A 2.00 * 10^2 g sample of water at 60.0°C is heated to steam at 140.0°C. How much heat is absorbed? ?

(4.184 J/(g·°C)) x (200 g) x (100.0 - 60.0)°C =

33472 J to warm the water to its boiling point

(2257 J/g) x (200 g) = 451400 J to vaporize the water

(2.000 J/(g·°C)) x (200 g) x (140.0 - 100.0)°C =

16000 J to heat the steam to 140.0°C

33472 J + 451400 J + 16000 J = 500872 J = 501 kJ total absorbed

m= 200gm = 0.2Kg

S(water) =4185.5 J/kg/K

L(vaporization) =2260000 J/Kg

S(steam) =1996 J/ kg/K

T1= 60C

T2=140C

H = mS(water)*(100-T1)+mL+ mS(steam) *(T2-100)

=5014520.8J

=501.45 KJ