Require How Much Capacity of Electric Heater

What is the liquid? How long do you have to raise the temperature?

liquid is water. I have to maintain the temp for 24*365. The water temp should be constant

The problem here is to find an approximation for your heat loss - if the tank was perfectly insulated, and nothing at a different temperature was added, the temperature would not change, even without a heater.

Is there any flow of water (i.e. incoming water at a lower temperature)? What is the material of the tank walls? What is the approximate surface area of the tank? What is the lowest outside temperature outside the tank that you expect to have during a year?

Since you seem not in a position to give the requirements for assessment therefore I prefer to give one solved example below, try to calculate your needs in accordance therewith:

Example;

A cubic water tank has a surface area of 6.0 m square and is filled to 90% capacity six times daily. The water is heated from 20 degree C to 65. The losses per square meter of tank surface per 1 degree C temperature difference are 6.3 watts. Calculate the load and the efficiency of the tank. Specific heat of water is 4200 Joule/kg/degree C and one kWh = 3.6 MJ.

If l is the side of the tank , the total surface area of tank=6xl square.

6xl sq. = 6 or l = 6/6 = 1 m cube.

Since l meter cube of water weighs 1000 kg, mass of water to be heated daily = 5.4 x 1000 = 5400 kg

Heat required to raise the temperature of water = 5400 x 4200 (65-20) = 1020 MJ = 1020/3.6 = 283.3 kWh

Daily loss from the surface of the tank = 6.3 x 6 x(65-20)x 24/1000 = 40.8 kWh

Energy supplied per day = 283.3 + 40.8 = 324.1kWh

Loading in kW = 324.1/24 = 3.5kw

Efficiency of the tank = 283.3 x 100/324.1 = 87.4 %