how much heat necessary to boil 40.0g of H2O at 100 degrees C, then raise the temp of H2O vapor from 100.0 to 110.0 degrees C. ?
The specific heat of water vapor = 2.080 J/g degrees C. The heat of vaporization of water = 2,257 J/g.
- hcbiochemLv 74 months ago
Heat to vaporize the water:
q = (40.0 g ) X 2257 J/g = 9.03X10^4 J
Heat to raise temp of steam:
q = m c (T2-T1)
q = 40.0 g (2.080 J/gC) (110.0-100.0) = 823 J
Total = 9.11X10^4 J = 91.1 kJ