I insulated the brewing container with an old bed sheet and on top of that added a layer of aluminium foil, and filled the container with 25 liters of water, and measured the temperature to be at 19,2°C and left the container in a room at around 24,5°C.
After an hour, the temperature had risen to 19,3°C so the peltier will definitely work. Of course I will have to keep the brew at around 15°C but if there was a temperature rise of 0,1°C over an hour with delta between the room and the container of 5,3°C, then delta of 10°C should still be very well within same type of figures. Should be linear function?
Calculating the energy transfer
On this I already did some quick work and calorie is the perfect unit of measure for this. If 25 liters of water rises 0,1°C in temperature it requires 25 liters (kg) times 1000 (to grams) times 0,1 (rise in temperature) times 4,2 joules (a calorie) of energy.
25×1000×0,1×4,2 equals 10 kJ of energy.
And since 1 Joule is one Watt of power for one second, and if peltier is 3% efficient, then 180 Watts of input to the peltier will produce 5,40 Watts of actual usable energy (heat transfer), then it will be 5,4 Joules per second, which in an hour is 19 440 Joules, so the peltier should be about double what is needed.
And since I need delta of 10°C and if it is linear function then it will be about just perfect to keep the brew at 15°C. Not including the heat produced by the yeast, which may be significant. Could probably be calculated by taking the amount of sugars and the end alcohol content and dividing that over time? Could be significant.
3 and a half hour mark
19,8°C and 9 kJ per hour.