--
when you believe the only tool you have is a hammer.
All problems look like nails.
Seems counter-intuitive. The more time you spend heating it up, the
longer time period over which it is losing heat to the environment.
Heat loss is basically the integral of temperature difference over
time (because the rate of heat transfer is essentially proportional
to temperature difference). Of course there are lots of complicating
factors like air currents and convection, but in the absence of
evidence to the contrary, I'd put my money on heating it as fast as
possible being the most energy-efficient thing.
- Logan
>Heat loss is basically the integral of temperature difference over
>time (because the rate of heat transfer is essentially proportional
>to temperature difference). Of course there are lots of complicating
>factors like air currents and convection, but in the absence of
>evidence to the contrary, I'd put my money on heating it as fast as
>possible being the most energy-efficient thing.
With radiation loss increasing with T^4, it might pay to keep
an electric stove burner cooler.
Nick
Interesting point. My gut feeling is that the reflector ("drip pan")
will make up for a lot of that and a lot of the radiated heat will end
up in the cookware anyway (plus the temperature difference between the
burner and the cookware might not be that large because the "contact
patch" between them is big). But on the other hand, T^4 is a REALLY
fast-growing function...
- Logan