Question

Calculate the amount of energy (in kJ) necessary to convert 457 g of liquid water from 0°C to water vapor at 192°C. The molar heat of vaporization (Hvap) of water is 40.79 kJ/mol. The specific heat for water is 4.184 J/g · °C, and for steam is 1.99 J/g · °C. (Assume that the specific heat values do not change over the range of temperatures in the problem.)

Answer 1

The process involves follwoing steps :

Heating water from 0^o to 100^o --------> phase transition at 100^o water to steam ------> heating steam from 0^o to 192^o

energy exchANGE   = msT

Total energy needed = 1step + 2nd step + 3rd step

= [457 x 4.184] x 100  +[(457/18) x 40.79 ] + [457 x 1.99 ] X 92

= 191.2088 + 1035.612 + 83.667 = 1310.48 KJ