The answer is in this formula: Energy [kWh] = (Tdiff * 4180 * Litters) / 3600000
If you have 1 kg of water (=1 liter) and you wish to warm it by 1 degree Celsius, you need to use 0,00116111 kWh of energy.
If you have 1 kg of water (=1 liter) and you wish to warm it by 90 degree Celsius (e.g. from 10°C to 100°C – ad in warming up 1 liter of cold water in an electric kettle), you need to use 0,1045 kWh of energy.
This means 1 kWh — 1 tick of your energy meter – can in a theory be used to make 9,56 times 1 liter of hot boiling water.
If you have 10 liters (10 kgs) and you wish to warm by 1 degree, you need to use 0,116111 kWh of energy. If you have 100 liters (100 kgs) and you wish to warm it from 10 °C to 80°C, you need to use 8.12 kWh of energy.
One 3V 100AH LFP battery can release some 300Wh of energy in a matter of seconds. This means it can make some 2.8 liters of water to start boiling immediately. This is also the reason why using the water to cool down the short circuit in a battery installation is not efficient.
A solar panel with 300Wp during a sunshine time of 6 hours will produce some 1.4kWh of energy (at 80% efficiency). This is enough to warm 25 liters of water from 10°C to 60°C. To warm 100 liters of water the same way, 4 solar panels of 300Wp would do.