You start with 20 g of ice (H2O) at 260 K in a thermally isolated container. If 15500 J is added by heat, what will be the final temperature and state of the H2O? Show your work.

(Cice = 2.1 J/gK, Cwater = 4.18 J/gK, Cwatervapor = 1.7 J/gK, ΔHfus(H2O) = 334 J/g,
ΔHvap (H2O) = 2261 J/g, Melting Point = 273 K, Boiling Point = 373 K)

1 answer

I'll get you started. K = 260 so that is -13 C.
How much heat is required to move the ice from -13 to zero? That's
q1 = mass ice x specific heat ice x (Tfinal - Tinitial) = 20 x 2.1 x [0 - (-13)] = ? J.
q2 to melt ice is mass ice x heat fusion = 20 x 2.1 = ? J.
q3 to move liquid water from zero C to 100 C = formula for q1
q4 to boil water = formula for q2 but heat vap
Starting with 15,500 J I would subtract q1, then q2, then q3 and continue until you have used almost all of the 15,500 and the next subtraction leaves you with a negative number which, of course, can't be. Then use the next formula in the series above to determine final T. Post your work if you get stuck.