Question

At 1 atm, how much energy is required to heat 93.0 g H2O(s)93.0 g H2O(s) at −10.0 ∘C−10.0 ∘C to H2O(g)H2O(g) at 137.0 ∘C?137.0 ∘C? Use the heat transfer constants found in this table.

Answer 1

Answer

Q1= energy needed to warm the ice from -10 degree C to 0 degree C

Q1= mass of ice*specific heat of ice *temperature change

    = (93 g)*(2.087 J/gC)*(10 degree C)= 1940.91 J

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

Q2= mass of H2O* latent heat of fusion of water

      = (93 g)*(333.6 J/g)= 31024.8 J

Q3= energy needed to warm the water from 0 to 100 degree C

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

     = (93 g)*(4.184 J/gC)*(100)= 38911.2 J

Q4=energy needed to phase change of H2O( from liquid to vapour )

     = mass of H2O*heat of vaporization of water

     = (93 g )*(2257 J/g)= 209901 J

Q5= energy needed to raise the temperature of steam from 100 to 137degree C

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

      = (93 g )*(2.0 J/gC)*(137-100) = 6882J

Q= Q1+Q2+Q3+Q4+Q5

Q=1940.91 + 31024.8 + 38911.2 + 209901 + 6882= 288659.9 J = 288.6 kJ