Well, you can use the first principle of thermodynamics. Namely,

Qtotal = Qrise + Qloss

Qrise = m*C*(Tf - Ti)

m - mass of the air in your house

C - specific heat of air under constant pressure (look on the Internet for the value)

Tf and Ti are the final and the initial temperature, respectively (21C and you didn't provide the initial temperature; let assume you start from 0C).

Anyway, since you don't know the mass of air in your house, you can express the mass as: m = rho * VH, where rho = 1.2 kg/m3 is air density and VH is the volume of your house that you need to figure out (you can assume it's a cube with a = 10 m, for example, and then VH = 10*10*10 = 1000 m3 ).

Now, when you say power you need to be specific in what time interval you need to heat up the house to 21C; in 1 hour, 1 day, 10 minutes, 1 year, etc. Assuming it's in 1 hour, then

Qtotal = 1.2*VH*C*(21-0) + 2.52*107

Note that 7 kWh = 2.52*107 J

 Thank you very much for the answer!

But in this case the total value still depends on the house capacity. I guess there is no other way to calculate total power needed, withouth assuming the house dimensions?

I think the question might lacking some important details. For example, if the total leakage (I presume *heat* leakage, rather than just air leakage) is a fixed steady 7 kW (and not a function of internal and external temperatures), then i would argue that you can maintain any temperature in the house that you like, providing you provide steady-state 7 kW of heat to avoid any further net heat transfer.

John is right! If all losses of the house envelope are correctly determined to 7 kW, you have to provide the heat source which can cover these losses. It can be a boiler fueled by conventional fuel, or biomass, or a heat pump, or solar collectors. You have to choose capacity bigger than your losses at least 30%, standard sizes are 12, 18, 25 kW or bigger. You have to check the figure 7 kW, it seems rather small for a family house in continental climate.