How long will it take for 6.6G of 1.102 wort at 79°F to cool to 68°F when sitting in an 8.5G plastic cylindrical fermenter in an ambient temperature of 66°F?
Several hours. Overnight or “a shift” would be good.
I could probably work out the math but I don’t feel like it.
6 hours enough?
Most likely. Honestly, a few degrees isn’t going to hurt anything anyway.
Putting a fan on it will help, but it will still likely take hours to get within 2 degrees of ambient. Those last few degrees are slow.
In reality there are way too many factors (air currents, humidity, material, geometry/wall thickness) to get an answer here, plus we’d have to take an indefinite integral while I’m on still on my first cup of coffee, but we can go all Fermi approximation on it. Short answer, to get to 2°F above ambient will take a very long time.
Desired heat transfer we do know exactly: 6.6 gal * 3.8 L/gal * 1.1 kg/L * 11°F * 1.8°C/°F * 4.2 kJ/kg°C = 2300 kJ
Newton’s law of heat transfer to keep things simple: Q’ = hA∆T. Figure a best-guess of 5 W/m^2K and 12"x 13.5" WxD gives hA = 1.6 W/K. Our average ∆T is 7.5°F = 4.2°C so Q’ = 6.7 W and 2,300,000 J / 6.7 W = ~95 hours. Assuming constant flux is going to be over by quite a bit due to the long tail, so approximate it as 95*e^-1 = 35 hours.
But as Dave points out you don’t need to exponentially decay all the way out to the exact target temperature, so in reality you’ll be within a degree or so after maybe 24 hours.
Wrapping the fermenter with a damp towel and running a fan (to use use latent heat of vaporization of water) will also help to cut the cooling time. You will have to keep the towel wet/damp as it dries out.
10 minutes using an immersion cooler. Maybe less.
Are you one of those people who make up math questions?
Wow, thanks for taking the time to post this!
Haha definitely not, but the OP does sound far too similar to an exam question for my liking!
If Mark has seven candy bars in one hand and six candy bars in the other hand, what does Mark have?
Diabetes… Mark has diabetes…