Question

Twenty grams of aluminum metal at 83.2 C were mixed with 49.3 g of water at 19.7 C. Assuming no heat loss, what was the final temperature of the Aluminum water metal? specific heat for Al= 0.215 specific heat for H2O=1.00

Answer 1

heat lost by Al + heat gained by water = 0
(mass Al x specific heat Al x (Tfinal-Tinitial) + (mass H2O x specific heat H2O x (Tfinal-Tinitial) = 0
Substitute and solve for Tf of the Al and H2O.