300gallons 4800kwh per year

In terms of energy consumption, the amount of energy required for tilapia production can vary depending on factors such as the type of production system, water quality, and feeding regime. However, on average, it can take around 0.5-1 kWh/kg of tilapia(full) produced.
Effective flesh yield is 350grams from 1kg of full grown fish, after deboning etc. Hence 1kwh/0.35kg = 2.86kwh/kg effective flesh. And about 4kg tomatoes for this same 2.86kwh energy .

Therefore, the energy consumption for producing 15-20 kg of tilapia in 300 gallons of water over an 8-month cycle would be approximately 7.5-20 kWh.
----------
how much kwh to raise 20c to 30c for 300 gallons of water and keep it there for 24 hours

To raise the temperature of 300 gallons of water from 20°C to 30°C and keep it at that temperature for 24 hours, we need to calculate the total amount of energy required for both heating and maintaining the temperature. The specific heat capacity of water is 4.186 joules per gram per degree Celsius (or 1 BTU per pound per degree Fahrenheit). Therefore, the energy required to heat 300 gallons of water (approximately 1135.6 liters) by 10°C would be:
(1135.6 kg) x (10°C) x (4.186 kJ/kg/°C) = 47,543 kJ
To convert this to kWh, we divide by 3,600 seconds per hour and 1,000 watts per kilowatt, which gives us:
47,543 kJ / (3,600 s/hour x 1,000 W/kW) = 13.2 kWh
So, it would take approximately 13.2 kWh of energy to raise the temperature of 300 gallons of water from 20°C to 30°C and keep it there for 24 hours.
thus 0.55kwh x 365 x24 = 4800kwh per year
13.2 * 240 days over 8 months = 3168kwh /8 months