How much heat (in kJ) is evolved in converting 1.00 mol of steam at 145.0 degrees C to ice at -45.0 degrees C? The heat capacity of steam is 2.01 J/g* degrees C and of ice is 2.09 J/g*degrees C .

Answer 1

q1 = heat removed to convert steam at 145 C to steam at 100 c.
q1 = mass steam x specific heat x delta T. Delta T is of course 45.

q2 = heat removed to condense steam at 100 to liquid water at 100 C.
q2 = mass steam x heat vaporization.

q3 = heat removed to cool water at 100 C to zero C.
q3 = mass water x specific heat x delta T.

q4 = heat to convert liquid water at zero C. to ice at zero C.
q4 = mass water x heat fusion.

q5 = heat removed to convert ice at zero C. to ice at -45 C.
q5 = mass ice x specific heat x delta T.

Total heat is q1 + q2 + q3 + q4 + q5.
Note: One mole of water is 18.015 grams. I see specific heats are in J/g so it's easier to convert mole water to grams first than to convert each of the specific heats to J/mol. Wben you finish with the total heat, convert J to kJ.

Answer 2

36.4 kJ

Answer 3

How much heat is evolved in converting 1.00 mol of steam at 145.0 ∘C to ice at -50.0 ∘C? The heat capacity of steam is 2.01 J/(g⋅∘C) and of ice is 2.09 J/(g⋅∘C).
Express your answer in units of kilojoules Assume the system is at atmospheric pressure.