For many years drinking water has been cooled in hot climates by evaporating it from the surface of canvas bags or porous clay pots. how many grams of water can be cooled from 40 ∘c to 20 ∘c by the evaporation of 43 g 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 : The mass of water cooled can be 1234.4 grams.

Explanation :

First we have to calculate the heat released during the reaction.

where,

= enthalpy change or heat of vaporization = 2.4 kJ/g = 2400 J/g

q = heat released = ?

m = mass of water vapor = 43 g

Now put all the given values in the above formula, we get:

Thus, the heat released during the reaction is 103200 J

Now we have to calculate the initial mass of water.

where,

m = initial mass of water = ?

q = heat absorbed = heat released = 103200 J

c = specific heat =

= final temperature =

= initial temperature =

Now put all the given values in the above formula, we get:

Therefore, the mass of water cooled can be 1234.4 grams.