Question

How much heat (in kJkJ) is required to warm 13.0 gg of ice, initially at -14.0 ∘C∘C, to steam at 108.0 ∘C∘C? The heat capacity of ice is 2.09 J/g⋅∘CJ/g⋅∘C and that of steam is 2.01 J/g⋅∘CJ/g⋅∘C.

Answer 1

Q1= energy needed to warm the ice from -14 degree C to 0 degree C

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

    = (13 g)*(2.09 J/gC)*(14 degree C)= 380.38 J

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

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

      = (13 g)*(334 J/g)= 4342 J

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

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

     = (13 g)*(4.186 J/gC)*(100)= 5441.8 J

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

     = mass of H2O*heat of vaporization of water

     = (13 g )*(2260 J/g)= 29380 J

Q5= energy needed to raise the temperature of steam from 100 to 108degree C

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

      = (13 g )*(2.01 J/gC)*(108-100) = 209.04 J

Q= Q1+Q2+Q3+Q4+Q5

Q=380.38 + 4342 + 5441.8 + 29380 + 209.04 = 39753.2 J = 39.8 kJ