> following situation for the difference between water and ambient temperature >> >> (4h 2.5°C) (5h 2.9°C) (6h 3.1°C) (7h 3.2°C) (8h 3.2°C) (9h 3.2°C) (10h >> 3.1°C) (11h 3.1°C) (12h 3.1°C) (13h 3.0°C) >> (14h 2.9°C) (15h 2.8°C) (16h 2.7°C) (17h 2.6°C) (18h 2.5°C) >> >> were the temperatures are taken at the beginning of the hour. >> >> How do you explain this? >> > > The ambient temperature is falling. The reactor is well insulated so it > takes longer to cool off than the room does. That is all there is to it. > > You can simulate this easily with the following steps: > > 1. Fill a glass with warm water. > 2. Measure the temperature difference between the water and air. > 3. Move the glass to the refrigerator, and measure the difference between > the water and the air in the refrigerator. > 4. The second temperature difference will be much larger, because the > water does not instantly cool off. Does that mean there is a source of heat > in the water? No. > > If you do not understand this, you are not qualified to do calorimetry. > > I pointed this out before. I will not point it out again, and I will not > discuss this with you again. > > - Jed > >
