Question

How much heat is evolved in converting 1.00 mol of steam at 130.0 ∘C to ice at -45.0 ∘C? The heat capacity of steam is 2.01 J/(g⋅∘C) and of ice is 2.09 J/(g⋅∘C).

Answer 1

Q = heat change for conversion of vapour at 130 ^oC to vapour at 100 ^oC + heat change for conversion of steam at 100 ^oC to water at 100 ^oC + heat change for conversion of water at 100^oC to water at 0 ^oC + heat change for conversion of water at 0^oC to ice at 0^oC + heat change for conversion of ice at 0 ^oC to ice at -45 ^oC

Amount of heat released , Q = mcdt + mL + mc'dt + mL' + mc"dt"

                                              = m(cdt + L + c'dt' + L' + c"dt" )

Where

m = mass of steam = 1.00mol x 18 (g/mol) = 18.0 g

c = Specific heat of steam = 2.01 J/g degree C

c' = Specific heat of water = 4.186 J/g degree C

c" = Specific heat of ice= 2.09 J/g degree C

L = Heat of Vaporization of water = 2260 J/g

L'= Heat of fusion of ice = 334.9 J/g

dt = 130-100 = 30^oC

dt' = 100 -0 =100 ^oC

dt'' = 0-(-45) =45 ^oC

Plug the values we get Q = m(cdt + L + c'dt '+ L' + c"dt" )

                                      = 57.02x10³ J

                                      = 57.02 kJ