Question

a 15.7 g aluminum block is warmed to 53.2 degrees celsius and plunged into an insulated beaker containing 32.5 g of water initially at 24.5 degrees celsius. the aluminum and the water are allowed to come to thermal equilibrium. assuming no heat is lost, what is the final temperature of water and aluminum?

Answer 1

heat lost by Al + heat gained by H2O = 0
[mass Al x specific heat Al x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Solve for Tfinal.