Quote from: JerryM on March 31, 2013, 12:14 PM
what this is saying is that if your pan rim temp is anything like (irrespective of your Kw) then you will see the same effect as i do.
I respectfully disagree. It is (IMHO) very important to differentiate between
heat and
temperature. Heat is a measure of energy; temperature is a measure of the average speed of motion of the molecules making up the object of interest. If you have achieved a rim temperature of between 185
oC and 220
oC and you add
anything to that pan, then if the temperature of what you add is lower than 185
oC and 220
oC, the temperature of the pan (and of its contents) will drop. The speed with which it regains its former temperature is a function of the thermal inertia of your pan (which also affects the speed at which the temperature drops), the thermal inertia of the contents (-- ditto --) and the heat output of your burner. Note : the
heat output, not the temperature (so long as the latter is sufficiently high). The higher the heat output of the burner, the faster your pan and contents will regain their former (and desired) temperature. To clarify this, imagine transferring your pan from your burner to the hottest part of an oxy-acetylene flame : a far far higher temperature than your standard (or propane) gas burner can achieve, yet the pan will never regain its former temperature (it may melt, at the point of contact twixt flame and pan) but the pan and its contents will slowly cool. Now imagine transferring that same pan from your burner to a vast vat of molten tin ("vast" meaning at least 100 times the volume of your pan) : the molten tin will be at about 230
oC, and your pan will not only not cool, it will actually increase in temperature until it is at the same temperature as the molten tin. The oxy-acetylene flame is at a temperature of about 3000
oC, yet the pan cools; the tin at a temperature of only 230
oC, yet the pan gets hotter. Heat (energy), not temperature, is what makes the difference.
** Phil.