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I was reading an article on fermentation temperature control where it says,

However, since yeast growth and fermentations are exothermic and therefore generate heat, figure that the temperature within the fermenter can be as much as 8 °F (4 ºC) higher than outside of the fermenter during the early days of fermentation. So beers that are fermenting in refrigerators set at 65 °F (18 ºC) are most likely fermenting at about 72 °F (22 ºC).

4 ºC is a huge difference!

Right now I have a carboy in a fridge that's temperature controlled to 19 ºC (on the lower side of the optimal range for WLP351). The sensor strip is taped to the glass of the carboy, and consistently reading 19.2 ºC.

I'm wondering if next time I should set the temperature control lower in the first 12 hours, but I realize I have no idea how that translates to change in different sections within the carboy—near the glass, the bottom, the top (where ale yeasts float, correct?), and so on. For example even if it's 19.2 ºC on the outside of the glass, the liquid touching the inside of the glass is probably much warmer due to convection.

Is there a gradient diagram or some rough formula that would help me determine the effective fermentation temperature (where the yeast actually is) from the outside glass temperature? Or, do I have to find a way to stick a (sanitized) sensor strip into the carboy.

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    It's true the temperature in the fermenter will be higher, but there shouldn't be any significant gradient within the liquid itself. As carbon dioxide leaves the beer it tends to create a great deal of fluid flow inside the carboy (just look at it on the second or third day and you can see it churning around), effectively evening out any temperature differences. This is increasingly not the case the larger your fermenter is, but I've only heard of temperature gradients in huge (100bbl+) commercial tanks.
    – Franklin P Combs
    Mar 31, 2015 at 16:21
  • Ah, that's reassuring. Actually, I think this indirectly answers the question. My question was based on the premise that there would be a significant (measurable) temperature gradient. If that's not the case, then there's no question here, except finding the liquid's temperature from an outside-the-glass measurement, which I think is answered in this prior duplicate—insulating the sensor from the air. @FranklinPCombs would you like to move your comment to an answer? (I'd like to flag this question as a duplicate, afterwards.)
    – Andrew Cheong
    Mar 31, 2015 at 16:36
  • No need. Flag away
    – Franklin P Combs
    Mar 31, 2015 at 16:40

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