How much energy would be released if the steam at 140.0°C were cooled to water at 60.0°C? ?

mass m is in grams

step 1 : cool the steam down to 100°C

E1 = m(100-140)*2.01)

step 2 : convert the steam into water @ 100°C

E2 = -m*2260

step 3 : cool the water down to 60°C

E3 = m(60-100)*4.19)

E = E1+E2+E3 = -m(40*2.01+2260+40*4.19) = 2.509*m kjoule 

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