Question

Question #: 9 How much energy is required to warm 90.1 g (5.00 mol) of H2O(s), initially at –19.4 oC, to H2O( l) at 100.0 oC? melting point = 0.00 oC boiling point = 100.0 oC ΔHfus = 6.02 kJ/mol ΔHvap = 40.7 kJ/mol Cs of H2O(s) = 2.09 J/g oC Cs of H2O(l) = 4.18 J/g oC Cs of H2O(g) = 2.01 J/g oC

Answer 1

The energy required for the whole process has to be calculated as follows:

H2O(s)   ------->     H2O(s)   -------->      H2O(l)    ------> H2O(l)   ----------> H2O(g)

-19.4                            O                                  O                  100                          100 all in ^oC

          2.09J/g                         deltaH fus               4.18J/g                  deltaH vap            energy required

Energy for step I = 2.09 x 90.1 = 188.309J

energy for step II = 6.02x1000 x 5 =30100J

energy for step III = 4.18 x 90.1 =376.618J

energy for step Iv = 40.7 x 1000 x 5 = 203500J

Thus the totyal energy required for all the processes = 234164.927 J =234.16 kJ