If 40.0 g of water at 70.0°c is mixed with 40.0 g of ethanol at 10.0°c, what is the final temperature of the mixture?

When the heat lost by water = heat gained by ethanol

∴(  M* C * ΔT )w  = (M*C*ΔT ) eth

when Mw mass of water = 40 g 

C specific heat of water = 4.18

ΔT = (70- Tf) 

and M(eth) mass (ethanol) = 40 g 

C specific heat of ethanol = 2.44 

ΔT = (Tf - 10 ) 

by substitution:

40* 4.18 * (70 - Tf) = 40 * 2.44 * (Tf-10) 

∴Tf = 47.9 °C 

212 degrees F, and 100 degrees C.

Explanation:

If the temperature is held constant (which requires some heat input, since evaporation cools things) the liquid will all evaporate. If the temperature is much above 212 F, the water will boil. That means that it wont just evaporate from the surface but will form vapor bubbles, which then grow, inside the liquid itself.   :)