Is there a formula or tool that you can use to determine how long it will take to "x" volume of wort (or equal amount of water as a proxy) from a starting liquid temp to a lower target liquid temp based upon a "y" ambient air temp?
IOW if I have 5 gal of wort sitting at 65*F, want to cool it to 40*F and the ambient cooling air temp is 35*F, how long will it take to cool the wort to the target?
Maybe newton's law of cooling. Its not going to satisfy the hyperanal überhomebrewer savant, (it does have significant limitations), but perhaps we could use it:
(dT/dt) = -k(T - Ta)
T2 = Ta + (T1 - Ta)e^-kt <---- solution
T1 = initial temp of liquid
T2 = temperature of the liquid at time t (must keep units of time lol)
Ta = ambient temp fridge, room, should be constant
What you want to determine:
Ta = 35f
T1 = 65f
T2 = 40f
t = time difference between T2 and T1
k: constant
You'll have to make observations to get your constant by measuring T2 after a known period of time.
So lets say the t between T2 is 52f after 30 minutes (caveat, not measured data):
Ta = 35f
T1 = 65f
T2 = 52f
t = 30 minutes
k = solve
T2 = Ta + (T1 - Ta)e^-kt
52 = 35 + (65 - 35)e^-30k
52 = 35 + 30e^-30k
17 = 30e^-30k
(17/30) = e^-30k
(30/17) = e^30k
1.76 = e^30k
ln(1.76) = ln(e^30k)
ln(1.76) = 30k
k = (ln(1.76))/30
k = 0.0189
or just use
k = (-1/t) * ln ((T2 - Ta)/(T1 - Ta))
k = (-1/30) * ln ((52 - 35)/(65 - 35)) = 0.0189
Ok so you have your constant k which is 0.0189, can move onto solving for t with a desired final temp (T2):
Ta = 35f
T1 = 65f
T2 = 40f
t = solve
k = 0.0189
T2 = Ta + (T1 - Ta)e^-kt
40 = 35 - (65 - 35)e^(-0.0189t)
40 = 35 - (30)e^(-0.0189t)
5 = 30e^(-0.0189t)
5/30 = e^(-0.0189t)
30/5 = e^(0.0189t)
6 = e^(0.0189t)
ln(6) = 0.0189t
1.792 = 0.0189t
t = 1.792/0.0189
t = ~95 minutes
Ive not used this but Ill check on this with a starter I need to cold crash. I have another starter of the same volume to make right afterwards so I can gather more data for accuracy and precision. Its obviously going to be problematic when using different volumes, vessels and ambient temps, but this is about as
practical an answer for your question.