How many grams of water can be cooled from 41 C to 18 C by the evaporation of 54g of water?

Question

How many grams of water can be cooled from 41 C to 18 C by the evaporation of 54g of water? (The heat of vaporization of water in this temperature range is 2.4 kJ/g. The specific heat of water is 4.18 J/g K.

Answer 1

The heat removed by evaporation is the heat of vaporization times the mass that evaporates.  If that is hard to understand, look at the units of the quantities.   Energy per gram, times grams, give energy.   So it is with the other half.   The specific heat (an archaic term, really, also called heat capacity) of the water is the heat needed to change the temperature by one degree (energy per gram per degree).  So with a certain amount of energy available, divide energy by specific heat to get (E/E/g*d) or degrees*grams.  Now divide that by degrees (41-18)  to get the answer in grams.

Answer 2

Somebody is throwing a monkey wrench at you on this one.  If you evaporate water, you require heat to do this.  In theory, you could extract the heat from the cooling water, and the calc. is straightforward:  Heat to evaporate (54 x 2400)=heat extracted (w x 4.18 x 23).    In practice, you can't do this perfectly, since you have a non absolute-zero heat sink (that should blow the mind out of the Chem 101 people).