FoodBanter.com - View Single Post

(Yvon Thoraval) wrote:

> ok, in that case, take a time constant of 20 mn (saying tau), as
> a rule of thumb, then, having T1 the romm temperature where the
> bottle was, take T2 the fridge temperature, the bottle
> temperature will follow :
>
> Tbottle = T1 + (T2 - T1).e^( - time / tau)
>
> time being expressed in the same unit as tau...
>
> for time in hours (tau = 1/3) gives :
>
> Tbottle = T1 + (T2 - T1).e^( - 3.time)
>
> giving, for time :
>
> time (in hours) = (ln((T2 - T1)/(Tbottle - T1)))/3
>
> as far as i remember well algebraic computation ;-)

OK, I don't understand a single word, but let me put in another
way round. A bottle with a given temperature Tb, put into a
surrounding at a temperature Ts, in a certain lapse of time will
warm (or cool) to the midway temperature between Tb and Ts. In the
same time lapse again, it will again warm/cool midway this
temparature and Ts, and so on.

Thus said, for a standard (750ml) bottle and the surrounding media
being air (= fridge), this fixed time span is 40 to 45 minutes.
If the surrounding media is water (= ice bucket), this fixed time
span shortens to 12 to 15 minutes, water being a much better heat
conductor than air.

M.