Question

how much heat is released when 10.0 g of steam (water vapor ) at 105.0 C is cooled to liquid water at 25 C? S(water) = 4.18 J/g.C. ... S(steam) = 2.01 j/ g.C

the heat of fusion of water is 6.02 KJ/ mol. The heat of vaporization of water is 40.7 KJ/mol

Answer 1

Answer

Q1= energy needed to decrease the temperature of steam from 105 to 100degree C

      =  mass of H20* specific heat of water vapour *temperature change

      = (10 g )*(2.01 J/gC)*(105-100) = 100.5 J

Q2=energy needed to phase change of H2O( from vapour to liquid  )

     = mass of H2O*heat of vaporization of water

     = (10 g )*(2260 J/g)= 22600 J

Q3= energy needed to cool the water from 100 to 25 degree C

Q3= mass of water *specific heat of water *temperature change

     = (10 g)*(4.186 J/gC)*(100-25)=3139.5 J

Qtoal = Q1+Q2+Q3 = 100.5+22600+3139.5 = 25840 J= 25.8 kJ