I have 1 litre of water at 45 °C. How long would it take to evaporate when ambient temperature is 25 °C?
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)