Question

At 1 atm, how much energy is required to heat 63.0 g of H2O(s) at –18.0 °C to H2O(g) at 159.0 °C?

Answer 1

Q = heat change for conversion of ice at -18 ^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 159 ^oC

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

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

Where

m = mass of water = 63.0 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’’ = 159-100 = 59^oC

dt' = 100 -0 =100 ^oC

dt = 0-(-18) =18 ^oC

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

                                      = 200.03x10³ J

                                      = 200.03 kJ