All things other than color being equal, and the temperature of the
surrounding room being less than the temperature of the hot teapot,
then the the shiny teapot should stay hotter longer than the black one.
The reason for this is that the black one radiates energy away in
the form of infrared light (and lower frequency electromagnetic waves)
at a faster rate than the shiny one. The rate at which a hot object
radiates energy is related to its shininess, and here's the reason why:
Heat travels from hot objects to cold ones. If two objects are at
the same temperature, there is no net heat flow. If the room were at
the same temperature as a teapot, the rate at which radiant energy is
absorbed by the teapot has to be the same as the rate at which it is
emitted, in order for the teapot's temperature to remain the same. In
this case, the black teapot absorbs and radiates energy at a faster
rate than the shiny one.
Cool down the room, and the rate at which infrared and other light
gets absorbed by the teapots is less, but the rate at which the teapots
radiate energy does not change. The black one therefore radiates energy
away faster and cools off faster.
Here it is assumed that the thermal conductivities are the same,
the amount of hot water in the teapots is the same, the shapes of the
teapots are the same, and convection currents remain the same.
(republished on 07/27/06)