The evaporation rate $E_r$ can be approximated by:
$$E_r = C \times A \times (x_s - x) = 0.013$$ kg/h
$C$... evaporation coefficient $ = 25 + 19 \times v$
$A$... water surface $= 0.03 $ m$^2$
$x_s$... saturation humidity $= 0.027$ kg/kg (at 30 °C)
$x$... absolute humidity $= 0.011 $ kg/kg (at 30 °C and 40 % rel. humidity)
$v$... air speed above water surface $= 0.08$ m/s
If you mesure the air temperature above the water surface - which I approximated with 30 °C - you can get a way better approximation for your case. Also having the rel. humidity you can then find the values for $x_s$ and $x$ in a Mollier diagram (h,x-diagram).
PS: You will waste a lot of energy when you want to obtain the water temperature at 45 °C. Heating the air instead should be more effective.
Reference: Link (in German)