What you have to do is configure you system the way you want it to be over the time you are interested and, during a time when the ambient is stable, measure the rate of change of temperature over a period of time.
A gallon of water weighs 8.34 pounds a changes temperature by 1 degree Fahrenheit for each BTU of heat removed or added. Thus if you have 5.5 gallons of water at 90 °F and it takes 10 minutes for the temperature to drop 1°F your heat loss was (1 °F)*(5.5 gal)*(8.34 lbs/gal)*(1 BTU/lb•°F)/(1/3 hr) = 137.61 BTU/hr. As this happened when the temperature difference between ambient and water was (95 - 64) = 31 °F your thermal conductivity is 137.61/31 = 4.43903 BTU/hr•° and you heat loss rate is dQ/dt = 4.43903*∆T. The temperature change rate in the tank was 1°/(1/3 hr*31°F) = 0.0967742 °F/hr•°F. This is the conductivity divided by the thermal mass. Thus the rate of change of temperature when the difference between ambient and tank is ∆T is d∆T/dt = 0.0967742 ∆T. The solution to this differential equation is
in which ∆T0 is the initial temperature difference (31°F), t is the time in hours and 1/0.0967742 = 10.33 hours is the 'time constant' of this system which depends on the insulating properties of the tank. If the tank is well insulated the time constant is long. If it is not, it is short. When t = 10.33 hours ∆T = ∆T0*exp(-1) = 0.368*∆T0 = 11.4 °F so that the water temperature in the tank is 64 + 11.4 = 75.4. When t = 2*10.33 (two time constants) ∆T = ∆T0*exp(-2) = 4.2 °F and the water temperature is 64 + 4.2 = 68.2. Thus the temperature decreases 'exponentailly' with time falling fastest when the tank is hottest and eventually slowly creeping down to the ambient.
If you know how to do curve fitting in Excel, IGOR or some other analysis program you will get a more accurate value for the time constant by recording ∆T over a period of an hour or so and using to program to get an exponential fit to the data.
Note that 0.0967742 °F/hr•°F is the conductivity divided by the thermal mass you can use the value 0.0967742 found for 5.5 gal of water by multiplying by (5.5/gal). This is a little dicey as, if you fill the fermenter fuller there is more wort area in contact with the walls and so more conduction. Also if dealing with wort you can divide the value obtained in an experiment with water by the specific gravity of the wort.
This all assumes that conditions stay static. Obviously a change in the ambient or the addition of a large fan blowing on the fermenter is going to change the rate of heat flow.
