He's asking about heat transfer. I'll just point out that in working with W/W heat pumps we calculate the heat transfer as 500*gpm*∆T when water is used. That comes from gpm*8.34 being the pounds of water through the exchanger per minute and 60 minutes in the hour. The specific heat of water is 1 BTU/lb•°, Hence the formula. When an antifreeze is used, whatever it's composition, the manuals say to use 485*gpm*∆T. This says that the effective specific heat of the antifreeze mix is 0.97 that of water, whatever the mix is and, as that can't be right, it says that it is pretty close to this for any mix that is likely to be used. As the difference in transfer is only 3% between water and mix I'd assume that it is even less between different mixes. IOW, I'd assume the performance of your system will not change appreciably. This is not your problem. The problem is that a change in ∆T from 11 °C to 10 °C means a 10% reduction in heat transfer (much larger than the probable transfer difference from changing coolants) and that at an 11° rise the heat leaking into the fermenter from the ambient is equal to the heat that can be transferred out to the bath. To solve the problem you clearly need to either
1)Get the bath colder
2)Minimize heat incursion
3)Implement some combination of 1) and 2)
