Question

At 1 atm, how much energy in kJ is required to heat 85.0 g H 2 O ( s ) at − 10.0 ∘ C to H 2 O ( g ) at 167.0 ∘ C? Use the heat transfer constants found in this table.

Quantity	per gram	per mole
Enthalpy of fusion	333.6 J/g	6010. J/mol
Enthalpy of vaporization	2257 J/g	40660 J/mol
Specific heat of solid H2O (ice)	2.087 J/(g·°C) *	37.60 J/(mol·°C) *
Specific heat of liquid H2O (water)	4.184 J/(g·°C) *	75.37 J/(mol·°C) *
Specific heat of gaseous H2O (steam)	2.000 J/(g·°C) *	36.03 J/(mol·°C) *

Answer 1

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

    = (85 g)*(2.087 J/gC)*(10 degree C)= 1773.95 J

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

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

      = (85 g)*(333.6 J/g)= 28356 J

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

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

     = (85 g)*(4.184 J/gC)*(100)= 35564 J

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

     = mass of H2O*heat of vaporization of water

     = (85 g )*(2257 J/g)= 191845 J

Q5= energy needed to raise the temperature of steam from 100 to 167degree C

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

      = (85 g )*(2.0 J/gC)*(167-100) = 11390 J

Q= Q1+Q2+Q3+Q4+Q5

Q=1773.95 + 28356 + 35564 + 191845 + 11390 = 268928.95 J = 269 kJ