Question

Calculate the enthalpy change (in kJ) associated with the conversion of 25.0 grams of ice at -4.00 °C to water vapor at 109.0 °C.
The specific heats of ice, water, and steam are 2.09 J/g-K, 4.18 J/g-K, and 1.84 J/g-K, respectively.
For H2O, ΔHfus = 6.01 kJ/mol and ΔHvap = 40.67 kJ/mol.

Answer 1

A. heat to move ice from -4.00 to zero C.
q = mass ice x specific heat ice x (Tfinal-Tinitial)
B. heat to melt ice at zero C to liquid water at zero C.
q = mass ice x heat fusion.
C. heat to move liquid water from zero C to 100 C.
q = mass water x specific heat H2O x (Tfinal-Tinitial)
D. heat to convert H2O @ 100 C to steam @ 100 C.
q = mass water x heat vaporization
E. heat to move steam @ 100 C to 109 C.
q = mass steam x specific heat steam x (Tfinal-Tinitial)
Total q = delta H = q1 + q2 + q3 + q4 + q5.
Note that specific heats are given in J/g*K but heat fusion and heat vaporization are given in kJ/mol. I would suggest you change kJ/mol to J/g so as to keep all of the units the same.
Post your work if you get stuck. I'll check your answer and your work but only if you show your work.