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.
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