Explanation:
The given data is as follows.
          Temperature of dry bulb of air = 
           Dew point =  = (10 + 273) K = 283 K
 = (10 + 273) K = 283 K
At the dew point temperature, the first drop of water condenses out of air and then,
         Partial pressure of water vapor  = vapor pressure of water at a given temperature
 = vapor pressure of water at a given temperature 
Using Antoine's equation we get the following.
             
             
                                = 0.17079
                     = 1.18624 kPa
 = 1.18624 kPa
As total pressure  = atmospheric pressure = 760 mm Hg
 = atmospheric pressure = 760 mm Hg 
                                    = 101..325 kPa
The absolute humidity of inlet air = 
                   
                  = 0.00735 kg  / kg dry air
/ kg dry air
Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg  / kg dry air.
/ kg dry air.
Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.
                  0.07 kg - 0.00735 kg
               = 0.06265 kg  for every 1 kg dry air
 for every 1 kg dry air
Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.
                 
                   = 6.265 kg
Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.