homework and exercises - What temperature is an evaporating cup of water at "room temperature"?

If I put 100 g of water in a cylindrical cup and I know it takes 2 weeks (lets say 1.2 × 10⁶ seconds) at some RH/temp to completely evaporate in my room with a constant temperature, can I determine it's temperature?

I would know how much heat the water absorbed using the latent heat of vaporization ΔH_vap = 2257 kJ kg^-1 (at 100 °C...not sure how/if to correct), and the specific heat: C_P = 4.1855 J kg^-1 K^-1 (at 15 °C).

So if it took t to evaporate m of water, it should be:

$$\frac{m\,\Delta H_{vap}}{t\,C_P} = \frac{0.1\,\mathrm{kg}\times 2.257 \times 10^6\mathrm{\ \frac{J}{kg}\ }}{1.2 \times 10^6\,\mathrm{s}\times 4.186\mathrm{\frac{J}{kg \cdot K}}} = \dots$$

But there's a left-over mass term...I could maybe square m...

The water pulled 225.7 kJ out of the room (lets say the room is sufficiently large to not change temperature due to that), so the average heat transfer was 0.188 W. My assumption is that because the surface area of the water is uniform (cylinder), that the rate of evaporation would be constant, which would mean constant heat...but given the same heat and a shrinking mass, the temperature would exponentially decrease: absurd. Where/what's the gap in my reasoning?

Does the 0.188 W of cooling need to be exactly balanced by 0.188 W of heating by the room, so they are actually at the same temperature? Would I need to consider something else (e.g. the thermal conductance between the room and the water)?