J.R. S. answered 10/17/20
Ph.D. University Professor with 10+ years Tutoring Experience
My approach is somewhat different from that provided by Dr. Bob. Here is the way I would approach this problem.
Heat lost by the hot gold + that lost by the hot silver must equal the heat gained by the cool water.
Heat lost or gained is q = mC∆T where m is the mass, C is the specific heat and ∆T is change in temperature.
Before applying this formula, we will first find the mass of the gold, silver and water.
mass of gold = (2.63 cm)3 x 19.3 g/cm3 = 351 g
mass of silver = (2.25 cm)3 x 11.39 g/cm3 = 120 g
mass of water = 119 ml x 1.00 g/ml = 119 g
heat lost by gold = (351 g)(0.1256 J/gº)(81.6 - Tf)
heat lost by silver = (120 g)(0.2386 J/gº)(81.6 - Tf)
heat gained by water = (119 g)(4.184 J/gº)(Tf - 20.3)
(351 g)(0.1256 J/gº)(81.6 - Tf) + (120 g)(0.2386 J/gº)(81.6 - Tf) = (119 g)(4.184 J/gº)(Tf - 20.3)
When I solve for Tf (final temperature) I get a value of 28.1ºC
You should go over this and do it yourself. Check my math please.