Gigi DiMarco <[email protected]> wrote:
> 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. > No, it isn't. That is why a gap opens between the room temperature and the calorimeter, and the gap persists until early morning. > 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. > No, it isn't. See Newton's law of cooling. > 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? > Yes, I can. If ambient stays stable, the reactor and water temperature will remain stable at 0.6 deg C above room > Where in the data do you see that 0.6 °C is the maximum? > It goes no higher after 1.4 hours. You can see this in other data sets as well, such as early in the morning with this data set. Whenever ambient remains stable for a few hours or more, the reactor temperature always settles 0.6 deg C warmer. > > Please don't be contemptuous and dismissive; it is not the case. If > someone does't understand calorimetry it is not me. > You do not understand Newton's law of cooling and you cannot tell the difference between ambient cooling and heat generation in a cell. In my opinion, you are terribly confused and totally unqualified to do calorimetry. - Jed
