Question

At 1 atm, how much energy is required to heat 57.0 g of H2O(s) at –16.0 °C to H2O(g) at 137.0 °C?

Answer 1

Q = heat change for conversion of ice at -16 ^oC to ice at 0 ^oC + heat change for conversion of ice at 0^oC to water at 0^oC + heat change for conversion of water at 0^oC to water at 100 ^oC +heat change for conversion of water at 100 ^oC to vapour at 100 ^oC+ heat change for conversion of vapour at 100 ^oC to vapour at 137 ^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 water = 150 g

c” = Specific heat of steam = 2.1 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’’ = 137-100 = 37^oC

dt' = 100 -0 =100 ^oC

dt = 0-(-16)=16 ^oC

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

                                     = 57((2.09x16)+334.9 + (4.186x100) + 2260 + ( 2.1x37) )

                                    = 178.1x10³ J

                                    = 178.1 kJ